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This is an experimental version of the lessons, designed to make it easier if you are
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Practice answers are also moved to just below the Practice problems, to make the answers
easier to check when working “on screen.”
To use the onscreen version: when you see the red stars *****
stop scrolling down. Work the problem on your paper. Then scroll down. The answer will
be below.
The regular “printed book” version which can be printed is also available on the
download page. Calculations In Chemistry
Modules 17 and 18
Ideal Gases, Gas Labs, Gas Reactions
Module 17 – Ideal Gases ................................................................................................... 1
Lesson 17A:
Lesson 17B:
Lesson 17C:
Lesson 17D:
Lesson 17E:
Lesson 17F:
Lesson 17G:
Lesson 17H: Gas Fundamentals................................................................................................. 1
Gases at STP ........................................................................................................... 6
Complex Unit Cancellation................................................................................ 14
The Ideal Gas Law and Solving Equations ...................................................... 22
Choosing Consistent Units................................................................................. 28
Density, Molar Mass, and Choosing Equations .............................................. 34
Using the Combined Equation .......................................................................... 45
Gas Law Summary and Practice ....................................................................... 54 Module 18 – Gas Labs, Gas Reactions............................................................................ 1
Lesson 18A:
Lesson 18B:
Lesson 18C:
Lesson 18D: Charles’ Law; Graphing Direct Proportions...................................................... 1
Boyle’s Law; Graphs of Inverse Proportions ................................................... 11
Avogadro’s Hypothesis; Gas Stoichiometry.................................................... 16
Dalton’s Law of Partial Pressures ..................................................................... 30 For additional modules, visit www.ChemReview.Net ©2011 ChemReview.net v. f1 Page OnScreen 17 i Module 17 — Ideal Gases
Timing: Ideal gas law calculations are covered in Module 17. The laws discovered by
Boyle, Charles, Avogadro, and Dalton are addressed in Module 18. Kinetic Molecular
Theory and Graham’s law calculations are covered in Module 19. If you need to solve
calculations for Module 18 or 19 topics before those using the ideal gas law, complete
Lessons 17A and 17D, then Modules 18 and/or 19.
Prerequisites: It will help with the Module 17 calculations if you have completed Modules
2, 4, 5, 8 and Lesson 11B. Those lessons can be done quickly.
Pretest: If you believe you have previously mastered ideal gas law calculations, try the
problems in the last lesson in this module. If you can do those, you may skip this module. ***** Lesson 17A: Gas Measurements and Fundamentals
Gas Quantities and Their Units
Chemistry is most often concerned with matter in 3 states: gas, liquid, and solid. The gas
state is in most respects the easiest to study, because by most measures, gases have similar
and highly predictable behavior. Gas quantities can be measured using 4 variables:
• Pressure, volume, temperature, and moles of gas molecules. The symbols for these variables are P, V, T, and n. Gas Pressure
In an experiment, a glass tube about 100 cm (1 meter) long is sealed at one end and then
filled with mercury: an element that is a dense, silvercolored metal (symbol Hg) and is a
liquid at room temperature. The open end of the filled tube is covered, the tube is turned
over, and the covered end is placed under the surface of additional mercury in a partially
filled beaker. The tube end that is under the mercury in the beaker is then uncovered.
What happens? In all experiments at standard atmospheric pressure, the same result. The
top of the mercury in the tube quickly falls from the top of the tube until it is about 76
centimeters above the surface of the mercury in the beaker. There, the mercury descent
stops. The result is a column of mercury inside the glass tube that is about 76 cm high.
What is in the tube above the mercury column? A bit of mercury vapor, but no air. The
space above the mercury in the tube is mostly empty: close to a vacuum.
What happens if the top of the tube is snapped off? The mercury inside the tube behaves
the same as a straw full of liquid when you take your finger off the top. The liquid mercury
in the tube falls quickly until it reaches the same level as the mercury in the beaker.
However, as long as the tube is sealed and is longer than 76 cm, the top of the mercury in
the tube in will remain about 76 cm above the top of the mercury in the beaker.
Why? The device made in this experiment is a mercury barometer. It measures the
pressure of the air outside the tube. ©2011 ChemReview.net v. f1 Page OnScreen 17 1 A barometer is a kind of balance, like a playground “teetertotter.” The pressure of the
dense column of mercury in the tube, pressing down on the
top of the pool of mercury in the beaker, is balanced by the
pressure of the 20milehigh column of air, the atmospheric
pressure, pressing down on the pool of Hg outside the tube.
We could construct our barometer using water as the liquid.
However, for water to balance the air, since water is about
13.5 times less dense than mercury, our tube would need to
be about 13.5 times higher, about 30 feet high, roughly three
stories on a typical building.
If the air pressure outside the mercury column increases, the
mercury is pushed higher in the tube. If the surrounding
atmospheric pressure is lowered, the mercury level in the
tube falls.
When a weather forecast states that “the barometric pressure
is 30.04 inches and falling,” it is describing the height of the
mercury column in a barometer (76 cm is about 30 inches). In
meteorology, a rising barometer, or a high pressure system, is usually associated with fair
weather. Falling barometers and low pressure systems are often associated with clouds,
storms, and precipitation. Measuring Pressure
In the branches of science, pressure is measured in a variety of units.
Chemistry defines standard pressure as a gas pressure of exactly 760 mm (76 cm) of
mercury as measured in a barometer. This is also known as exactly one atmosphere of
pressure. Normal atmospheric pressure at sea level on a fairweather day is about one
atmosphere.
The SI base units for pressure are termed pascals in honor of the 17th Century French
mathematician and scientist Blaise Pascal, whose experiments with gases led him to
propose the concept of vacuum. (Pascal’s contemporary, the scientist and mathematician
René Descartes, disagreed with the vacuum concept, writing that Pascal had “too much
vacuum in his head.”)
The following table of pressure units should be memorized. These equalities will be used
frequently to convert among pressure units. Pressure Units
Standard Pressure ≡ 1 atmosphere ( ≡ means “is defined as equal to”) ≡ 760 mm Hg (mercury) ≡ = 101 kilopascals (kPa) = 1.01 bars 760 torr (not exact; not a definition) Any two of those measures can be used in a conversion factor for pressure units. ©2011 ChemReview.net v. f1 Page OnScreen 17 2 Practice A
Memorize the pressure unit definitions, then use the equalities in these calculations. Do
one. Check your answer. Then do another. Answers are below and on the next pages.
1. The lowest atmospheric pressure at sea level was recorded in 1979 during Typhoon Tip;
a pressure of 870. millibars. What is this pressure in kPa?
2. If 2.54 cm ≡ one inch (exactly), and standard pressure is defined as exactly 760 mm Hg,
what is standard pressure in inches of mercury?
3. Standard pressure in English units is 14.7 pounds per square inch (psi). If a bicycle tire
has a pressure of 72 psi, how many atmospheres would this be?
4. 25.0 torr is how many kPa? ***** ANSWERS
Practice A
1. WANTED = ? kPa
SOLVE: DATA: 870. millibars ? kPa = 870. millibars ● 2. WANTED = ? inches Hg DATA: 10─3 bar ● 101 kPa = 87.0 kPa
1 millibar
1.01 bars 760 mm Hg (exact)
2.54 cm ≡ one inch (exact) ? inches Hg = 760 mm Hg ● 1 cm ● 1 inch = 29.92 inches Hg (exactly equals std. pressure;
10 mm
2.54 cm
all of the numbers are exact.) ©2011 ChemReview.net v. f1 Page OnScreen 17 3 3. WANTED = ? atmospheres DATA: 72 psi
1 atm. = std. pressure = 14.7 psi SOLVE: ? atmospheres = 72 psi ● 4. ? kPa = 25.0 torr ● 1 atm
760 torr 1 atm = 4.9 atm.
14.7 psi ● 101 kPa
1 atm = 3.32 kPa *****
Gas Volumes
Gas volumes are measured in standard metric volume units: Liters (dm3) and mL (cm3). Gas Temperature
Temperature is defined as the average kinetic energy of particles.
Kinetic energy is energy of motion, calculated by the equation
Energy of Motion = 1/2 (mass) (velocity)2, or KE = 1/2 mv2 Since the chemical particles of a substance have a constant mass, this equation means that
when molecules move twice as fast, they must have “2 squared,” or four times as much
kinetic energy. Their absolute temperature is four times higher.
One of the implications of this equation is that, though particles cannot have zero mass,
they can have zero velocity: they can (in theory) stop moving.
If all of the molecules in a sample had zero velocity, the temperature of the sample would
be ─273.15 ºC, which is defined as absolute zero. Absolute zero is the bottom of the
temperature scale. Nothing can be colder than absolute zero.
The Kelvin (or absolute) temperature scale simplifies the mathematics of calculations
based on gas temperatures. The Celsius scale defines 0 degrees as the melting and freezing
temperature of water, and 100 degrees as the boiling temperature of water at standard
pressure. The Kelvin scale keeps the same size degree as Celsius, but defines 0 as absolute
zero.
The equation relating the Kelvin and Celsius scales is K = oC + 273 (use 273.15 when
other measurements are quite precise). This equation must be memorized. The SI unit for
temperature is the kelvin. It helps to recall that a temperature in kelvins is always 273
degrees higher than the temperature in degrees Celsius.
When is the Kelvin scale needed? In any gaslaw calculation where a gas temperature
changes, temperature in kelvins must be used.
In measurements, the abbreviation for kelvins is a capital K without a degree symbol. In
equations, the symbol for kelvins is a capital T (a lower case t is often used as a symbol for
temperature in degrees Celsius). ©2011 ChemReview.net v. f1 Page OnScreen 17 4 When measuring gases, it is preferred to record gas volumes at standard temperature,
which is defined as the temperature that a mixed icewater bath will always adjust to at
standard (typical room) pressure: zero degrees Celsius, which is 273 K. Practice B
1. Commit to memory the equation relating kelvins to degrees Celsius, then complete the
following chart (boiling points are listed at standard pressure):
In Kelvin In Celsius Absolute Zero ________ __________ Water Boils ________ __________ Nitrogen Boils ________ ─196 oC 1074 K __________ Water Freezes (std. P) ________ __________ Std. Temperature ________ __________ Table Salt Melts 2. In a problem involving gases, you calculate a temperature for the gas of ─310 degrees
Celsius. Your answer is…..? ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 5 ANSWERS
Practice B
1. In kelvins
Absolute Zero
Water Boils
Nitrogen Boils
Table Salt Melts
Water Freezes
Std. Temperature In ºCelsius 0K
373 K
77 K
1074 K
273 K
273 K ─273 ºC
100. ºC
─196 ºC
801 ºC
0 ºC
0 ºC Note that when using K = oC + 273 you are adding or subtracting, and the significant figures are
determined by the highest place with doubt:. All of those numbers have doubt in the one’s place. 2. In a problem involving gases, you calculate a temperature for the gas of 310 degrees
Celsius. Your answer is…..?
Mistaken. ─310 ºC is below absolute zero (─273 ºC). There
cannot be a temperature colder than absolute zero.
***** Lesson 17B: Gases at STP
Prerequisites: You should be able to do the calculations in this lesson if you have
completed Modules 2, 4, 5, and 8, plus Lessons 11B and 17A.
***** Standard Temperature and Pressure (STP)
In experiments with gases, it is preferred to measure gas volumes at standard temperature
and pressure, abbreviated STP. The following values must be committed to memory.
Standard Temperature ≡ 0oC = 273.15 K = the temperature of an icewater mixture
Standard Pressure ≡ 1 atmosphere ≡ 760 mm Hg ≡ 760 torr
= 101 kilopascals (kPa) = 1.01 bars Molar Gas Volume At STP
Gases have remarkably consistent behavior. One important example involves gas volumes:
if two samples of gas have the same number of gas molecules, and they are at the same
temperature and pressure, the two samples will in most cases occupy the same volume. ©2011 ChemReview.net v. f1 Page OnScreen 17 6 This rule is true in most cases even if the gases have different molar masses or different
molecular formulas. It is even true for different mixtures of gases.
In addition, if we know the temperature, pressure, and number of moles of gas particles,
the gas volume will be predictable.
The example that we use most often concerns gases measured at the preferred temperature
and pressure: STP. In most cases, for one mole of any gas or any mixture of gas molecules,
the volume at STP will be 22.4 liters.
The equality one mole of gas = 22.4 liters of gas at STP provides us with a conversion
factor that can be used to solve gas calculations at STP.
Our rule will be
The STP Prompt
In calculations, if a gas is at STP, write in the DATA: 1 mole gas = 22.4 L gas at STP
In a calculation, if you see STP, or see a T and P that are equivalent to STP, write the STP
prompt.
For calculations involving gases at STP, we will use the STP prompt if we can solve by
conversions. If we need equations to solve, we may not need the conversion. When in
doubt, write the conversion. With practice, you will gain intuition as to when the prompt
equality is needed and when it is not. Attaching P and T to V
Note that in this equality, “at STP” is attached only to the gas volume. Because gas volumes
vary with temperature and pressure, all gas volumes must have pressure and temperature
conditions stated if the volume is to be a measure of the number of particles in the sample.
In calculations, gas volumes (L, dm3, mL) must have a P and T attached for the volume to
be a measure of the moles of gas in the sample.
The rule is: Gas volumes must be labeled with a T and P if the T and P are known. NonIdeal Behavior
When gases approach a pressure and temperature at which they condense (become liquids
or solids), the ideal gas assumptions of the STP prompt begin to lose their validity. At
pressures above standard pressure, variations from ideal behavior can also become
substantial. However, unless a problem indicates nonideal behavior, if the conditions are
at STP, you should assume that the STP prompt applies. Practice A
1. Write values for standard pressure using 5 different units.
2. Write values for standard temperature in 2 different units. ©2011 ChemReview.net v. f1 Page OnScreen 17 7 ANSWERS
Practice A
1. Standard P = 1 Atmosphere ≡ 760 mm Hg ≡ 760 Torr = 101 kilopascals (kPa) = 1.01 bars 2. Standard T = 0 ºC or 273 K *****
Calculations For Gases At STP
The STP prompt is often used in conjunction with two other prompts.
Grams Prompt: If grams of known chemical formula are part of any unit in the
WANTED or DATA, write in the DATA,
(Molar Mass) grams formula = 1 mole formula
Avogadro Prompt: If any part of the WANTED or DATA involves a number of
particles or molecules or 10xx of a substance formula, write in the DATA,
1 mole anything = 6.02 x 1023 (molecules or particles) anything
Together, these three prompts allow you to solve most STP gas calculations using
conversions. You will also need to recall that if a problem asks for
• molar mass, you WANT grams per 1 mole; • density, you WANT a mass unit (g or kg) over 1 volume unit (L, dm3, mL). Try this problem in your notebook, then check your answer below.
Q1. 2.0 x 1023 molecules of NO2 gas would occupy how many liters at STP? ***** ©2011 ChemReview.net v. f1 (Stop here and try the problem. The answer is below). Page OnScreen 17 8 Answer
WANT: ? L NO2 gas at STP = DATA: (first, write the unit you WANT) 2.0 x 1023 molecules of NO2 gas
6.02 x 1023 of anything = 1 mol anything (Avogadro Prompt) 1 mol any gas = 22.4 L any gas at STP (STP Prompt) Strategy: Want a single unit? Start with a single unit, and find moles first.
SOLVE:
? L NO2(g) STP = 2.0 x 1023 molec. NO2 ● 1 mol ● 22.4 L gas STP = 6.02 x 1023 molec. 1 mol gas = 7.5 L NO2(g) at STP
In problems, the words grams, STP, and molecules are prompts about the relationships that
you will need in your DATA to solve a problem.
Note the difference between these calculations and stoichiometry. The above problem
involved only one substance. If stoichiometry steps are needed, you will see DATA for two
substances involved in a chemical reaction.
An effective technique in learning physical science (and math) is to work the examples in
your textbook. Use these steps: cover the answer, read the example, and try to solve. If
you need help, peek at the answer, and then try the example again.
Try that method, this time solving for a ratio unit.
Q2. Determine the density of O2 gas at STP, in grams per milliliter. ***** (Stop here and try the problem. The answer is below). When solving for a ratio, if a calculation involves only one substance, the rule “want a ratio,
start with a ratio” will usually solve faster than “solve for the top and bottom units
separately.” Use that hint if needed. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 9 Answer
WANTED: ? g O2 gas
mL at STP DATA: 32.0 g O2 = 1 mol O2 (grams prompt) 1 mol any gas = 22.4 L any gas at STP
SOLVE: (STP Prompt) (Want a ratio? Start with a ratio. See Lesson 11B.) =
? g O2 gas
mL gas STP 32.0 g O2 ●
1 mol gas
●
1 mol O2
22.4 L gas STP 10─3 L =
1 mL 0.00143 g O2
mL at STP Your answer may have those 3 conversions, rightside up, in any order. Note that STP was
attached to the gas volume unit (mL). Note also that
• the conversions for molar mass and particles per mole are valid whether the substance
is a gas, liquid, or solid. • The STP prompt, however, only works for gases, and only works at STP. Practice B
Make certain that you can write all 3 of the above prompts from memory. Try that now in
your notebook. Then do all of Problems 1 to 3 below. After #3, do every other problem,
and more if you need more practice.
1. Write the units WANTED when you are asked to find
a. Molarity e. Density b. Molar Mass f. c. Volume g. Gas Pressure d. Mass h. Temperature Speed or Velocity 2. Calculate the g/L of SO2 gas at STP.
3. The density of a gas at STP is 0.00205 g•mL─1. What is its molar mass?
4. If 250. mL of a gas at STP weighs 0.313 grams, what is the molar mass of the gas?
5. Calculate the number of molecules in 1.12 L of CO2 gas at STP.
6. Calculate the volume of 15.2 grams of F2 gas at 273 K and standard pressure (in mL).
7. If 0.0700 moles of a gas has a volume of 1,760 mL, what is the volume of one mole of the
gas, in liters, under the same temperature and pressure conditions?
8. Calculate the density of Rn gas at STP, in kg•L─1. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 10 ANSWERS
Practice B
1 Moles
1 L solution a. Molarity grams
1 mole b. Molar Mass
c. Volume L, mL, dm3, cm3
kg, g, ng d. Mass WANTED: any mass unit (kg or g)
any volume unit (L, mL, dm3) f. any distance (cm, miles)
any time (sec, hour) Speed g. Gas Pressure atm., torr, mm Hg, kPa, bars
h. Temperature ? g SO2 gas
L SO2 gas at STP DATA: 2. e. Density 64.1 g SO2 = 1 mol SO2 ºC or K (Write ratio units WANTED as fractions)
(grams prompt) 1 mol any gas = 22.4 L any gas at STP
SOLVE: ? (Want a ratio? Start with a ratio. Since grams is on top in the answer, you could start with
grams on top as the given ratio, but these conversions can be in any order.)
= 64.1 g SO2 ● 1 mol gas
g SO2 gas
= 2.86 g SO2(g)
L SO2 gas at STP
1 mol SO2
L SO2 gas at STP
22.4 L gas STP g
mol WANT: ? DATA: 3. (STP Prompt) 0.00205 g gas = 1 mL gas at STP (write the unit WANTED for molar mass: g/mol)
( g • mL─1 = g / mL ) 1 mol any gas = 22.4 L any gas at STP (STP Prompt) (Note that the grams prompt only works if you know a substance formula.)
SOLVE: (the conversions below may be in any order, so long as they are rightside up.) ? g = 0.00205 g gas ● 1 mL ● 22.4 L any gas at STP = 45.9 g
mol
1 mL gas at STP
10─3 L
1 mol gas
mol ©2011 ChemReview.net v. f1 Page OnScreen 17 11 g
mol WANT: ? DATA: 4. 0.313 g gas = 250. mL gas at STP
1 mol any gas = 22.4 L any gas at STP SOLVE: (STP Prompt) (These conversions below may be in any order.) 0.313 g gas
g=
● 1 mL ● 22.4 L any gas STP = 28.0 g
mol
250. mL gas at STP 10─3 L
1 mol gas
mol ? Reminders
• Attach temperature and pressure conditions, if known, to gas volumes. • In the interest of readability, most unit cancellations in these answers are left for you to do.
However, in your work, always mark your unit cancellations as a check on your conversions. WANT: ? molecules CO2 gas = DATA: 5. 1.12 L CO2 gas at STP
6.02 x 1023 anything = 1 mol anything (Avogadro Prompt) 1 mol any gas = 22.4 L any gas at STP (STP Prompt) SOLVE:
? molecules CO2(g) = 1.12 L CO2(g) STP ● 1 mol gas ● 6.02 x 1023 molec. = 3.01 x 1022 molec. CO2
22.4 L gas STP
1 mole WANT: ? mL F2 gas at STP = DATA: 6. (these conditions are STP) 15.2 g F2 gas (single unit given) 38.0 g F2 gas = 1 mol F2 gas
1 mol any gas = 22.4 L any gas at STP
SOLVE: (Want a single unit?) ? mL F2(g) STP = 15.2 g F2(g) ● 1 mol F2 ● 22.4 L gas STP ● 1 mL
38.0 g F2 ©2011 ChemReview.net v. f1 1 mol gas = 8.96 x 103 mL F2(g) STP 10─3 L Page OnScreen 17 12 7. WANTED: ? L gas at given P and T
1 mol (Strategy: If the problem is asking for a unit per one unit, it is asking for a ratio unit.
This problem does not specify STP, but you can solve for the requested unit.
Compare units: You WANT liters and moles. You are given mL and moles.) DATA: 0.0700 mol gas = 1,760. mL gas at given T and P SOLVE: (Since answer unit moles is on bottom, and answer unit liters is not in the data, you might
start with moles on the bottom in your given ratio.) ? L gas at given T and P = 1,760 mL gas at given T and P ● 10─3 L = 25.1 L gas at given T and P
1 mole
0.0700 mole
1 mL
mole 8. Hint: The grams prompt applies to kilograms, too. If needed, adjust your work and try again. **** * WANTED: ? kg Rn gas
L at STP DATA: 222 g Rn = 1 mol Rn (kg = g prompt) 1 mol any gas = 22.4 L any gas at STP (STP Prompt) SOLVE: ( kg • L─1 = kg / L . Write ratio units WANTED as fractions) (Start with a ratio. In your given, you may want one unit where it belongs in the answer, but
your conversions may be in any order that cancels to give the WANTED unit.) =
? kg Rn gas
L Rn gas at STP ● 222 g Rn ● 1 kg = 9.91 x 10─3 kg Rn
22.4 L gas STP
1 mol Rn
103 g
L Rn gas at STP
1 mol gas ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 13 Lesson 17C: Cancellation of Complex Units
To solve gas law and heat problems will require working with complex units: those with
reciprocal units or more than one unit in the numerator or denominator. Let’s review some
rules for working with complex fractions. Why Not To Write “A/B/C”
When solving complex fractions, the notation for the fractions must be handled with care.
Why? Try these problems.
A. 8 divided by 2 =
4 B. 8 divided by 4 =
2 C. 8/4/2 = ***** Answer
A. 2 divided by 2 = 1 B. 8 divided by 2 = 4 C. Could be 1 or 4, depending on which is the fraction: 8/4 or 4/2.
The format A/B/C for numbers or units is ambiguous unless you know, from the context
of the problem or from prior steps, which is the fraction.
Let’s try problem C again, this time with the fraction identified by parentheses. Use the
rule of algebra: perform the operation inside the parentheses first.
D. (8/4)/2 = E. 8/(4/2) = ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 14 Answer:
D. 2/2 = 1 E. 8/2 = 4 With the parentheses, the problem is easy. In part C, for 8/4/2 without the parentheses,
you cannot be sure of the right answer. For this reason, writing numbers or units in a
format A/B/C should be avoided.
In these lessons, we will either use a thick underline, or group fractions in parentheses, to
distinguish a numerator from a denominator. Multiplying and Dividing Fractions
1. By definition: 1 divided by (1/X) = the reciprocal of 1/X = 1
1
X =X Complex fraction Rule 1: Simplify the reciprocal of a fraction by inverting the fraction.
The reciprocal of a value is “1 over the value.”
Recall that 1/X can be written as X─1. Another way to simplify the above is to apply
the rule: when you take an exponential term to a power, you multiply the exponents.
(X─1)─1 = (X+1) = X .
Q1. Apply Rule 1 to the following problem, then check your answer below.
1/(B/C) = ***** Answer:
The reciprocal of a fraction simplifies by inverting the fraction.
1/(B/C) simplifies by inversion (flips the fraction over) to C/B.
In symbols: ©2011 ChemReview.net v. f1 1=C
B
B
C In exponents: (B/C) ─1 = (B ● C─1) ─1 = B─1 ● C Page OnScreen 17 15 Calculations in chemistry may involve fractions in the numerator, denominator, or both.
To handle these cases systematically, use the following rule.
Complex fraction Rule 2: When a term has two fraction lines ( either _____ or / ),
separate the terms that are fractions. To do so, apply these steps in this order.
a. If a term has a fraction in the denominator, separate the terms into a reciprocal of the
fraction ( 1/fraction in the denominator ) multiplied by the remaining terms.
b. If there is a fraction in the numerator, separate and multiply that fraction by the
other terms in the numerator or denominator.
c. Then simplify: invert any reciprocal fractions, cancel units that cancel, and
multiply the terms.
Let’s learn to apply these rules using examples.
Q2. Simplify: A/(B/C) ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 16 Answer: A/(B/C) = A
B
C = A● 1
B
C =A●C =
B ^Rule 2a A●C
B ^Rule 2c. Since there is a fraction in the denominator, separate that fraction into a reciprocal
times the other terms, then invert the fraction and multiply.
You can also solve using exponents: A ● (B ● C─1)─1 = A ● B─1 ● C
Q3. Simplify (A/B)/C = ***** Answer
Note these three different but equivalent ways of representing this problem, and
then how the answer differs from the previous example.
(A/B)/C = A
B
C = A ●1
B
C Rewrite for clarity^ = A
B●C ^Rule 2b Since there is a fraction in the numerator, separate that fraction from other terms.
Q4. Simplify (A/B)/(C •(D/E)) = ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 17 Answer
(A/B)/(C •(D/E)) A=
B
C• D
E = Rewrite for clarity^ A● 1
B
D
C
E
^Rule 2a = A● 1 ● 1
B
C
D
E
^Rule 2b = A ●E
B●C ●D ^Rule 2c In solving problems, when you are writing terms with two or more fractions, you will need
to develop a systematic way to distinguish fractions in the numerator and denominator in
cases where confusion may occur. Practice A:
1. meters
meters
sec 2. (D/E)─1 = X/(Y/Z) = 3. Learn the rules, then simplify these. 5. = 4. (meters/sec)
sec = meters/sec/sec = ***** ANSWERS
Practice A
1. X/(Y/Z) = X • 1 = X • Z
Y
Y
Z ©2011 ChemReview.net v. f1 2. (D/E)─1 = 1=
D
E E or (D─1 • E)
D Page OnScreen 17 18 3. meters
meters
sec = meters • 1=
meters
sec = meters • sec = sec
meters = meters • 1 = meters
sec
sec
sec2 4. (meters/sec)
sec 5. meters/sec/sec = ©2011 ChemReview.net v. f1 Cannot be evaluated unless you know the numerator or denominator. Page OnScreen 17 19 * ****
Dividing Complex Numbers and Units
For some derived quantities and constants, units are complex fractions. Examples that we
will soon encounter include
The Gas Constant = R = 0.0821 atm • L
mole • K
The specific heat capacity of water = cwater = 4.184 joule/gram•K
(Joule is a unit that measures energy.) In those units, the dot between two units means that
the two units are multiplied together in either the numerator or denominator. In
4.184 joule/gram•K , grams and kelvins are both in the denominator.
4.184 joule/gram•K = 4.184 joule/(gram•K) = 4.184 joule • g─1• K─1
(If your course at this point uses the unit─1 notation frequently in calculations, you should
complete Lesson 27F after this lesson.)
When multiplying and dividing terms with complex units, you must to do the math for
both the numbers and the units. When units are found in fractions, unit cancellation follows
the rules of algebra reviewed in the section above. Try this problem.
Q. Solve for the number and the unit of the answer. =
360 joules
18.0 K ● 0.50 joules
gram·K ***** Answer
Note how the units are rearranged in each step.
=
360 joules
18.0 K ● 0.50 joules
gram·K 360 joules ●
1
= 360 joules ● gram•K = 40. g
18.0 K
0.50 joules
18.0 K
0.50 joules
gram·K Solving uses Rule 2a: if there is a unit that is a fraction in the denominator, separate the
fraction into 1/fraction in the denominator multiplied by the other terms. Then, invert
the reciprocal, cancel units that cancel, and multiply.
Mark the unit cancellation in the final step above.
©2011 ChemReview.net v. f1 Page OnScreen 17 20 Cancellation Shortcuts
When canceling numbers and units in complex fractions, you can often simplify by first
canceling separately within the numerator and denominator, and then canceling between the
numerator and denominator. However, when using this shortcut, you must remember that
canceling a number or unit does not get rid of it: it replaces it with a 1. Try this example.
Q. First cancel units in the denominator, then
between the numerator and denominator: 360 joules
12 K ● 0.20 joules
gram·K = *****
Answer
360 joules
12 K ● 0.20 joules
gram·K = 360 joules
12 K ● 0.50 joules
gram·K = 360
●
12 ● 0.20 1
1
gram = 150 g In the third term, the 1’s are important. 1/grams and 1/1/grams are not the same.
When in doubt about unit cancellation, skip the shortcuts and use systematic Rules 12c. Practice B: Apply the rules above to simplify these. 1. calories
=
calorie • gram
gram • ºC 3. 2. atm ● L
=
(mole) • atm ● L
mole ● K (mole) • atm ● L • (K) =
mole ● K
liters ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 21 ANSWERS
Practice B
=
calories
calorie • gram
gram • ºC 1. 2. atm ● L
(mole) • atm ● L
mole ● K 3. = 1
● calories
calorie
gram
gram • ºC atm ● L
mole ● 1
=
atm ● L
mole ● K = gram • ºC ● calories
calorie
gram = ºC atm ● L ● mole ● K = K
mole
atm ● L (mole) • atm ● L • (K) = (mol) • ( atm • L ) • (K) ● 1
mole ● K
mol • K
L
liters = atm ***** Lesson 17D: The IdealGas Law  and Solving Equations
In calculations for a gas at STP, conversions using the STP prompt will often be the fastest
way to solve. However, if gas conditions are at temperatures or pressures that are not STP,
you will need to use gas equations to solve.
The most frequently used gas equation is the ideal gas law (often memorized as pivnert).
For all “ideal” gases, PV = nRT
where
•
•
•
• P and V are pressure and volume in any units;
T is temperature in kelvins
n represents the number of moles of gas; and
R is a number with units called the gas constant. In this equation, P, V, T, and n are variables. They can have any values, depending on the
conditions in the problem.
The gas constant (R) is not a variable. It does not change as you vary P, V, T and n. ©2011 ChemReview.net v. f1 Page OnScreen 17 22 When using R to solve problems, R must have the same units as those used that are used to
measure P and V in the problem. The number and units used for R will change depending
on which P and V units are used. However, just as 12 fluid ounces is the same as 355 mL
(our sodacan equality), the different numbers and unit combinations used for R do not
change the constant quantity that R represents.
Using algebra, if you know any three of the four variables in the ideal gas law, and you
know a table value for the constant R, you can solve for the fourth variable.
One of the interesting implications of the ideal gas law is that if any of the three variables in
PV=nRT are the same for samples of different gases, the fourth variable will have the same
value no matter what are the chemical formulas for the gases. When using PV=nRT, the
formula for the gas is not needed, and no matter what the substance formula for the gas is,
it will not change the answer. NonIdeal Behavior
For most gases, the ideal gas law is a good approximation in predicting behavior, so long as
the gas is not either at pressures substantially above one atmosphere or close to conditions
of pressure and temperature where it condenses to form a liquid or solid. In reality, all real
gases display “nonideal” behavior if the temperature is low enough and/or the pressure is
high enough. However, most gases that are not approaching a condensation point will
display close to ideal behavior and comply approximately with the predictions of the ideal
gas law. Solving Problems Which Require Equations
Calculations in chemistry can generally be put into three categories: those that can be
solved with conversions, those that require equations, and those requiring both.
We will use the gas laws to develop a system for solving calculations that require equations.
This system will be especially helpful if you take additional science courses.
Let’s start with an easy problem, one in which we know the correct equation to use. This
relatively easy example can be solved several ways, but we will solve with a method that
has the advantage of working with more difficult problems. Please, try the method used
here.
For this problem, do the steps below in your notebook.
Q. A table of R values may usually be consulted when solving gas problems.
However, R can also be easily calculated for any set of units, based on values you
know for standard temperature and pressure (measured in several different units)
and the volume of one mole of gas at STP.
For example, one mole of any gas occupies a volume of 22.4 liters at a standard
pressure of one atmosphere and standard temperature (273 K).
Use this data and the ideal gas law equation to calculate a value for the gas constant
(R), using the units specified in this problem. Steps For Solving Using Equations
1. It says to use the ideal gas law. Write the equation from memory.
2. Next, make a data table that includes each symbol in the equation. ©2011 ChemReview.net v. f1 Page OnScreen 17 23 DATA: P=
V=
n=
R=
T= 3. Put a ? after the symbol WANTED in the problem.
4. Read the problem again, and write each number and unit after a symbol. Use the units
to match the symbol to the data (one mole goes after n, 273 K goes after T, etc.).
Do those steps, then check below. *****
At this point, your paper should look like this.
PV = nRT
DATA: P = 1 atm.
V = 22.4 L
n = 1 mole
R= ?
T = 273 K 5. Write SOLVE, and, using algebra, solve the fundamental, memorized equation for the
symbol WANTED. Do not plug in numbers until after you have solved for the
WANTED symbol in symbols.
6. Plug in the numbers and solve. Cancel units when appropriate, but leave the units that
do not cancel, and include them after the number that you calculate for the answer.
Do those steps, then check your work below. ***** On your paper, you should have added this:
SOLVE: PV = nRT
? = R = PV
nT ©2011 ChemReview.net v. f1 = (1 atm)(22.4 L)
(1 mol)(273 K) = 0.0821 atm ● L
mol ● K Page OnScreen 17 24 Be certain that your answer includes both numbers and units.
A check inside the cover of your class textbook may show that this answer is accurate to
within one doubtful digit for one of the values listed for R.
Values for R vary depending on the units used to measure pressure and volume. If you
need a value for R and you do not have access to a table, use the following rule as applied
in the problem above.
To calculate a value for R using required units, use PV=nRT and the values for P, V,
and T that are true for those units and one mole of gas at STP.
***** Summary: To solve an equation when you know which equation is needed.
1. Write the memorized equation.
Memorize equations in one fundamental format, then use algebra to solve for the
symbols WANTED.
For example, memorize: K = oC + 273 Do not memorize oC = K ─ 273 If Celsius is WANTED, knowing kelvins, write the memorized equation above,
then solve the equation for Celsius.
2. Make a data table. On each line, put one symbol from the equation.
3. Based on units, after each symbol, write the matching DATA in the problem.
4. Solve your memorized equation for the WANTED symbol before plugging in
numbers and units. Symbols move with fewer mistakes than numbers with their
units.
5. Put both numbers and units into the equation when you solve. If units cancel
correctly, it is a check that the algebra was done correctly. If the units do not cancel
properly, check your work. ©2011 ChemReview.net v. f1 Page OnScreen 17 25 Practice
1. Assign each of these gas measurements with a symbol from PV = nRT.
a. 0.50 moles c. 11.2 dm3 b. 202 kPa d. 373 K e. 38 torr 2. Solve PV = nRT for
a. n = b. T = c. V = Use your calculated value for R (R = 0.0821 atm•L/mol•K) and the ideal gas law to solve
the following problem. Employ the method used in this lesson.
3. If the pressure of one mole of gas at STP increases to 2.3 atmospheres but the volume of
the gas is held constant, what must the new temperature be? ***** ANSWERS
1. a. 0.50 moles n 2. If PV = nRT: b. 202 kPa P (a) n = PV
RT c. 11.2 dm3 V (b) T = PV
nR d. 373 K T e. 38 torr P (c) V = nRT
P If you cannot do this algebra correctly every time, find a friend or tutor who can help you to review the
algebra for this and the following lessons in this module. It will not take long to master. Gas laws are not
difficult if you can do this algebra but are impossible if you cannot. ©2011 ChemReview.net v. f1 Page OnScreen 17 26 3. PV = nRT
DATA: P
V
n
R 2.3 atm.
22.4 L
1 mol
0.0821 atm • L
mol • K
T=? SOLVE: =
=
=
= PV = nRT ? = T = PV = (2.3 atm)(22.4 L)
1
=
nR
(1 mole)
( 0.0821 atm • L )
mol • K (2.3 atm)(22.4 L) ● mol • K
= 630 K
1 mol
0.0821 atm • L 630 K is a large rise in temperature from 273 K. It is reasonable because for one mole of gas at 22.4 liters
volume to cause a pressure of 2.3 atm. (rather than 1 atm, as at STP), the gas must be very hot, with
molecules moving much faster on average than at standard temperature.
The unit cancellation: From Lesson 16C, use Rule 2a: Separate a fractional unit on the bottom into a
reciprocal, and Rule 2c: A fraction on the bottom flips over when you bring it to the top.
Mark the unit cancellation in the final step above.
When solving with equations, always include the unit cancellation. If the units cancel correctly, the
numbers were probably put in the right place to get the right answer.
***** ©2011 ChemReview.net v. f1 Page OnScreen 17 27 Lesson 17E: Choosing Consistent Units
Consistent Units
When solving most equations in science, numbers must have units attached, and the units
must cancel to result in the unit WANTED. In order for units to cancel, the units must be
consistent. This means
For each fundamental or derived quantity used in an equation, you must choose
one unit to measure the quantity, then convert all data for that quantity to that unit.
For example, as one part of solving calculations using PV=nRT , you must pick a
pressure unit (such as atmospheres, pascals, or torr), and convert all DATA that involves
pressure to the unit you choose.
In some cases, an equation will require certain units.
For example, gas law equations using a capital T require temperature to be measured in
an absolute temperature scale. In the metric system, this means kelvins. A DATA
temperature that is not in kelvins must be converted to kelvins. If degrees Celsius is
WANTED, kelvins must be found first.
When using PV = nRT , consistent units are a factor
• when you must pick an R value to use, and • when you must convert DATA to match the units of R. Let’s take these cases one at a time. Choosing an R
The gas constant R is one quantity, but it can be
expressed in different units. Some of the equivalent
values for R are in the table at the right. R= 0.0821 atm•L/mol•K = 8.31 kPa•L/mol•K = 62.4 torr •L/mol•K Note that in these values, the only unit that differs
is the pressure unit.
In most problems that require the use of R, you will be given a table of R values such as the
above (you generally will not need to memorize values for R), but you will need to select
which R to use.
If you need to choose an R, the rule is:
Pick the R which has units that most closely match the units in the DATA.
Applying that rule, try the following problem in your notebook.
Q. A sample of an ideal gas at 293 K and 202 kPa has a volume of 301 mL. How many
moles are in the sample? (Use one of the gas constant values above.) ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 28 The WANTED unit has the symbol n, 293 K is a T, 202 kPa is a P, and 301 mL is a V. What
equation relates n, T, P, and V?
*****
PV = nRT Complete the DATA table using those symbols. *****
DATA: P = 202 kPa
V = 301 mL
n = ? mol = WANTED
R= ?
T = 293 K Which R value should you choose? ***** R = 8.31 kPa•L/mol•K is not an exact match with the DATA units, but because it
uses kPa it is the closest. Add that R value to the DATA table.
One more change is needed. The units for P, T, n, match the chosen R units, but our best
match for R uses liters, while the supplied V is in mL. The volume unit must be consistent:
we must pick mL or L, or the units will not cancel and the equation will not work. Since we
don’t see an R value that uses mL, what’s the best option? ***** Convert the supplied DATA to liters. Do so in the DATA table and complete the
problem. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 29 DATA: SOLVE: P = 202 kPa
V = 301 mL = 0.301 L
n = ? moles = WANTED
R = 8.31 kPa•L/mol•K
T = 293 K
PV = nRT (by inspection, see Lesson 12A) Are all of the units now consistent? for the WANTED symbol then plug in numbers and units. ***** ? = n = PV
RT = (202 kPa)(0.301 L)
(293K) 8.31 kPa • L
mol • K = (202 kPa)(0.301 L) •
mol • K =
(293K)
8.31 kPa • L = 0.0250 mol
Check that the units cancel properly. They must. If you have done the problem
correctly, they will. Converting to Consistent Units
In many science calculations, we are given only one version of a constant. In such cases,
the constant can be converted to other units, but it is usually easier to convert the DATA to
the units of the constant.
In solving with equations, once you have chosen the consistent units, a systematic approach
is to write those units in the DATA table after each symbol, such as
DATA: P = ? atm. =
V=?L=
T=?K= Then list the data as supplied, and, in the DATA table, convert the supplied data to the
consistent unit written after the variable symbol.
Try that approach in this problem.
Q. Find the temperature in degrees Celsius of 0.0100 moles of an ideal gas that has a
volume of 125. mL and a pressure of 2.00 atm. (USE R = 8.31 kPa•L/mol•K) Decide the equation that will be used, then make a DATA table using the symbols in the
equation. After each symbol, write the chosen consistent unit, then an = sign, as done
above. In this problem, choose as consistent units the units in the supplied R. Since this R
will decide the remaining units, put R and its value in the DATA table first. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 30 The wanted unit has the symbol T and the data is in moles (n), mL (V) and pressure (P).
The equation relates T, n, V and P?
PV = nRT
DATA: R = 8.31 kPa•L/mol•K (required in this problem) P = ? kPa = 2.00 atm
In this problem, P must be in kPa to be consistent with the required R units. Convert
the P data to kPa, then finish the table. ***** P = ? kPa = 2.00 atm ● 101 kPa
1 atm = 202 kPa V = ? L = 125 mL
V must be in L to be consistent with the R units. Convert V, then finish the table and
solve. ***** V = ? L = 125 mL = 0.125 L (by inspection: mL is 1000x L) n = ? mol = 0.0100 mol
T must always be in K but ºC is WANTED
For temperature, we want Celsius, but the equation only works in kelvins. When using
equations, the rule is: solve for the WANTED quantity in consistent or required units
first, then convert if needed to other WANTED units.
SOLVE: PV = nRT for the WANTED symbol then plug in numbers and units. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 31 T = PV =
nR (202 kPa) (0.125 L)
1
=
(0.0100 mol)
( 8.31 kPa • L )
mol • K
= 304 K (202 kPa) (0.125 L) ●
mol • K =
(0.0100 mol)
8.31 kPa • L If the WANTED unit is what it should be after unit cancellation, there is a good chance
that the numbers were put in the right place to get the right answer.
Finally, since the WANTED unit Celsius, use the equation that relates K and ºC.
K = ºC + 273 so ºC = K ─ 273 = 304 K ─ 273 = 31 ºC To summarize:
To solve with equations, first convert to consistent units, solve in consistent units,
then convert the consistent WANTED unit to other units if needed. Practice: Use one of the following values. R = 0.0821 atm•L/mol•K = 8.31 kPa•L/mol•K = 62.4 torr •L/mol•K
1. If 0.0500 mole of an ideal gas at 0ºC has a volume of 560. mL, what will be its pressure
in torr? ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 32 ANSWER
1. PV = nRT
DATA: P in torr = WANTED
(the only R value that uses torr) R = 62.4 torr • L
mol • K V = ? L to match R unit = 560. mL = 0.560 L (see Lesson 12A) n = 0.0500 mol
T must be in K : K = ºC +273 = 0 ºC + 273 = 273 K ***** SOLVE: PV = nRT = 1,520 torr
? = P = nRT = nRT ● 1 = (0.0500 mol)( 62.4 torr • L )(273 K) ● 1
V
V
mol • K
0.560 L
Mark the unit cancellation in the last step.
To help with unit cancellation, a good rule is
When solving an equation in symbols results in a fraction that include terms with fractional units (such as R
above) on the top or bottom, rewrite the equation with the symbols in the denominator separated into a
reciprocal, then plug in numbers into this separated format (see Lesson 16C).
However, you may do the math for the numbers and units in any way you choose, provided you do both
numbers and units.
***** ©2011 ChemReview.net v. f1 Page OnScreen 17 33 Lesson 17F: Density, Molar Mass, and Choosing Equations
The ideal gas law has one constant (R) and 4 variables (P, V, T, and n). The PV = nRT
equation may be linked to other equations to find other variables. The key to this relating of
equations is to find a variable which the equations share.
In chemistry, the variable that most relationships share most often is the unit for counting
the number of particles involved in a chemical process: moles. Grams, Molar Mass, and Ideal Gases
Knowing any four of the quantities in the equation PV=nRT, the fifth can be calculated.
For the grams, moles, and molar mass of any substance, if you know any two of those
quantities, conversions will find the third.
Because moles is a common factor in both of those two relationships, PV=nRT can be linked
to calculations involving grams and molar mass.
In calculations involving grams and molar mass with the ideal gas law, a useful strategy is to
solve in two parts, solving for the common variable (moles) in one part, then using that
answer to solve the other part.
• Find moles using whichever relationship, either PV=nRT or the grams to moles
conversion, provides enough data to find moles. If the final WANTED variable is in
one part, you will usually need to solve the other part first. • Then use the found moles as DATA in the other part to solve for the variable
WANTED. Keeping those two steps in mind, try this example in your notebook.
Q. A sample of gas has a volume of 5.60 liters at 2.00 atm pressure and standard
temperature. If the gas sample has a mass of 2.00 grams, what is the molar mass of
the gas? (R = 0.0821 atm•L/mol•K = 8.31 kPa•L/mol•K = 62.4 torr •L/mol•K) ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 34 Answer
It says to use R. So far, we only know one relationship involving R.
PV = nRT
Knowing the specific equation needed, use those symbols for the DATA table.
DATA: P = 2.00 atm.
V = 5.60 L
n= ?
R = 0.0821 atm • L
mol • K (consistent units: P uses atm.) T = 273 K (std. T)
This problem has additional WANTED and DATA:
WANT: Molar Mass = ? g
mol DATA: 2.00 g gas The final WANTED unit is a grams to moles ratio. The data supplies grams. IF the
moles of the gas can be found, dividing grams by moles finds the molar mass.
In the PV=nRT part of the problem, 4 of the 5 quantities are known; algebra will find
the 5th, the moles needed to find the molar mass.
The strategy? Solve for the linked variable first in the part of the calculation that does
not include the final WANTED unit, then use that answer to solve the relationship that
includes the final WANTED unit.
SOLVE: PV = nRT
? = n = PV
RT You want to find moles.
= (2.00 atm)(5.60 L)
=
( 0.0821 atm • L ) (273 K)
mol • K 0.5000 moles (In calculations that use an answer from one part to solve a later part, carry an extra sf
until the final step.)
Use the answer for that part as DATA to find the WANTED.
WANTED: ? g=
2.00 g
mol
0.5000 mol ©2011 ChemReview.net v. f1 = 4.00 g
mol Page OnScreen 17 35 Practice A: Use the above methods. Answers are below.
(R = 0.0821 atm•L/mol•K = 8.31 kPa•L/mol•K = 62.4 torr •L/mol•K)
1. If a 1.76 gram sample of uranium hexafluoride gas is at 380. torr and 600. K, what
volume will it occupy?
2. A sample of neon gas has a volume of 8.96 L at 298 K and 2.00 atm. How many grams
of Ne are in the sample? ***** ANSWERS
Practice A
1. If a 1.76 gram sample of uranium hexafluoride gas is at 380. torr and 600. K, what
volume will it occupy?
Hints: The data will include 1.76 g UF6
It says to use R. Write the only relationship that we know so far which uses R. ***** PV = nRT Make your data table to match the equation: ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 36 DATA: n= ? V = ? WANTED
Plus: P = 380 torr R = 62.4 torr • L
mol • K
T = 600. K (P uses torr) 1.76 g UF6
(g prompt) 352.0 g UF6 = 1 mol UF6
Strategy: Looking at the PV=nRT data table, if all of the quantities but one are known, we can solve
with algebra. Missing 2 variables, we can’t. But the plus data gives us a way to find moles of
the gas. The rule is: first solve the part without the final WANTED variable. ***** ? mol UF6 = 1.76 g UF6 ● 1 mol UF6 = 5.000 x 10─3 mol UF6
352.0 g UF6
Adding those moles to the PV=nRT data table, the V WANTED can be found.
= 0.493 L
? = V = nRT = nRT ● 1 = ( 5.000 x 10─3 mol )( 62.4 torr • L )(600 K) ● 1
P
P
mol • K
380 torr 2. A sample of neon gas has a volume of 8.96 L at 298 K and 2.00 atm. How many grams
of Ne are in the sample?
WANT:
DATA: ? g Ne
PV = nRT
P = 2.0 atm.
V = 8.96 L
R = 0.0821 L • atm
mol • K
T = 298 K (It says to use a relationship with R)
n= ?
(Pick an R value that has units that match
the units used in the problem.) Plus: 20.2 g Ne = 1 mol Ne (grams prompt) Strategy: You want grams. If you know the moles, you can find the grams by molar mass conversion.
Using PV=nRT, you can find the moles. Complete that part first. ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 37 mol • K
= 0.7324 mol Ne
? = n = PV = PV ● 1 = 2.00 atm ● 8.96 L ●
RT
TR
298 K
0.0821 L • atm
Note how the term with the fractional unit (R) was separated to simplify cancellation.
Use the common variable from one part to solve the other part. ***** ? g Ne = 0.7324 mol Ne ● 20.2 g Ne = 14.8 g Ne
1 mol Ne ©2011 ChemReview.net v. f1 Page OnScreen 17 38 *****
Density, Molar Mass, and Ideal Gases
The relationship between the ideal gas law, molar mass, and gas density is more
mathematically complex than the relationship involving grams and moles. For a complex
relationship, it is often best to derive an equation, then memorize it.
The relationship using density and molar mass for ideal gases can be derived as follows.
Molar Mass (MM) ≡ grams , so we can write grams = moles ● MM
moles
Density (D) ≡ mass ; using grams for mass,
volume Dgrams = (1) grams
Any V unit (2) moles = Dgrams
V
MM (3) Substituting equation (1) into (2),
Dgrams = moles ● MM , which can be rewritten as
V
Since PV = nRT can be rewritten as P = moles ● RT ,
V Substituting with equation (3) gives P = Dgrams ● RT
MM (4) The density can be measured using any mass unit as long as the in mass/mol ratio of the
molar mass, the mass is measured in the same unit.
Equation (4) is a form of the ideal gas law that uses density and molar mass in place of
moles and volume. You can either work out the above derivation every time an ideal gas
problem involves density..., OR you can memorize these two equations:
The ideal gas law: PV = nRT and P = Dgrams • RT
MM If an ideal gas problem mentions gas density or molar mass at conditions not at STP, it is
usually a prompt to use the form of the ideal gas law that includes density and molar mass. When the Equation To Use Is Not Clear
So far, we have learned only two gaslaw equations, but we are about to learn a few more.
Other science courses will have many equations to memorize. How do you decide which
equation to use when?
Correctly choosing which to use will require a system. The following system is easy to use.
It will work in chemistry and in other science courses. ©2011 ChemReview.net v. f1 Page OnScreen 17 39 A System for Finding the Right Equation
When you are not sure which equation to use, OR whether to use equations or conversions to
solve, do these steps.
1. As you have been doing, first write the WANTED unit and/or symbol.
2. List the DATA with units, substances, and descriptive labels. Add any prompts.
3. Analyze whether the problem will require conversions or an equation.
• Try conversions first. Conversions often work if most of the data in pairs and ratio
units. • If the data is mostly in single units, you will likely need an equation. Recent
lessons in your course will likely indicate the equations that may be needed. • Watch for hints at the need for an equation. For example, the mention of R in a gas
problem is likely a prompt that you will need a form of the ideal gas law to solve. 4. If a lecture or textbook is frequently using certain equations, learn them before you do
the practice problems. Write them at the top of your paper at the start of each
assignment, quiz, or test.
5. If you recognize which equation is needed, write the equation, then make a DATA table
listing every symbol in the equation. Fill in the table with the DATA from the problem.
6. If conversions don’t work and you suspect you need an equation, but you are not sure
which equation is needed, try this:
Label each item of WANTED and DATA with a symbol based on its units. Use symbols
that are used in the equations for the topic you are studying.
For example,
• 25 kPa would be labeled with a P for pressure; •
• 293 K would get a T for Kelvin scale temperature;
20oC gets a lower case t for degrees Celsius; • Liters or mL or cm3 or dm3 would be assigned a V for volume; • 25 grams would be labeled with a lower case m for mass; • 18.0 grams per mole  label MM for molar mass; • 2.5 moles per liter, in a problem about aqueous solutions, would be labeled M
for molarity; • 7.5 grams per liter, a mass over volume, would be labeled D for density. Label the WANTED unit, as well as each item of DATA, with a symbol.
7. Compare the symbols listed in the WANTED and DATA to your written, memorized list
of equations for the topic. Find the equation that uses those symbols. Write the
equation, as memorized, below the data.
If no equations match exactly, see if the problem’s symbols can be converted to give the
symbols for a known equation (for example, degrees Celsius can be converted to
kelvins, and grams can be converted to moles if you know a substance formula).
©2011 ChemReview.net v. f1 Page OnScreen 17 40 8. Watch for the variables that make similar equations different. For example, in the two
forms of the ideal gas law, the first uses volume and moles. The second uses molar
mass and density instead of volume and moles. The symbols that you assign to the
WANTED and DATA will identify which equation to use.
9. SOLVE the equation for the WANTED variable in symbols before plugging in numbers.
Symbols move quickly. If instead of moving symbols, you move numbers and their
units and their labels, more errors will tend to occur. Practice B
Using the method above, try these in your notebook. If you get stuck, read a portion of the
answer and try again.
1. Assign symbols used in gas law equations to these quantities.
a. 122 g/mol
2. Solve b. 202 kPa P = Dgrams • RT
MM c. 13.5 g/mL for a) D = d. 30˚C
b) MM c) T 3. If the density of a gas at 27oC and standard pressure is 1.79 g/L, what is its molar
mass?
4. Write values for standard pressure in 5 different units.
5. If one mole of any gas occupies 22.4 liters of volume at STP, calculate the value of R in
units of kPa • L/mol • K. (Do not use a table value for R.)
6. What is the density of hydrogen gas (H2) in grams per liter at STP? ***** ANSWERS
Practice B
1. a. 122 g/mol MM ©2011 ChemReview.net v. f1 b. 202 kPa P c. 13.5 g/mL D d. 30˚C t Page OnScreen 17 41 2. Solve P = Dgrams • RT
MM for b) MM = Dgrams • R • T
P a) D = P ● MM
RT c) T = P ● MM
Dg ● R 3. Use conversions or equations? To decide, list the data. If the data is mostly pairs or ratio units, try
conversions. If conversions won’t work (and on this problem they don’t), use the units and words in the
problem to assign symbols, and see if a known formula fits the symbols.
WANTED: ?
DATA: g
mole
27 ºC MM
t T = 27 ºC + 273 = 300. K Std. P = 1 atm.
1.90 g = 1 L
(Strategy: (3 sf – doubt in one’s place when adding) P
Dgrams If the gas were at STP, we could use conversions to solve, but 27ºC is not standard
temperature.
We have memorized two gas equations that use pressure and temperature:
PV = nRT and P = Dgrams • RT
MM The second equation fits the data. (R is a constant that you are normally allowed to look up
in a table even if it is not supplied (and if not, it is easy to calculate)).
If an equation fits the symbols, try the equation. Solving for the wanted MM:) *****
©2011 ChemReview.net v. f1 Page OnScreen 17 42 MM = density • RT = 1.79 g
P
L • 1 • 0.0821 atm • L • 300. K = 44.1 g
1 atm
mol • K
mol Terms with fractional units (density and R) were separated for cancellation. 4. 1 Atmosphere ≡ 760 mm Hg ≡ 760 Torr = 101 kilopascals (kPa) = 1.01 bars 5. WANTED: R in kPa· L
mol· K
(R is in both “ideal gas law” equations. To decide which to use, list and symbol the data.)
DATA: std P = 101 kPa in the units wanted for R
22.4 L V 1 mol P n std. temp = 273 K T
Strategy: Looking at all 5 symbols labeling WANTED and DATA, which equation works? ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 43 SOLVE: PV = nRT
? = R = PV
nT = (101 kPa)(22.4 L) = 8.29 kPa • L
(1 mol)(273 K)
mol • K Using less rounded values for P, T, and V, the accepted value is 8.31 kPa • L / mol • K 6. What is the density of hydrogen gas (H2), in grams per liter, at STP?
g H2 gas
L H2 gas at STP WANTED: ? DATA: 2.016 g H2 = 1 mol H2 (grams prompt) 1 mol any gas = 22.4 L any gas at STP
Strategy: (STP Prompt) Analyze your units. Equalities lend themselves to conversions. You want grams over liters.
You know grams to moles and moles to liters equalities. ***** SOLVE:
? =
g H2 gas
L H2 gas at STP 2.016 g H2 ● 1 mol gas
= 0.0900 g H2(g)
1 mol H2
L H2 at STP
22.4 L gas STP ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 44 Lesson 17G: Using the Combined Equation
Many gaslaw calculations involve the special case where the gas is trapped (the moles of
gas in a sample does not change) while P, V and/or T are changed.
If the number of gas moles is held constant, we can rewrite PV = nRT as
PV = nR = (constant moles)(gas constant R) = (a new constant)
T
When you multiply two constants, the result is new constant. While R is always a constant,
the new constant above will only be true for the number of moles in a problem.
However, if gas conditions are changed while the moles of gas is held constant, the
equation above means that the ratio “P times V over T” will keep the same numeric value
no matter how you change P, V, and T.
Another way to express this relationship: so long as the number of particles of gas does not
change, if you have an initial set of conditions P1, V1, and T1, and you change to a new set
of conditions P2, V2, and T2, the ratio PV/T must stay the same. Expressed in the elegant
and efficient shorthand that is algebra,
nR = P1V1 = P2V2
T2
T1 when the moles of gas is held constant. The boxed equation means that IF 5 of the 6 variables among P1, V1, T1, P2, V2, and T2 are
known, the 6th variable may be found using algebra, without knowing n or needing R.
We will call this relationship the combined equation (because it combines three historic gas
laws). For problems in which the moles of gas particles do not change, when one set of
conditions is changed to new conditions, the quickest way to solve is usually to use
P1V1 = P2V2
T1
T2 where P1, V1, and T1 are starting gas measurements,
and P2, V2, and T2 are the final gas measurements.
and the moles of gas stay the same. This equation is often memorized by repeated recitation of “P one V one over T one equals
P two V two over T two.”
It may help to remember this rule:
When a problem says a sample of gas is sealed or trapped or has constant moles, and the
sample has a change in conditions, see if the combined equation symbols fit the data.
For a problem in which the equation needed is known, apply the system used for previous
equations.
• Write the fundamental equation. • Set up a data table that contains the symbols in the equation. ©2011 ChemReview.net v. f1 Page OnScreen 17 45 In a problem that requires the combined equation, start with
P1V1 = P2V2
T2
T1
DATA: (when the moles of gas are held constant) P1 = P2 = V1 = V2 = T1 = T2 = Then,
• put the numbers and their units from the problem into the table. • Label the symbol WANTED with a ?. Add the units WANTED if they are specified. • SOLVE the fundamental equation in symbols for the WANTED symbol, using
algebra, before you plug in numbers. • Then plug in numbers and units, and solve. Do unit cancellation as well as number
math. Cover the answer below, and then try the method on this problem, using the combined
equation to solve:
Q. If a sample of gas in a sealed but flexible balloon at 273K and 1.00 atm. pressure has
a volume of 15.0 liters, and the pressure is increased to 2.5 atmospheres while the
temperature is increased to 373 K, what will be the new volume of the balloon? ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 46 P1V1 = P2V2
T1
T2
P1 = 1.0 atm. P2 = 2.5 atm. V1 = 15.0 Liters V2 = ? T1 = 273 K (std T) DATA: T2 = 373 K SOLVE: ? = V2 = P1V1 T2
P2 T1 = (1.0 atm) (15.0 L) (373 K) = 8.2 L
(2.5 atm) (273 K) If the units do not cancel to give the WANTED unit, check your work. Practice A: Check answers at the end of this lesson as you go. 1. Solve the combined equation for
a. T2 = b. P1 = c. V2 = 2. A gas cylinder with a volume of 2.50 liters is at room temperature (293 K). The pressure
inside the tank is 100. atmospheres. When the gas is released into a 50.0 liter container,
the gas pressure falls to 2.00 atmospheres. What will be the new temperature of the gas
in kelvins, and in degrees Celsius? ***** ANSWERS
Practice A
1. a. T2 If P1V1 = P2V2 , then T
PVT
2= 221
T1
T2
P1 V1 ©2011 ChemReview.net v. f1 Page OnScreen 17 47 b. P1 = P2 V2 T1
T2 V1 c. V2 = P1 V1 T2
T1 P2 As a check, note the patterns above.
• The P’s and V’s, if in the same group, have the same subscript, but a T grouped with them will
have the opposite subscript. • In the fractions, there is always one more term on the top than on the bottom. • If you substitute consistent units for all of the symbols, the units cancel correctly. If you have trouble solving for any of the six symbols, find a friend or tutor to help you learn the algebra. 2. If the moles of gas particles do not change, and one set of conditions is changed to new conditions, use
P1V1 = P2V2
T2
T1
P2 = 2.00 atm. DATA: P1 = 100. atm.
V1 = 2.50 L V2 = 50.0 L T1 = 293 K T2 = ? t2 = ? ***** SOLVE: (In gas problems, to find Celsius, solve for kelvins first.)
? = T2 = P2 V2 T1
P1 V1
To find Celsius: ©2011 ChemReview.net v. f1 = (2.00 atm) (50.0 L) (293 K) = 117 K
(100. atm) (2.50 L)
K = ºC + 273 ºC = K ─ 273 = 117K ─ 273 = ─ 156 ºC Page OnScreen 17 48 ***** Consistent Units
In gas law equations, a capital T means absolute temperature. Because the combined
equation is derived from PV=nRT , when using metric units, both of those equations require
that
• temperatures must be converted to kelvins before substituting in the equation. • If a temperature that is not kelvins is WANTED, kelvins must be solved first, and
then converted to other temperature units. For P and V, each variable must be converted to a consistent unit to solve. In most
combined law calculations, since there is no R value with complex units that we are
required to match, it does not matter which V or P units you choose.
• Volume may be in mL or L; and • Pressure can be converted to kPa, atmospheres, torr, bars, or other pressure units. However, you must choose one unit for P and for V, and all DATA for that quantity must be
converted to that unit. It usually simplifies calculations if you convert DATA to the
WANTED unit, but any unit will work as long as units are consistent.
Keeping that in mind, try this problem.
Q. An aerosolspray can contains 250. mL of gas under 4.5 atm. pressure at 27ºC. How
many liters would the gas occupy at 50.5 kPa and std. T? ***** Answer: When one set of gas conditions is changed to new conditions, but moles are held
constant, try
P1V1 = P2V2
T1
T2
DATA: (If you need an equation, write it first. The more
often you write it, the longer it is remembered.) P1 = 4.5 atm. P2 = 50.5 kPa = 0.500 atm. V1 = 250. mL = 0.250 L V2 = ? liters (L) WANTED T1 in K = 27 ºC + 273 = 300. K T2 in K = 273 K ©2011 ChemReview.net v. f1 Page OnScreen 17 49 Unit Conversions:
For T1 , must use K. K = ºC +273 = 27 ºC + 273 = 300. K For P1 , choose one of the units in the DATA: kPa or atm. Either choice will result in
the same answer. If you choose atm,
P2 = ? atm = 50.5 kPa ● 1 atm
101 kPa = 0.500 atm. ( or P1 = 455 kPa) Since liters is WANTED, and the initial volume is in mL, pick a unit and convert the
other to it. It’s best to pick the answer unit if it is specified.
? L = V1 = 250. mL = 0.250 L (automatic: See Lesson 12A) Solve for the WANTED symbol, then substitute the DATA. ***** SOLVE: ? = V2 = P1V1 T2
P2 T1 = (4.5 atm) (0.250 L) (273 K) = 2.0 L
(0.500 atm) (300. K) Practice B
1. A sealed sample of hydrogen gas occupies 500. mL at 20. ºC and 150. kPa. What would
be the temperature in degrees Celsius if the volume of the container is increased to 2.00
liters and the pressure is decreased to 0.550 atm.? ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 50 ANSWERS
Practice B
P1V1 = P2V2
T2
T1 1. For trapped gas moles changing from original to new conditions, use
DATA:
P1 = 150 kPa P2 = 0.550 atm. = 55.6 kPa V1 = 500. mL V2 = 2.00 L = 2.00 x 103 mL T1 = 20 ºC + 273 = 293 K T2 = ? t2 = ? Needed Unit Conversions:
For T1 , K must be used. K = ºC + 273 = 20. ºC + 273 = 293 K For P1 , kPa or atm. could be used as units. If you choose kPa,
? kPa = 0.550 atm ● 101 kPa = 55.6 kPa
1 atm ***** SOLVE: (In gas problems, solve in kelvins first, then Celsius).
? = T2 = P2 V2 T1
P1 V1 = (55.6 kPa) (2,000 mL) (293 K) = 434 K
(150. kPa) (500. mL) To find Celsius, first write the memorized equation, then solve:
K = ºC + 273 ©2011 ChemReview.net v. f1 ºC = K ─ 273 = 434 K ─ 273 = 161 ºC Page OnScreen 17 51 *****
Simplifying Conditions
In a calculation involving a change from initial to final conditions for a constant moles of
gas, the combined equation is used to solve. However, if temperature does not change,
converting to kelvins is not necessary. Why?
If P1V1 = P2V2
T1
T2 , and T1 = T2 , then P1V1 = P2V2 ; use
T1
T2 P1V1 = P2V2 If the “before and after” temperatures are the same, T is not needed to solve.
If any two symbols have the same value in the problem, the values will cancel in the
combined equation because they are the same on both sides. This will simplify the
equation and the arithmetic. Try this example in your notebook.
Q. A sample of chlorine gas has a volume of 22.4 liters at 27ºC and standard pressure.
What will be the pressure in torr if the temperature does not change but the volume
is compressed to 16.8 L?
* * ** *
Answer
P1V1 = P2V2
T2
T1
DATA: (Write needed fundamentals before you get immersed in details) P1 = Std. P  use 760 torr to match P2 = ? in torr V1 = 22.4 L V2 = 16.8 L T1 = 27ºC = T2 T2 Solve for the WANTED symbol first, then substitute the DATA. ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 52 SOLVE: P1V1
T1 = P2V2 ;
T2 ? = P2 in torr = P1V1
V2 P1V1 = P2V2 = (760 torr) (22.4 L) = 1,010 torr
(16.8 L) Practice C
1. A 0.500 liter sample of neon gas in a sealed metal container is at 30. ºC and 380. mm Hg.
What would be the pressure of the gas in kPa at standard temperature? ***** ANSWERS
Practice C
1. For trapped gas changing to new conditions, use P1V1 = P2V2
T1
T2 DATA:
P1 = 380. mm Hg = 50.5 kPa
V1 = 0.500 L = V2
T1 = 30. ºC + 273 = 303 K P2 = ? kPa
(a sealed metal container will have a constant volume)
T2 = 273 K Needed Unit Conversions:
For the pressures, you must change to consistent units: either kPa or mm Hg. If you choose kPa,
? kPa = 380. mm Hg ● = 50.5 kPa
101 kPa
760 mm Hg *****
©2011 ChemReview.net v. f1 Page OnScreen 17 53 P2 = P1 V1 T2
T1 V2 SOLVE: = (50.5 kPa) (273 K) =
(303 K) 45.5 kPa ***** Lesson 17H: Gas Law Summary and Practice
Gas Law Summary
If you have not already done so, master flashcards that include the following rules.
1. Pressure Units
Standard Pressure ≡ 1 atmosphere ≡ 760 mm Hg (mercury) ≡ 760 torr
= 101 kilopascals (kPa) = 1.01 bars
2. If temperature changes in a gas problem, you must use Kelvin temperature.
K = ºC + 273
3. To be a measure of moles, gas volumes must be labeled with a P and a T.
4. The STP prompt.
If a gas is at STP, write as DATA: 1 mole gas = 22.4 liters gas at STP
For gas calculations at STP, try the STP prompt and conversions first. Conversions
solve faster than equations.
5. The ideal gas law:
where R= PV = nRT and P = Dgrams • RT
MM the gas constant = 0.0821 atm•L/mol•K = 8.31 kPa • L /mol•K = 62.4 torr •L/mol•K 6. The combined equation. If the gas conditions are
changed but the moles of gas does not change, use P1V1 = P2V2
T1
T2 7. To solve with equations, first convert to consistent units, solve in consistent units, then
convert the consistent WANTED unit to other units if needed.
8. For solving all kinds of calculations, use a system.
• List the WANTED and DATA; • Try solving with conversions first. If conversions do not work, • Add symbols to the WANTED and DATA, and then write a memorized
fundamental equation that uses those symbols. • Convert all DATA to consistent units. • Solve for the WANTED symbol before plugging in numbers. • Cancel units as a check of your work. ©2011 ChemReview.net v. f1 Page OnScreen 17 54 9. For numbers, units, or symbols: 1/1/X = X ; A/(B/C) = (A · C)/B
***** Practice: The problems below involve a variety of gas relationships. Using the rules
above, you will have a system to solve. Try the odd problems, and some evens in a later
study session. Problem 7 is more challenging.
(R = 0.0821 atm•L/mol•K = 8.31 kPa•L/mol•K = 62.4 torr •L/mol•K)
If you get stuck, read a part of the answer to get a hint, then try again.
1. Solve for the value of R in torr · L /mol · K . Use known data at STP. Do not use a
supplied value for R.
2. The gas in a sealed flexible balloon has a volume of 6.20 liters at 30.ºC and standard
pressure. What will be its volume at ─10. ºC and 740. torr? 3. How many gas molecules will there be, per milliliter, for all gases at STP?
4. If gas in a sealed glass bottle has a pressure of 112 kPa at 25ºC, and the temperature of
the gas is increased to 100.ºC, what will be the pressure?
5. If the density of a gas is 0.0147 g/mL at 20.ºC and standard pressure, what is its molar
mass?
6. In 70.4 grams of UF6 gas, at a volume of 4.48 L and a pressure of 202.6 kPa, what is the
temperature in degrees Celsius?
7. Hydrofluoric acid can be used to etch glass. If 994 mL of HF gas at SP and 30.ºC is
dissolved in water to make 250. mL of HF acid solution, what is the [HF]? *****
ANSWERS
1. WANTED: R in torr· L
mol· K (Hint: Use measurements at STP) R is used in two memorized ideal gas equations. Add symbols to decide which equation to use.
DATA: std P = 760 torr in the units wanted for R
V 22.4 L at STP = 1 mol n std. T = 273 K
Strategy: The symbols above are R, P, V, n, T. Use PV = nRT *****
©2011 ChemReview.net v. f1 Page OnScreen 17 55 SOLVE: 2. WANTED:
DATA: ? = R = PV
nT ? V at end = (760 torr)(22.4 L)
(1 mol)(273 K) = 62.4 torr ● L
mole ● K V2 6.20 L initial V1 30.ºC + 273 = 303 K initial T1 std P = 760 torr initial, using the P units in the problem
─10.oC + 273 = 263 K final
740 torr final
Strategy: P1 T2 P2 The WANTED and DATA symbols match P1V1 = P2V2
T1
T2 ***** SOLVE: 3. WANTED:
DATA: ? = V2 = P1V1 T2
P 2 T1 = (760 torr) (6.20 L) (263 K) = 5.53 L
(740. torr) (303 K) ? molecules gas
mL gas at STP
6.02 x 1023 molecules = 1 mole
1 mol any gas = 22.4 L any gas at STP ©2011 ChemReview.net v. f1 (“molecules” calls the Avogadro prompt)
(STP Prompt)
Page OnScreen 17 56 Strategy: Molecules and mL are wanted. The conversions use molecules, moles, and liters. The
wanted is a ratio; all the data is in equalities. Try conversions. ***** SOLVE:
? molecules gas = 6.02 x 1023 molecules ● 1 mole gas ● 10─3 L = 2.69 x 1019 molecules
mL gas at STP
1 mole
mL gas at STP
22.4 L gas STP
1 mL 4. WANTED:
DATA: ? P  problem uses kPa units, and wants P at end = P2
P = 112 kPa
t = 25oC initial ;
t = 100.oC final Strategy: = P at start = P1
K = 25oC + 273 = 298 K = T1
K = 100.oC + 273 = 373 K = T 2 There are 2 P’s and 2 t’s, but no V’s. However, for a change in a sealed glass bottle, V1 will
equal V2. ***** ©2011 ChemReview.net v. f1 Page OnScreen 17 57 SOLVE:
P1V1 = P2V2 ;
T2
T1 5. WANTED: ? P1 = P2 ; ? = P = P1T2
2
T1
T2
T1 g
mol = 112 kPa • 373 K = 140. kPa
298 K MM DATA: m 0.0147 grams gas = 1 mL gas NOT at STP V
20 oC = 20 + 273 = 293 K mass/Vol is Density T Std P = use 1 atm
Strategy: If the gas were at STP, we could use conversions, but it is not. Try adding symbols.
Choose any P unit since none is specified. Pick one using your units for R.
The D and MM symbols fit P = Dgrams • RT MM
SOLVE: ***** ? = Dg = MM =
mol Dg • RT = 0.0147 g • 1 • 0.0821 atm L • 293 K = ???
mol K
P
mL 1 atm Oops! The volume units don’t cancel. I must have done something wrong. Hmm. My
R uses liters, but the density was in mL. Let’s add a conversion so that L and mL cancel.
? g = MM = density • RT = 0.0147 g • 1 • 0.0821 atm L • 293 K • 1 mL = 354 g
mol
P
mL 1 atm
mol K
10─3 L
mol (It is better to convert all the data to the consistent units needed to cancel your R before you plug into the
equation, but the above, though messy, works. If the units cancel properly, it’s probably right.) ©2011 ChemReview.net v. f1 Page OnScreen 17 58 6. WANTED:
DATA: ? ºC t 70.4 g UF6
352.0 g UF6 = 1 mol UF6
4.48 L gas
202.6 kPa (grams prompt) V
P R = 8.31 kPa • L / mol • K
(P uses kPa)
The symbols are t , V , P , R . You are missing n , but you can solve for moles from the data. Use? ***** PV = nRT n = ? mol UF6 = 70.4 g UF6 • 1 mol UF6
= 0.200 mol UF6
352.0 g UF6 1
T = PV = (202.6 kPa) (4.48 L)
nR
(0.200 mol)
( 8.31 kPa • L )
mol • K = (202.6 kPa) (4.48 L) ● mol • K
(0.200 mol)
8.31 kPa • L = 546 K ºC = K ─ 273 = 546 ─ 273 = 273 ºC 7. WANTED:
DATA: ? mol HF
L HF soln. t 994 mL HF gas at SP and 30.ºC (= 0.994 L HF gas) V
250. mL HF soln. (= 0.250 L HF soln.)
(This problem has two volumes, one for the gas and one for the aqueous solution. Label your
volume units as gas or soln. so that you do not use the solution volume in the gas equation.) Strategy: First, analyze the answer unit.
On top: since V, P, and T are known, moles of gas can be found with PV=nRT.
For the bottom unit, mL, and therefore liters of the solution, is known.
Any R can be used, provided P and V are converted to the R units chosen. *****
©2011 ChemReview.net v. f1 Page OnScreen 17 59 SOLVE: Find moles. PV = nRT DATA: P
V
n
R =
=
=
= 1 atm.
(you can use standard pressure in any unit, so long as the P unit matches R)
= 0.994 L HF gas
?
(Any R can be used, as long as the units are consistent )
0.0821 atm • L
mol • K
T in K = 30.ºC + 273 = 303 K ? = n = PV = 1 ● PV
RT
R
T = mol ● K
● (1 atm)(0.994 L gas)
( 0.0821 atm ● L )
(303 K) = 0.03996 mol HF Separate a complex unit in the denominator, and carry an extra sf until the final step.
Using the answer for that part as DATA to find the WANTED unit. ***** WANTED: ? mol HF = 0.03996 mol HF =
L HF soln
0.250 L HF soln. 0.160 mol HF
L HF soln. = [HF] ##### ©2011 ChemReview.net v. f1 Page OnScreen 17 60 Module 18 — Gas Labs, Gas Reactions Module 18 — Gas Labs, Gas Reactions
Lessons 18A and 18B address graphing of experimental data and writing conclusions for
lab reports which include graphs of data. Those lessons use gas laws as examples, but they
may also be of benefit at any time that you are asked to interpret graphs of experimental
results in chemistry, physics, applied math, or engineering. Lesson 18A: Charles’ Law; Graphing Direct Proportions
Timing: Lesson 18A should be done when you are asked to graph experimental data,
discuss the meaning of a direct proportion, or work with Charles’ law for gases.
Prerequisites: Lessons 17A on gas fundamentals and 17D on the ideal gas law.
***** Charles’ Law: A Direct Proportion
For the special case of gas measurements when volume and temperature are varied, but
moles and pressure are held constant, the ideal gas law PV = nRT can be rewritten as
V = nR = (constant moles)(gas constant R) = (a new constant) = c
T
P
(constant Pressure)
Multiplying and dividing constants results in a new constant, a number with units that we
will term c. The above equation can then be written as
V =c
T for an ideal gas, when moles and pressure are held constant. This is Charles’ law, discovered in the early 19th century by Jacques Charles, a French
scientist and hot air balloonist.
In the above equation, V and T are variables and c is a constant.
Charles’ Law means that in a gas sample, if moles and pressure are held constant, the
numeric value of the ratio V/T will stay the same if you change V or T.
A second way to express this relationship: as long as no particles of gas enter or escape, and
pressure is held constant, if an initial set of conditions V1 and T1 is changed to a new set of
conditions V2 and T2, the V/T ratio will be the same under both conditions. In equation
form:
V2
so long as pressure and moles are held constant.
nR = c = V1 =
T1
T2
P
In this second form of Charles’ law, if any three of the four V and T measurements are
known, the fourth can be found using algebra, without finding or knowing c.
A third way to write Charles’ law, with moles and pressure held constant, is
since V = c
T , we can write V = cT ©2011 ChemReview.net v. f1 , where c is constant. Page OnScreen 18 1 Module 18 — Gas Labs, Gas Reactions Using Charles’ law in the form V = cT , it is easy to see that
• if the absolute temperature (T) is doubled, since c is constant, V must double. • If the volume is cut by half, it can only be because T in kelvins has been reduced by
half. Any relationship that fits the general equation A = cB , where A and B are variables and c
is constant, is called a direct proportion. The c in the equation is the proportionality
constant. Since V and T can vary but c is constant, Charles’ law V = cT is a direct
proportion.
This leads to a fourth way to represent Charles’ law in equations: since V = cT is true,
V T . The is a symbol for “is proportional to.” The equation V T is read, “V is
directly proportional to T.”
Directly proportional simply means if one of the two variables changes by any multiple
(quadruple, cut by 2/3, up by 40%), the other variable must change by the same multiple.
Using this fourth equation, Charles’ law can be translated into words as, “when moles and
pressure of a gas are held constant, volume is directly proportion to absolute temperature.”
To summarize, Charles’ law can be written in four equivalent ways:
Charles’ law: V = c or
T V1 = V2
T1
T2 or V = cT or V T when P and n are constant. These four ways of writing Charles’ law should be memorized. The form of these four
equations can be used to describe any relationship between two variables that is a direct
proportion, and many direct proportions are encountered in science. If any one of the forms
of a equations representing a direct proportion is true, all four will be true. The Loss of Ideal Behavior
One implication of Charles’ law is that as an ideal gas gets colder and colder, its volume
becomes smaller and smaller. At absolute zero, the gas volume would be zero.
In reality, when taken to lower temperatures and/or higher pressures, at some point all
gases condense into liquids or solids. A gas loses most of its volume when it condenses,
but that lower volume then stays close to constant as the liquid or solid is further cooled or
compressed.
In general, the ideal gas law best predicts gas behavior at high temperatures and a
pressures at or below standard pressure (1 atm.). As gases are placed under higher
pressure or approach a temperature and pressure at which they condense, most gases lose
their close to ideal behavior. Under those conditions, they no longer obey the ideal gas law
or laws such as Charles’ law that are derived from the ideal gas law.
**** * ©2011 ChemReview.net v. f1 Page OnScreen 18 2 Module 18 — Gas Labs, Gas Reactions Practice A: Answer all of the following without a calculator. Check answers at the end of this lesson.
1. Which variables must be held constant for Charles’ law to predict gas behavior?
2. Write four equivalent ways of expressing Charles’ law.
3. A sample of gas is contained in a large syringe: a glass cylinder with a tightly sealed
but moveable piston. The pressure of the gas inside the syringe is equal to the
atmospheric pressure outside the syringe, which is held constant.
If the gas volume in the syringe is 30. mL at 200. K,
a. What would be the volume of the gas
i. at 400. K? ii. at 300. K?
b. What must the temperature be if the volume of the gas is changed to
i. 15 mL? ii. 90. mL?
4. A sample of trapped gas molecules in a glass container (a constant volume) is attached
to a pressure gauge. Pressure measurements are taken at a variety of temperatures.
a. Rewrite PV=nRT to conform to the conditions of this experiment, with the 2 variable
terms on the left, and the constants grouped on the right.
b. Will the P versus T relationship be a direct proportion? Why or why not?
c. The following measurements are recorded:
1) At 150. K, P = 74 kPa 2) At 200. K, P = 101 kPa 3) at 250. K, P = 125 kPa Within experimental error, are these data consistent with a direct proportion? Why
or why not?
d. What pressure would you predict at a temperature of 450. K? ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 3 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice A
1. In Charles’ Law, pressure (P) and particles of gas (n) are held constant. V = c or
T 2. 3a. i. V1 = V2
T1
T2 or V = cT or V T when P and n are constant. at 400. K? 60. mL. The absolute temperature doubled; the volume must double. ii. at 300. K? 45 mL. The Kelvin temperature rose 50%; the volume must rise 50%. 3b. i. 15 mL? 100. K. If the volume is cut in half, the temperature in kelvins is halved. ii. 90. mL? 600. K. If the volume is tripled, the absolute temperature is tripled. ©2011 ChemReview.net v. f1 Page OnScreen 18 4 Module 18 — Gas Labs, Gas Reactions 4. a. P = nR
T
V (Variables on the left, constants on the right.) b. Yes. The Part a equation can be written P/T = c . If either P or T doubles, the other variable must
double. c. Yes. The ratio P/T is about the same for all 3 points, which is one test for a direct proportion. d. In each case the value of P in kPa is onehalf T , so P would be predicted to be about ½ of 450 =
~ 225 kPa. ©2011 ChemReview.net v. f1 Page OnScreen 18 5 Module 18 — Gas Labs, Gas Reactions ***** Graphing Direct Proportions
Many relationships can be investigated by changing one variable and measuring a second
variable, while holding all other variables constant. In science experiments, you are often
asked to graph the results of those experiments.
With experimental data, a frequent goal is to find a graph that is an approximate straight
line. The data can then be interpreted based on the equation for a line: y = mx + b .
As noted above, whenever a relationship can be expressed by an equation in the form:
A = (a constant) • B , where A and B can vary, then A and B are directly proportional. It is
also true that a graph of A versus B will produce a straight line which, if extended, passes
through the origin. Why?
The equation for a straight line on a graph is y = mx + b , where
• m is the constant slope of the graph (the rise over the run), and • b is the yintercept, the value of y at x = 0 (which is the value of y where the
graphed line crosses the yaxis). If the value of the yintercept (b) is zero, y = 0 when x = 0, and the straight line will pass
through the origin (0,0).
In the case of data for to variables that falls on a straight line on a graph that passes through
the origin,
• the equation for the line is y = mx + 0, which simplifies to y = mx . This equation
matches the form of a direct proportion: A = (a constant) • B . • The variables plotted on the y and x axis are directly proportional, and the slope m
is the proportionality constant of the direct proportion y = mx. If the value of the yintercept (b) of a straight line is not zero, the y versus x relationship is
said to be linear, but it is not directly proportional.
To summarize,
Direct Proportion = Line Through Origin
When two variables that are directly proportional are graphed, the data falls on a straight
line thru the origin.
When graphed data for two variables falls close to a straight line through the origin, the
variables are directly proportional, and the values for the data and relationship can be
predicted by the equation yaxis variable = (constant slope of line)(xaxis variable) .
In science, a fundamental goal is to develop equations that measure how changes in one
quantity will affect other quantities. Those equations can be written with relative ease
when plots of experimental data fall on a straight line. ©2011 ChemReview.net v. f1 Page OnScreen 18 6 Module 18 — Gas Labs, Gas Reactions Graphing Charles’ Law
Charles’ law is a direct proportion. The form of Charles law V = cT fits the general
equation form: y = (a constant) times x.
If in a Charles’ law experiment, pressure and moles of a gas sample are held constant, the
temperature of the gas is varied and recorded in kelvins , and the resulting volumes are
recorded, the V versus T data can then be graphed.
Since V = cT matches the form y = mx + 0 , if V is plotted on the
yaxis and T on the xaxis, the data falls on a line that looks like
the graph at the right, a straight line through the origin. V Since V = cT matches the form y = mx, the value of the
proportionality constant (c) will equal the slope (m) of the line through the origin. T ***** Explaining the Graph of a Direct Proportion
If any one of these statements is true, you may write that all of these statements are true.
1. When data for one variable Y is graphed on the yaxis, and data for the other variable X
is graphed on the xaxis, the data falls close to a straight line thru the origin.
2. A variable Y is directly proportional to a variable X.
3. Y = m X , where m is constant and is the slope of the line of Y plotted versus X. 4. Y X and X Y 5. Y1 = Y2 = c = m = the constant slope of a line fitting a graph of Y versus X.
X1 X2
6. For any measure of Y and X, the ratio Y over X will be constant. Writing Lab Reports Based on Graphed Data
For a laboratory experiment, if a graph of data fits (allowing for experimental error) on a
straight line that would pass through the origin, all of the above statements about the two
graphed variables can be discussed in the conclusion to your lab report.
Often, you will find the relationship between the two graphed variables addressed in your
textbook. Reviewing what the text has to say about the relationship between the two
variables may improve your understanding of an experiment. In the conclusion of a lab
reports, it is often appropriate to discuss what an experiment’s results should have been
ideally, and why your results may differ from theoretical results. ©2011 ChemReview.net v. f1 Page OnScreen 18 7 Module 18 — Gas Labs, Gas Reactions Practice B: First learn the six rules in the summary above, then do these problems to test
your knowledge. Check answers frequently.
1. A sample of trapped gas molecules in a container with a constant volume is attached to
a pressure gauge. Measurements of pressure are taken at a variety of temperatures.
a. What are the symbols for the two gas variables in this experiment?
b. What are the symbols for the gas variables held constant in this experiment?
c. Starting from the ideal gas law, group all of the constants in this experiment into a
single constant c, then write one equation with “ = c “ on the right and the variables
of this experiment on the left.
d. Write four mathematically equivalent equations that express the relationship
between P and T.
e. Express this relationship in words.
f. In this experiment, if pressure is plotted on the yaxis, and absolute temperature on
the xaxis, what will be the shape of the graph? 2. For the graph at the right, label each of the following statements as True or False.
a. C and D are directly proportional.
b. D C C c. D = mC , (where m is slope)
d. C = mD
e. D = (1/m) C
f. D C1 = C2
D2
D1 g. C1D2 = C2D1
3. The equation for converting degrees Celsius to Fahrenheit is: ºF = (9/5) ºC + 32 Sample data points for this relationship are: 32 ºF = 0ºC , 68 ºF = 20 ºC , 212 ºF = 100 ºC.
a. Will a graph of the data for ºF on the yaxis, versus ºC on the xaxis, fit on a straight
line? Why or why not?
b. Is the relationship between Fahrenheit and Celsius a direct proportion?
c. For a plot of ºF on the yaxis versus ºC on the xaxis, what will be the value of the
slope? What will be the value of the yintercept? ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 8 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice B
1. a. d. P and T b. V and n P = c and
T P1 = P2
T1
T2 c. Since P = nR and n, R, and V are all constant, P = c .
T
V
T and P = cT and PT (when V and n are held constant.) e. When volume and moles are held constant, gas pressure varies in direct proportion to absolute
temperature. f. A straight line through the origin. 2. a. C and D are directly proportional. True
b. D C True (if C D, D C) c. D = mC , (where m is slope) d. C = mD
C1 = C2
D2
D1 False ( x ≠ my ) D True ( y = mx )
True, since C = mD e. D = (1/m) C
f. C True; since C = mD, C/D = m = constant ©2011 ChemReview.net v. f1 Page OnScreen 18 9 Module 18 — Gas Labs, Gas Reactions g. C1D2 = C2D1 True; since part (f) is true, and (f) and (g) are the same algebraically. 3a. Straight line? Yes, because ˚F = (9/5) ˚C + 32 is in the form y = mx + b . However, the line will
not pass through the origin, since b ≠ 0. b. Direct proportion? No. A straight line represents a direct proportion only when the yintercept is
zero (the line passes through the origin). Here, the yintercept is + 32. Another way to check: the y
over x ratio would be constant for a direct proportion. For the data above, ˚F over ˚C is not constant. c. Slope? Since ˚F = (9/5)˚C + 32 is in the form y = mx + b, slope = (9/5) = 1.80
The yintercept? b = + 32
**** ©2011 ChemReview.net v. f1 Page OnScreen 18 10 Module 18 — Gas Labs, Gas Reactions Lesson 18B: Boyle’s Law – Graphs of Inverse Proportions
Prerequisites: Lessons 17A, 17D, and 18A.
***** Boyle’s Law
When moles and temperature are held constant, the terms in the ideal gas law are
PV = nRT = (constant moles)(gas constant R)(constant T)
Since multiplying constants results in a constant, the above equation can be written as
PV = c where c is constant when n and T are held constant. This is Boyle’s law, discovered in the mid1600’s by the English scientist Robert Boyle. For
any gas with ideal behavior, if moles and temperature do not change, the product P times V
will stay the same as you change P and V.
Another way to express this relationship: if you have an initial set of conditions P1 and V1,
and you change to a new set of conditions P2 and V2, the product P•V must stay the same,
as long as no particles of gas enter or escape, and temperature is the same for the initial and
final conditions.
In equation form: nRT = c = P1V1 = P2V2 when n and T are held constant. If 3 out of the 4 variables among P1, V1, P2, and V2 are known, the missing 4th variable may
be found using algebra, and without using R or knowing n, T, or c.
From the form PV = c, it can be seen that if the pressure is doubled, the volume must drop
by half for the product of P times V to remain constant. If the pressure is cut to onethird of
an original value, it can only be because the original volume has tripled. This type of
relationship is called an inverse proportion.
In words, for Boyle’s law, “if the number of particles and the temperature of a gas are held
constant, pressure will be inversely proportional to volume.”
Since PV = c can be rewritten as P = c(1/V) , Boyle’s law can also be written as
P 1/V This form is read as either “P is inversely proportional to V,” or as “P is directly proportional
to one over V.” Recall that 1/V can also be written as V─1 .
One implication of Boyle’s law is that as a gas is placed under higher and higher pressure,
its volume will approach zero. In reality, under increasing pressure, at some point all gases
lose their ideal behavior as they approach conditions at which they condense to liquids or
solids, and the gas behavior will no longer be predicted by gas laws.
Summary: Using flashcards, commit to longterm memory:
or P1V1 = P2V2 or P = c(1/V)
and applies when gas moles and temperature are held constant. Boyle’s law can be written PV = c or P 1/V *****
©2011 ChemReview.net v. f1 Page OnScreen 18 11 Module 18 — Gas Labs, Gas Reactions Practice A: Check your answers at the end of this lesson after each part. 1. Write four equivalent mathematical equations expressing Boyle’s law.
2. What two gas variables must be held constant for Boyle’s law to predict gas behavior?
3. For the relationship A 1/B ,
a. If A is tripled, what must happen to B?
b. If B goes from 400 to 200, A must go from 300 to _________? ***** ANSWERS
Practice A
1. Boyle’s law: PV = c or P1V1 = P2V2 or P = c(1/V) or P V─1 2. Two variables held constant for Boyle’s law: T and n 3. It may help to rewrite A 1/B as
a. If A is tripled? ©2011 ChemReview.net v. f1 AB = c B is reduced to 1/3 of its original value. Page OnScreen 18 12 Module 18 — Gas Labs, Gas Reactions b. If B goes from 400 to 200, A must go from 300 to 600 .
***** Graphing Inverse Proportions
When data for an inverse proportion such as PV= c is graphed,
the result is a portion of a hyperbola. Note that as xvalues
increase, yvalues decrease. As xvalues approach zero, yvalues
become large. This is the behavior of an inverse proportion.
PV=c can be written as P=c(1/V) . If data for P is plotted on the
yaxis and values calculated for 1/V are assigned to the xaxis, this
form matches the general equation for a line y = mx + b passing
through the origin (b=0, see Lesson 18A). P V
P A graph of the data for points representing of P and 1/V should
therefore fall on a straight line through the origin, where the slope
(m) of the line is the value of the constant c. The equation that
explains and predicts values for P and V is P = m(1/V) . 1/V If the value of the yintercept (b) is not zero, the y versus 1/x relationship will not be a
direct proportion, and the data on the y and xaxes will not be inversely proportional. Summary for a Inverse Proportion
If any one of these is true, all are true.
1. Variable Y is inversely proportional to a variable X;
2. Variable Y is directly proportional to 1/X;
3. Y is equal to a constant times 1/X;
4. Y = m (X─1) , where m is the constant slope of a line on a graph of Y versus 1/X
5. Y 1/X (and X 1/Y)
6. Y1 X1 = Y2 X2 = c = m = the constant slope of a line graphing Y versus 1/X
7. For any measure of Y and X, Y times X will be constant.
8. When Y is graphed versus 1/X, the points fall on a straight line thru the origin.
For labs in which the experimental data results in a graph of Y versus 1/X with points that
fall close to a straight line thru the origin, the above statements about the two variables can
be evaluated and discussed in the conclusion to your lab report.
* * * ** ©2011 ChemReview.net v. f1 Page OnScreen 18 13 Module 18 — Gas Labs, Gas Reactions Practice B: Check your answers after each part or three. 1. For the graph at the right, label each of the following
statements as True or False: E a. E and F are inversely proportional.
b. E = m(1/F) (where m is slope) 1/F c. F = m(E─1)
d. F = cE (where c is a constant) e. FE= (1/m)
f. FE= m g. E1F1 = E2F2
2. To cover a constant distance, the rate of travel (such as km per hour) is, by the definition
of the terms, inversely proportional to the time required. Higher rate values will mean a
smaller time values.
a. Write the relationship between rate, and time, and a constant distance by
completing these two equivalent equations.
(constant d) = and rate = b. Write the relationship between rate and time in two equivalent equations that use
the sign.
c. Express the equations above in words.
d. If the time for travel triples, the rate of travel _______________________.
e. To travel a fixed distance at various rates, if rate is plotted on the yaxis, and time on
the xaxis, what will be the shape of the graph?
f. If rate is plotted on the yaxis, what must be plotted on the xaxis to produce a graph
which is a straight line through the origin? ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 14 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice B
1. a. E and F are inversely proportional. True
b. E = m(1/F) (m is slope)
c. F = m(E─1) d. F = cE E True This is a form of y=mx+b True. If (b) is true, by algebra this is true. 1/F (c is constant) False. This would be a direct proportion. e. FE= (1/m) False. Part (b) is true, and therefore 1/m = 1/FE ≠ FE f. FE= m True. This is equivalent to (b). g. E1F1 = E2F2 True. If FE is equal to a constant (part F), this will be true. 2. a. Constant distance = (rate)(time) and rate = (constant d) / (time) or rate = (constant d) (1/time) b. To use the proportional sign, start with a form y = (constant) x , which means y x .
For this relationship, since rate = (constant d) (1/time), then rate 1/time ;
and since this equation can be rewritten as time = (constant d)(1/rate), then time 1/rate . ©2011 ChemReview.net v. f1 Page OnScreen 18 15 Module 18 — Gas Labs, Gas Reactions c. A few of several possibilities are, in traveling a constant distance, rate is inversely proportional to time;
rate is directly proportional to one over time; constant distance equals rate times time. d. If the time for the travel triples, the rate of travel is 1/3 as fast. e. Since rate and time are inversely proportional, the graph will be a section of a hyperbola. f. Since rate and time are inversely proportional, rate versus 1/time fits on a line through the origin. ***** Lesson 18C: Avogadro’s Hypothesis; Gas Stoichiometry
Timing: Do this lesson when you are assigned gas reaction calculations.
Prerequisites: Modules 2, 4, 5, 8, and 10, plus Lessons 11B, 17A, and 17D.
***** Avogadro’s Hypothesis
In 1811, the Italian scientist Amedeo Avogadro made a remarkable discovery: that in
chemical reactions, volumes of gases at the same temperature and pressure are used up and
formed in simple wholenumber ratios. To explain his results, he proposed that “equal
volumes of gases, at the same temperature and pressure, contain equal numbers of
molecules.” Avogadro’s hypothesis highlighted the importance of counting particles in
understanding chemical processes.
One implication of Avogadro’s hypothesis is that for ideal gases, the chemical formula and
molar mass of a gas have no effect on the volume that the gas will occupy.
• If samples the same gas, different gases, or mixtures of multiple gases have the
same pressure, temperature, and volume, they will contain the same number of gas
molecules. • For two samples of gas at the same temperature and pressure, if one sample has
twice as many molecules, it will have twice the volume. ©2011 ChemReview.net v. f1 Page OnScreen 18 16 Module 18 — Gas Labs, Gas Reactions Gas Volumes and Coefficients
We have previously found that coefficients of a balanced equation can be read as ratios of
• particles (molecules, ions, or formula units); • moles of particles (or any other multiple of particles); or • moles/liter of particles that are all in the same volume (such as gas particles in a
sealed glass container, or dissolved particles that react in the same aqueous
solution). Since coefficients are ratios of particles, and gas volume ratios at the same temperature and
pressure are also ratios of particles, based on Avogadro’s hypothesis we can add to the
above list. Coefficients of a balanced reaction equation that is all gases can also be read as
• volumes of gases, if the gases are measured at the same temperature and pressure. Apply this rule to the following problem, and then check your answer below.
Q. Carbon monoxide oxidizes to form carbon dioxide. The balanced equation is
2 CO(gas) + 1 O2(g) 2 CO2(g) Assuming that all of the gases are at the same temperature and pressure, starting
with ten volumes of CO;
a. How many volumes of O2 would be needed to completely use up the CO?
b. How many volumes of CO2 would be formed in the reaction? ***** Answer
a. The reaction coefficients are ratios of gas volumes for gases are at the same T and P. Ten
volumes of CO would require five volumes of O2 to be used up completely.
b. Ten volumes of CO plus five volumes of O2 would be completely used up. Ten
volumes of CO2 would be formed. ©2011 ChemReview.net v. f1 Page OnScreen 18 17 Module 18 — Gas Labs, Gas Reactions Practice A: Answers are at the end of this lesson. 1. For the reaction 2 H2(gas) + O2(g) 2 H2O(g) 20. liters of hydrogen gas and 20. liters of oxygen gas, both at standard pressure and
120˚C, are mixed and ignited. The resulting gases are adjusted to the original
temperature and pressure. (At standard pressure and above 100.˚C, all of the H2O will
be in the form of steam, a gas.)
a. Which is the limiting reactant?
b. For which of the reactants will some amount remain after the reaction?
c. After the reaction, which gases are present?
d. How many liters of each of the reactants and products are present after the reaction
is complete? ***** ANSWERS
Practice A
1. a. H2 b. O2. 20. L H2 uses up only 10. L O2. ©2011 ChemReview.net v. f1 c. O2 and H2O d. Zero H2, 10. L O2, 20. L H2O. Page OnScreen 18 18 Module 18 — Gas Labs, Gas Reactions Gas Stoichiometry
All reaction calculations can be solved using the same fundamental stoichiometry steps.
However, for gas reaction calculations, it will speed our work if we solve using these three
variations in the stoichiometry steps.
Method 1. IF the WANTED and given are both volumes of gases at the same temperature
and pressure (even if not at standard T and P (STP)),
• Method 2. IF all of the gas volumes in the WANTED and given are at STP,
• Method 3. solve based on Avogadro’s hypothesis: by inspection or with one
volumevolume conversion.
solve using standard 7step stoichiometry and the STP prompt. IF the WANTED and given are not gases at the same T and P, or not all gases
volumes in the DATA are at STP,
• solve using a rice table. We will do one calculation for each type. A method 1 calculation was covered in the
previous section. Let’s try method 2. Stoichiometry If All Gas Volumes Are At STP
In stoichiometry, if all gas volumes in the WANTED and DATA are at STP, you can solve
using the STP prompt and the standard steps of conversion stoichiometry.
Try the following problem in your notebook.
Q1. The unbalanced equation for the burning of propane gas is
C3H8(g) + O2(g) CO2(g) + H2O(g) How many grams of O2 gas are needed to burn 3.50 liters of propane at STP? ***** Answer
1. WANT: ? g O2 gas (Burning means reacting with O2 , so propane must be C3H8). 2. DATA: 3.50 L C3H8 at STP
1 mol gas = 22.4 L any gas at STP
32.0 g O2 = 1 mol O2 (WANTED # given formula)
(STP prompt)
(g prompt) For reactions, if WANTED formula ≠ given formula, use the stoichiometry steps.
©2011 ChemReview.net v. f1 Page OnScreen 18 19 Module 18 — Gas Labs, Gas Reactions Since the WANTED and DATA include only one gas volume and it is at STP, you can solve
using conversions.
3. Balance. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) 4. Bridge. 5 mol O2 = 1 mol C3H8 (the WANTED to given mol ratio) ***** Steps 57: When a single unit is WANTED, convert “final unit WANTED = # and unit of
given substance” to moles given, to moles WANTED, to units WANTED.
? unit WANTED = unit given ? g O2 = 3.50 L C3H8 gas STP ● >> mol given >> bridgemol WANTED > unit WANTED
1 mol gas ● 5 mol O2 ● 32.0 g O2 = 22.4 L gas STP 1 mol C3H8 25.0 g O2 1 mol O2 To use conversions to solve gas stoichiometry, it is not necessary that all of the gases be at
STP. The requirement is simply that all gas volumes in the WANTED and given be at STP.
In the problem above, the oxygen gas may or may not be at STP, but that does not affect the
WANTED mass of oxygen. In a sealed sample of a gas, even if P, V, and/or T are changed,
the number of gas particles (moles) does not change, and, since each substance particle has
a constant mass, the total mass of the gas will not change. Practice B
1. The unbalanced equation for the burning of ethanol is
C2H5OH(l) + O2(g) CO2(g) + H2O(g) If the density of liquid ethanol is 0.789 g/mL, how many liters of CO2 gas at STP will be
produced by burning 2.50 mL of ethanol? ©2011 ChemReview.net v. f1 Page OnScreen 18 20 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice B
1. WANT: ? L CO2 at STP 2. DATA: 2.50 mL C2H5OH (WANTED # given formula) 1 mol gas = 22.4 L any gas at STP (STP prompt) 0.789 g liquid C2H5OH = 1 mL C2H5OH
46.1 g C2H5OH = 1 mol C2H5OH (g prompt from density ) ***** For reactions, if WANTED formula ≠ given formula, use the stoichiometry steps.
Since the WANTED and DATA include only one gas volume, and it is at STP, you can solve using conversions.
3. Balance. C2H5OH(l) + 3 O2(g) 4. Bridge. 2 CO2(g) + 3 H2O(g) 2 mol CO2 = 1 mol C2H5OH (the WANTED to given mol ratio) ***** Steps 57:.
? unit WANTED = unit given >> mol given >> bridge > unit WANTED ? L CO2 STP = 2.50 mL C2H5OH ● 0.789 g C2H5OH ● 1 mol C2H5OH ● 2 mol CO2 ● 22.4 L CO2 STP
1 mL C2H5OH 46.1 g C2H5OH 1 mol C2H5OH 1 mol CO2 = 1.92 L CO2 STP
***** 2011 ChemReview.net v. f1 Page OnScreen 18 21 Module 18 — Gas Labs, Gas Reactions Stoichiometry If A Gas Volume Is Not At STP
The above two methods are for simple gas reaction situations in which either
• all reactants and products are gases at the same temperature and pressure, or • all of the gases involved in the reaction are at STP. For all other gas reaction calculations, our method to solve will be: use a rice table. A rice
table takes more steps than some methods, but it has the advantage of using the same steps
for every type of reaction amount calculation: whether solids or solutions, for gases
whether at STP or not, in problems where you may or may not need to find the limiting
reactant, and even for reactions that do not go to completion.
In Lesson 10G we learned a simple rule that can be used to decide whether to use
conversions or a rice table to solve. That rule is: To Solve Reaction Amount Calculations, Use Which Method?
If the reaction goes to completion and the given amount is clear, solve using conversion
stoichiometry. In all other cases, or whenever you are not sure how to proceed, solve
with a ricemoles table.
Another way to state this rule is: use a rice table for complex reaction calculations.
Stoichiometry involving gases not at STP is complicated because to find moles of a gas not
at STP, you need to use the ideal gas equation instead of conversions. Mixing equations
and conversions adds an element of complexity to the calculation, but the rice table will
provide us with a consistent steps to keep track of data and solve.
The general rules for a rice table are: write the WANTED unit, then go from
All supplied units to all moles to ricemoles table to WANTED units
Specifically, in gas stoichiometry, if Avogadro’s hypothesis cannot be used, and if the
WANTED or DATA includes a gas volume not at STP, the steps are
• If conversions can be used to find moles, do so. If the given amount is a gas that is
not at STP, use PV=nRT to find the moles of the gas. Enter those moles into the rice
table and solve the table. • If the WANTED unit can be found by conversions from the WANTED moles in the
table, do so. • If the WANTED amount is a gas not at STP, solve the rice table , then use PV=nRT
to change moles of WANTED gas in the End row to the units of gas WANTED. This method is best learned by example. Solve the following problem using the steps
above. If you get stuck, read the answer until unstuck, then try again.
Q2. A 0.972 gram sample of magnesium metal is reacted with an excess of
hydrochloric acid to produce hydrogen gas. The unbalanced equation is
Mg(s) + HCl(aq) MgCl2(aq) + H2(g) How many liters of H2 would be formed at standard pressure and 20.˚C?
(R = 0.0821 L·atm/mol·K) ©2011 ChemReview.net v. f1 Page OnScreen 18 22 Module 18 — Gas Labs, Gas Reactions ***** Answer
1. WANT: ? L H2 gas at SP and 20.˚C WANTED is a gas volume for a gas not at STP, and not all substances are gases at
the same T and P, so we use the rice steps to solve:
All supplied units > all moles > rice moles table > WANTED units
The only supplied substance data is g Mg. Convert to moles.
DATA: 0.972 g Mg (Start with a single unit) 1 mol Mg = 24.3 g Mg
? mol Mg (grams prompt) = 0.972 g Mg ● 1 mol Mg
24.3 g Mg = 0.0400 mol Mg For a rice table, the first row needs a balanced equation.
1 Mg(s) + 2 HCl(aq) 1 MgCl2(aq) + 1 H2(g) ***** Reaction
Initial
Change
End 1 Mg
0.0400 mol
― 0.0400 mol
0 mol 2 HCl 1 MgCl2 1 H2 excess   ― 0.0800 mol + 0.0400 mol + 0.0400 mol excess + 0.0400 mol + 0.0400 mol In the rice table, in a reaction that goes to completion, one reactant must be limiting
(totally used up). If there are two reactants, and one is in excess, the other will be
limiting. The key row is the Change row, in which the ratios used up and formed must
be the same as the reaction ratios (coefficients) in row 1.
Once the rice table is solved in moles, we can convert to other WANTED units. *****
©2011 ChemReview.net v. f1 Page OnScreen 18 23 Module 18 — Gas Labs, Gas Reactions WANTED: ? L H2 gas at SP and 20.˚C DATA: 0.0400 mol H2 To go from moles of a gas to liters not at STP, what equation is used? ***** PV =nRT
DATA: P = SP = 1 atm to match the P unit in the supplied R value
V = ? L H2 gas = WANTED
n = 0.0400 mol H2
R = 0.0821 L·atm/mol·K (from the End row)
(supplied) T = 20.˚C + 273 = 293 K
SOLVE:
? = V = nRT = nRT● 1 = (0.0400 mol)(0.0821 atm • L )(293 K) ● 1 = 0.962 L H2 gas
P
P
mol • K
1 atm
at SP and 20˚C Summary: Gas Stoichiometry
1. IF the WANTED and given are both volumes of gases at the same temperature and
pressure (even if not at standard T and P (STP)),
• Solve by inspection or with one volumevolume conversion, based on
Avogadro’s hypothesis, using the coefficients of the balanced equation as
wholenumber gas volume ratios. 2. If all of the gas volumes in the WANTED and given are at STP.
• Solve using standard 7step conversion stoichiometry and the STP prompt. 3. If method 1 or 2 cannot be used, write the WANTED unit, then solve with a rice
table and these steps:
• All supplied units > all moles > ricemoles table > WANTED units ©2011 ChemReview.net v. f1 Page OnScreen 18 24 Module 18 — Gas Labs, Gas Reactions ©2011 ChemReview.net v. f1 Page OnScreen 18 25 Module 18 — Gas Labs, Gas Reactions Practice C
1. In living cells, the sugar glucose is burned by this unbalanced equation.
C6H12O6 + O2 C O2 + H2O How many liters of CO2, measured at 740. torr and 30.˚C, can be formed by burning
0.250 moles of glucose? (R = 62.4 torr·L/mol·K)
2. Sodium metal can be reacted with excess water to produce hydrogen gas. The
unbalanced reaction equation is:
Na + H2O NaOH + H2 If 4.92 liters of H2 gas is produced at 101 kPa and 25˚C, how many grams of sodium
reacted? (R = 8.31 L·kPa/mol·K) ANSWERS
Practice C
1. 1. WANTED:
2. DATA: ? L CO2 gas at 740. torr and 30˚C
0.250 mol C6H12O6 (Want L, a single unit)
(This must be the singleunit given) Burning is a reaction. For reactions, if WANTED ≠ given, use stoichiometry steps.
Since WANTED is a gas volume for a gas not at STP, and not all substances are gases at the same T
and P, use the rice steps to solve:
All supplied units > all moles > ricemoles table > WANTED units
The only supplied substance amount is moles glucose, so the “convert DATA to moles” step is done.
For a rice table, the first row needs a balanced equation. ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 26 Module 18 — Gas Labs, Gas Reactions 1 C6H12O6 + 6 O2 6 CO2 + 6 H2O 1 C6H12O6 Reaction
Initial 0.250 mol Change 6 O2 6 CO2 6 H2O excess   + 1.50 mol l ― 0.250 mol End 0 mol excess + 1.50 mol “Burning” by definition means reacting something with excess O2 . in a reaction that goes to completion,
one reactant must be limiting (totally used up). If there are two reactants, and one is in excess, the other is
limiting. The glucose is therefore the limiting reactant that determines all of the other amounts in the
Change row..
Since an amount of CO2 is the WANTED unit, we can leave out the other parts of the rice table.
WANT: ? L CO2 gas at 740. torr and 30.˚C DATA: 1.50 mol CO2 gas at 740. torr and 30.˚C (from the End row) To go from moles of a gas to liters not at STP, what equation is used? ***** PV =nRT
DATA: P = 740. torr
V = ? L H2 gas = WANTED
n = 1.50 mol CO2
R = 62.4 torr•L/mol•K
T = 30.˚C + 273 = 303 K SOLVE:
1
= 38.3 L CO2 gas
? = V = nRT = nRT● 1 = (1.50 mol)(62.4 torr • L )(303 K) ●
P
P
mol • K
740. torr
at 740 torr and 30.˚C ©2011 ChemReview.net v. f1 Page OnScreen 18 27 Module 18 — Gas Labs, Gas Reactions 2. 1. WANT:
2. DATA: ? g Na
4.92 L H2 gas at 101 kPa and 25˚C
23.0 g Na = 1 mol Na (g prompt) For reactions, if WANTED ≠ given formula, use stoichiometry steps.
Since the only supplied amount is a gas volume for a gas not at STP, and not all substances are gases
at the same T and P, use the rice steps to solve:
All supplied units > all moles > ricemoles table > WANTED units
Job one is to convert all supplied units to moles.
At the first step, since 25˚C is not standard temperature, we cannot use the STP conversion to find
moles. To convert L H2 given to moles H2 can be done with the supplied data and
PV =nRT ***** DATA: P = 101 kPa
V = 4.92 L H2
n= ?
R = 8.31 kPa•L/mole•K
T = 25˚C + 273 = 298 K (101 kPa)(4.92 L)
SOLVE: ? = n = PV =
RT
( 8.31 kPa • L ) (298 K)
mol • K = 0.200 mol H2 gas given For a rice table, the first row needs a balanced equation.
2 Na + 2 H2O 2 NaOH + 1 H2 Solve this rice table backwards, from the known final moles of H2. ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 28 Module 18 — Gas Labs, Gas Reactions Reaction 2 Na 2 H2O 2 NaOH 1 H2 Initial 0.400 mol excess 0 0 ― 0.400 mol + 0.400 mol + 0.200 mol excess + + 1.50 mol + 0.200 mol Change ―0.400 mol End 0 mol Since the initial grams of Na are WANTED, convert from initial moles Na. ***** ? g Na = 0.400 mol Na ● 23.0 g Na = 9.20 g Na
1 mol Na
***** ©2011 ChemReview.net v. f1 Page OnScreen 18 29 Module 18 — Gas Labs, Gas Reactions Lesson 18D: Dalton’s Law of Partial Pressures
Timing: Do this lesson when you are assigned calculations which include Dalton’s law,
partial pressure, vapor pressure, or gas mixtures.
Prerequisite: Lessons 18A and 18C.
***** Vapor Pressure
Liquids and solids have a tendency to become gases. At the surface of a liquid or solid, the
vibrating molecules can break free of the liquid or solid and become part of the vapor
above the liquid or solid.
This tendency to vaporize increases with increasing temperature. At higher temperature,
the particles in a liquid and solid move faster, and they break free from the surface more
often. For the particles that enter the gas phase, the kinetic energy of their collisions with
the container walls creates gas pressure.
The vapor pressure above a liquid or solid is a characteristic of a substance at a given
temperature. If the gas that leaves the solid or liquid is contained so that it cannot escape,
the pressure caused by the vapor is predictable. Vapor pressure always rises with
increasing temperature. If the substance and its temperature is known, this vapor pressure
can be looked up in tables of chemical data.
A sample of vapor pressures for water is listed in the table at
the right. H2O vapor
pressure ˚C Boiling 20.˚C 17.5 torr A liquid boils at the temperature at which its vapor pressure
equals the atmospheric pressure above the liquid. A liquid can
be boiled at different temperatures by increasing or
decreasing the atmospheric pressure on the liquid. 25˚C 23.8 torr 30.˚C 31.8 torr 100.˚C 760 torr The “boiling” of a liquid is not the same as “evaporating.” Evaporation is a surface
phenomenon; measurable evaporation will occur from all liquids (and many solids) at any
temperature. However, a liquid “boils” only when gas bubbles can form anywhere in the
liquid, and not just at its surface. Gas Mixtures and Partial Pressure
Mixtures of gases with ideal behavior obey
Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is equal to the sum of the pressures that the
gases would exert alone.
The pressure exerted by each gas in the mixture is called its partial pressure. The partial
pressure is the pressure that a gas would exert in the given volume if it alone were present.
For calculations involving gas mixtures, Dalton’s law is useful in these forms which should
be committed to longterm memory. ©2011 ChemReview.net v. f1 Page OnScreen 18 30 Module 18 — Gas Labs, Gas Reactions Dalton’s Law Equations
To find the partial pressure of a single gas in a mixture:
Pgas one = (mole fraction gas one)Ptotal = (volume fraction gas one)Ptotal
For the total mixture:
Ptotal = Pgas one + Pgas two + … ( = Partial P gas one + Partial P gas two + … )
= (mole fraction gas one)Ptotal + (mole fraction gas two)Ptotal + …
= (volume fraction gas one)Ptotal + (volume fraction gas two)Ptotal + …
This string of equalities for the total mixture provides us with a number of relationships
that can be used to solve problems.
The mole and volume fractions in the above equations will always be numbers less than
one: decimals which total 1.00 for all of the fractions in a gas mixture.
Fraction = part
total Mole Fraction = moles of part
moles of total Equations 2 and 3 above are similar because, by Avogadro’s Hypothesis, for gases at the
same temperature and under the same pressure, their volumes are proportional to their
number of particles.
Use the above equations to solve the following problem.
Q. A mixture of 18.0 g of He gas and 16.0 g of O2 gas in a sealed glass bulb is at a
pressure of 120. kPa. What is the partial pressure of the He in the mixture? ***** WANT: ? PHe in kPa DATA: 18.0 g He
4.00 g He = 1 mole He (g prompt) 16.0 g O2
32.0 g O2 = 1 mole O2
120. kPa (total pressure) Strategy: There are 4 equations that involve partial and total pressure. Which one should
you use to solve? You know that
Ptotal = Partial Pressure He + Partial P O2 = PHe + PO2 = 120. kPa ©2011 ChemReview.net v. f1 Page OnScreen 18 31 Module 18 — Gas Labs, Gas Reactions Since that equation has one known and two unknowns, you will need to use one
of the other equations to gain more information. Which one? Check the data. ***** Should you use the equation using mole fractions, or volume fractions? We don’t
know the volumes, but from grams and the formula, we can find moles.
? Moles He = 18.0 g He • 1 mole He =
4.00 g He
? Moles O2 = 16.0 g O2 • 1 mole O2 =
32.0 g O2
Equation (2) uses mole fraction. 4.50 moles He
0.500 moles O2 Total moles = 4.50 moles He + 0.500 moles O2 = 5.00 moles total ***** ? = Partial Pressure He = (mole fraction He)(Ptotal) = 4.50 moles He • 120. kPa = 108 kPa
5.00 moles total Practice A
Memorize the forms of Dalton’s law, then use the equations to solve these.
1. By volume, air is a mixture of 78% N2 , 21% O2 , and 1% other gases. At standard
pressure, what is the partial pressure of the nitrogen gas in air, in mm Hg?
2. A mixture of Ne and Cl2 gases at standard pressure has a total mass of 24.3 grams. The
mass of the neon is 10.1 grams.
a. What is the mole fraction of the chlorine gas in the mixture?
b. What is the partial pressure of the chlorine gas (in kPa)? ©2011 ChemReview.net v. f1 Page OnScreen 18 32 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice A
1. WANT:
DATA: PN
2
The equation that uses partial pressure and volume fraction of a single gas is
To find the partial pressure of a single gas in a mixture:
Pgas one = (mole fraction gas one)Ptotal = (volume fraction gas one)Ptotal
List the DATA next to the symbols of the equation you need.
Pgas one = PN2 = ?
volume fraction N2 = 78% = 0.78 as the decimal equivalent fraction
Ptotal in mm Hg = Standard pressure in mm Hg = 760 mm Hg ***** SOLVE: PN = volume fraction N • P
2
2
total = 0.78 • 760 mm Hg = 590 mm Hg 2a. WANT: mol fraction Cl2 = mol Cl2
mol total
DATA: 10.1 g Ne 1 mol Ne = 20.0 g Ne 24.3 g (Ne + Cl2) 1 mol Cl2 = 71.0 g Cl2 Standard Pressure = 101.0 kPa ***** SOLVE: To find the top unit in the WANTED fraction, first find g Cl2.
g Cl2 = 24.3 g ─ 10.1 g = 14.2 g Cl2 ©2011 ChemReview.net v. f1 Page OnScreen 18 33 Module 18 — Gas Labs, Gas Reactions *****
? mol Cl2 = 14.2 g Cl2 • 1 mol Cl2 = 0.2000 mol Cl2
71.0 g Cl2
To find the bottom WANTED unit (total moles), the moles of Ne are needed.
? mol Ne = 10.1 g Ne • 1 mol Ne = 0 5000 mol Ne
20.2 g Ne
mol total = 0.2000 mol Cl2 + 0 5000 mol Ne = 0.7000 mol total *****
WANTED: 1b. WANT:
DATA: mol fraction Cl2 = mol Cl2 = 0.2000 mol Cl2 = 0.2857 = 0.286 mol Cl2
mol total
0.7000 mol total
mol total PCl
2
The equation that uses partial pressure and volume fraction of a single gas is
To find the partial pressure of a single gas in a mixture:
Pgas one = (mole fraction gas one)Ptotal = (volume fraction gas one)Ptotal
List the DATA next to the symbols of the equation you need.
Pgas one = PCl2 = ?
mole fraction Cl2 = 0.286 as the decimal equivalent fraction from part a. Ptotal in kPa = Standard pressure in kPa = 101 kPa
SOLVE: PCl = mol fraction Cl • P
2
2
total = 0.286 • 101 kPa = 28.9 kPa ©2011 ChemReview.net v. f1 Page OnScreen 18 34 Module 18 — Gas Labs, Gas Reactions *****
Gas Collected Over Water
Dalton’s law is often used when collecting bubbles of a gas over water or above an aqueous
solution. The resulting gas is a mixture which includes water vapor. The partial pressure
of water vapor at a known temperature can be looked up in a table. Dalton’s law will then
allow calculation of the pressure of a gas that is collected over water.
Ptotal over water = Pgas wanted + Pwater vapor from table
The partial pressure is the pressure that a gas would exert in the given volume if it alone
were present. This means that the WANTED partial pressure in the equation above can be
used with the ideal gas law to calculate other variables for that gas. Practice B
Use the table of H2O vapor pressure in the lesson above to solve these problems.
1. Approximately what would be the vapor pressure of water at 23˚C?
2. If the total pressure of the gas in a mixture of H2 gas and water vapor is 748 torr at 27˚C,
what is the partial pressure of the H2 gas?
3. A sample of oxygen gas (O2) is collected over water at 25˚C. If the volume of the gas is
6.0 liters and the total pressure of the gas mixture is 762 torr, how many moles of O2 are
in the sample? (R = 62.4 torr·L/mole·K)
4. 0.210 grams of lithium metal is reacted with dilute hydrochloric acid. The hydrogen
gas released is trapped in a tube above the aqueous acid solution. At the end of the
reaction, all of the lithium has reacted. The gas above the solution is a mixture of
H2 and water vapor, at 20˚C and 758 torr, with a volume of 0.370 liters.
a. Balance the reaction equation: Li(s) + HCl(aq) LiCl(aq) + H2(g) b. Based on the grams of lithium reacted, how many moles of H2 gas will be formed?
c. What will be the partial pressure of the H2?
d. The partial pressure of the hydrogen gas represents the pressure that it alone would
exert in the volume of the mixture. What would be the volume occupied by the
“dry” hydrogen gas (just the H2, with the water vapor removed).
i. At 20.˚C and its partial pressure? ii. At 20.˚C and standard pressure? (Use the “combined equation” and not R). ©2011 ChemReview.net v. f1 Page OnScreen 18 35 Module 18 — Gas Labs, Gas Reactions ANSWERS
Practice B
1. About 21 torr, estimated from the table. 2. Ptotal = Pgas one + Pgas two
Ptotal over water = Pgas wanted + Pwater vapor from table at 27˚C
Pgas wanted = Ptotal over water ― Pwater vapor = 748 torr ― about 27 torr = 721 torr 3. WANT:
DATA: ? moles O2 n “Over water” means a gas mixture is present. Use
Ptotal over water = Pgas wanted + Pwater vapor from table
Pwater vapor at 25˚C = 23.8 torr
25˚C t 6.0 liters O2 T = 25˚C + 273 = 298 K
V 762 torr = total pressure. Ptotal
R = 62.4 torr·L/mol·K
Strategy: We want moles of gas. R is given, so we likely need R. Which of the two equations
using R should be used? Label the data; see which fits. ***** ©2011 ChemReview.net v. f1 Page OnScreen 18 36 Module 18 — Gas Labs, Gas Reactions PV = nRT
But to use PV=nRT to find moles O2, we must know the pressure of only the O2.
Ptotal over water = Pgas wanted + Pwater vapor from table
Pwater vapor at 25˚C = 23.8 torr
Pgas wanted = Ptotal over water ― Pwater vapor = 762 torr – 23.8 torr = 738 torr for O2
This tells us the pressure the O2 gas would exert in the given volume if only the O2 were
present. That is what we need to know to use PV=nRT . ***** SOLVE: 4. a. Balance: b. WANT:
DATA: ? = n = PV
RT = (738 torr)(6.0 L)
( 62.4 torr • L ) (298 K)
mole • K 2 Li(s) + 2 HCl(aq) = 0.24 moles O2 gas 2 LiCl(aq) + 1 H2(g) ? moles H2 gas
0.210 g Li
6.94 grams Li = 1 mole Li
Since WANTED formula ≠ given formula, we need the steps of stoichiometry.
Since this part involves moles rather than a gas volume, try conversion stoichiometry
rather than a rice table. Bridge. Add to DATA the WANTED to given mole ratio: 1 mol H2 = 2 mol Li *****
©2011 ChemReview.net v. f1 Page OnScreen 18 37 Module 18 — Gas Labs, Gas Reactions SOLVE: ? moles H2 gas = 0.210 grams Li ● I mole Li ● 1 moles H2
6.94 g Li
2 moles Li = 0.0151 moles H2 c. What will be the partial pressure of the H2?
Ptotal = Pgas one + Pgas two
Ptotal over water = PH2 + Pwater vapor from table at 20˚C
PH2 = Ptotal over water ― Pwater vapor = 758 torr ― 17.5 torr = 740. torr d. i. At 20.˚C and its partial pressure? ? V = 0.370 liters – the volume of the mixture. ii. At 20.˚C and standard pressure (Do not use R).
WANTED: VH2 = L H2 (since the volume data is in liters) at 20.˚C and 760 torr V2 DATA: PH2 = 740. torr = P initial P1 0.370 L is the volume at 740. Torr = initial Volume
760 torr = Pressure at end. V1 P2 t1 = 20.˚C = t2
Strategy: Assign symbols, pick equation, then solve for WANTED symbol. ***** SOLVE: P1V1 = P2V2
T2
T1 ? = V2 = P1V1
P2 ©2011 ChemReview.net v. f1 = P1V1 = P2V2 ; = (740. torr) (0.370 L) = 0.360 L at 20.˚C and SP
(760 torr) Page OnScreen 18 38 Module 18 — Gas Labs, Gas Reactions ***** Summary: Gas Labs and Gas Reactions
1. Direct Proportions. If any one of these is true, all of these statements are true.
a. Variable Y is directly proportional to a variable X;
b. Y is equal to a constant times X;
c. Y = mX , where m is the constant slope of the line for Y versus X
d. Y X (and X Y)
e. Y1 = Y2 = c = m = the constant slope of a line graphing Y versus X
X1
X2
f. For any measure of Y and X, the ratio Y over X will be constant. g. When Y is graphed versus X, the data fits on a straight line thru the origin.
2. Inverse Proportions. If any one of these is true, all of these statements are true.
a. Variable Y is inversely proportional to a variable X;
b. Variable Y is directly proportional to 1/X;
c. Y is equal to a constant times 1/X;
d. Y = c (X─1) , where c will be the slope of a line on a graph of Y versus 1/X
e. Y 1/X (and X 1/Y)
f. Y1 X1 = Y2 X2 = c = m = the constant slope of a line graphing Y versus 1/X g. For any measure of Y and X, Y times X will be constant.
h. When Y is graphed versus 1/X, the points fall on a straight line thru the origin.
3. Avogadro’s Hypothesis: Equal volumes of gases, at the same temperature and
pressure, contain equal numbers of molecules.
4. Coefficients can represent particles, moles of particles, moles/liter of particles that are all
in the same volume, and volumes of gases measured at the same T and P.
5. For gas stoichiometry,
a. If WANTED and given are gas volumes at the same T and P, solve using coefficients.
b. If all gas volumes are at STP, do conversions using the STP prompt.
c. If the above two methods cannot be used, use PV = nRT and a ricemoles table.
6. Dalton’s Law of Partial Pressures: The total pressure of a mixture of gases is equal to
the sum of the pressures that the gases would exert alone. ©2011 ChemReview.net v. f1 Page OnScreen 18 39 Module 18 — Gas Labs, Gas Reactions 7. Dalton’s law equations.
To find the partial pressure of a single gas in a mixture:
Pone gas = (mole fraction gas)Ptotal = (volume fraction gas)Ptotal
For the total mixture:
Ptotal = Pgas one + Pgas two (= Partial Pressure gas one + Partial P gas two) + …
= (mole fraction gas one)Ptotal + (mole fraction gas two)Ptotal + …
= (volume fraction gas one)Ptotal + (volume fraction gas two)Ptotal + …
##### ©2011 ChemReview.net v. f1 Page OnScreen 18 40 Module 18 — Gas Labs, Gas Reactions NOTE on the Table of Atoms
The atomic masses in this Table of Atoms use fewer significant figures than most similar
tables in college textbooks. By “keeping the numbers simple,” it is hoped that you will use
“mental arithmetic” to do easy numeric cancellations and simplifications before you use a
calculator for arithmetic.
Many calculations in these lessons have been set up so that you should not need a
calculator at all to solve, if you look for easy cancellations first.
After any use of a calculator, use mental arithmetic and simple cancellations to estimate the
answer, in order to catch errors in calculator use. ##### ©2011 ChemReview.net v. f1 Page OnScreen 18 41 The Atoms –
The third column shows the atomic number:
The protons in the nucleus of the atom.
The fourth column is the molar mass, in
grams/mole. For radioactive atoms, ( ) is the
molar mass of most stable isotope.
Actinium
Aluminum
Americium
Antimony
Argon
Arsenic
Astatine
Barium
Berkelium
Beryllium
Bismuth
Boron
Bromine
Cadmium
Calcium
Californium
Carbon
Cerium
Cesium
Chlorine
Chromium
Cobalt
Copper
Curium
Dysprosium
Erbium
Europium
Fermium
Fluorine
Francium
Gadolinium
Gallium
Germanium
Gold
Hafnium
Helium
Holmium
Hydrogen
Indium
Iodine
Iridium
Iron
Krypton
Lanthanum
Lawrencium
Lead
Lithium
Lutetium
Magnesium Ac
Al
Am
Sb
Ar
As
At
Ba
Bk
Be
Bi
B
Br
Cd
Ca
Cf
C
Ce
Cs
Cl
Cr
Co
Cu
Cm
Dy
Er
Eu
Fm
F
Fr
Gd
Ga
Ge
Au
Hf
He
Ho
H
In
I
Ir
Fe
Kr
La
Lr
Pb
Li
Lu
Mg 89
13
95
51
18
33
84
56
97
4
83
5
35
48
20
98
6
58
55
17
24
27
29
96
66
68
63
100
9
87
64
31
32
79
72
2
67
1
49
53
77
26
36
57
103
82
3
71
12 (227)
27.0
(243)
121.8
40.0
74.9
(210)
137.3
(247)
9.01
209.0
10.8
79.9
112.4
40.1
(249)
12.0
140.1
132.9
35.5
52.0
58.9
63.5
(247)
162.5
167.3
152.0
(253)
19.0
(223)
157.3
69.7
72.6
197.0
178.5
4.00
164.9
1.008
114.8
126.9
192.2
55.8
83.8
138.9
(257)
207.2
6.94
175.0
24.3 Manganese
Mendelevium
Mercury
Molybdenum
Neodymium
Neon
Neptunium
Nickel
Niobium
Nitrogen
Nobelium
Osmium
Oxygen
Palladium
Phosphorus
Platinum
Plutonium
Polonium
Potassium
Praseodymium
Promethium
Protactinium
Radium
Radon
Rhenium
Rhodium
Rubidium
Ruthenium
Samarium
Scandium
Selenium
Silicon
Silver
Sodium
Strontium
Sulfur
Tantalum
Technetium
Tellurium
Terbium
Thallium
Thorium
Thulium
Tin
Titanium
Tungsten
Uranium
Vanadium
Xenon
Ytterbium
Yttrium
Zinc
Zirconium Mn
Md
Hg
Mo
Nd
Ne
Np
Ni
Nb
N
No
Os
O
Pd
P
Pt
Pu
Po
K
Pr
Pm
Pa
Ra
Rn
Re
Rh
Rb
Ru
Sm
Sc
Se
Si
Ag
Na
Sr
S
Ta
Tc
Te
Tb
Tl
Th
Tm
Sn
Ti
W
U
V
Xe
Yb
Y
Zn
Zr 25
101
80
42
60
10
93
28
41
7
102
76
8
46
15
78
94
84
19
59
61
91
88
86
75
45
37
44
62
21
34
14
47
11
38
16
73
43
52
65
81
90
69
50
22
74
92
23
54
70
39
30
40 54.9
(256)
200.6
95.9
144.2
20.2
(237)
58.7
92.9
14.0
(253)
190.2
16.0
106.4
31.0
195.1
(242)
(209)
39.1
140.9
(145)
(231)
(226)
(222)
186.2
102.9
85.5
101.1
150.4
45.0
79.0
28.1
107.9
23.0
87.6
32.1
180.9
(98)
127.6
158.9
204.4
232.0
168.9
118.7
47.9
183.8
238.0
50.9
131.3
173.0
88.9
65.4
91.2 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.
 Summer '09
 J

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