Chem17-18gases - Calculations In Chemistry ***** Modules 17...

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Unformatted text preview: Calculations In Chemistry ***** Modules 17 and 18 Ideal Gases, Gas Labs, Gas Reactions Module 17 – Ideal Gases ............................................................................................... 407 Lesson 17A: Lesson 17B: Lesson 17C: Lesson 17D: Lesson 17E: Lesson 17F: Lesson 17G: Lesson 17H: Gas Fundamentals............................................................................................. 407 Gases at STP ....................................................................................................... 411 Complex Unit Cancellation.............................................................................. 417 The Ideal Gas Law and Solving Equations .................................................... 422 Choosing Consistent Units............................................................................... 426 Density, Molar Mass, and Choosing Equations ............................................ 430 Using the Combined Equation ........................................................................ 437 Gas Law Summary and Practice ..................................................................... 443 For additional modules, visit www.ChemReview.Net ©2011 ChemReview.net v. f7 Page i Module 17 — Ideal Gases 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 particles. 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, silver-colored 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. f7 Page 407 Module 17 — Ideal Gases A barometer is a kind of balance, like a playground “teeter-totter.” 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 20-mile-high 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 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 fair-weather 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 must 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. Examples: 1 atm = 760 mm Hg ©2011 ChemReview.net v. f7 or 101 kPa = 760 torr Page 408 Module 17 — Ideal Gases Practice A Memorize the pressure unit definitions, then use the equalities in these calculations. Answers are at the end of this lesson. 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? 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? When using any equation which specifies a capital T. ©2011 ChemReview.net v. f7 Page 409 Module 17 — Ideal Gases 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). When measuring gases, it is preferred to record gas volumes at standard temperature (for gas laws) which is the temperature an ice-water mixture: zero degrees Celsius, which is 273 K. Practice B 1. Memorize the equation relating kelvins to degrees Celsius, then complete the following chart (assume boiling and freezing points are at standard pressure): In kelvins In Celsius Absolute Zero ________ __________ Water Boils ________ __________ Nitrogen Boils ________ ─196 oC Table Salt Melts 1074 K __________ Water Freezes (std. P) ________ __________ Std. Temperature ________ __________ 2. In a problem involving gases, if you calculate a temperature for the gas of ─310 degrees Celsius, your answer is…..? 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.) 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 ©2011 ChemReview.net v. f7 1 atm = 4.9 atm 14.7 psi ● 101 kPa 1 atm = 3.32 kPa Page 410 Module 17 — Ideal Gases 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 ice-water 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. 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. ©2011 ChemReview.net v. f7 Page 411 Module 17 — Ideal Gases 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. The rule is: Gas volumes must be labeled with a T and P if the T and P are known. In calculations, when you write gas volumes (such as L, dm3, mL), attach a P and T. Non-Ideal 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 non-ideal 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. 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 ©2011 ChemReview.net v. f7 Page 412 Module 17 — Ideal Gases 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? * * * * * (See How To Use These Lessons, Point 1, on page 1). 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 = 7.4 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. ***** 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. f7 Page 413 Module 17 — Ideal Gases 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, right-side 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. Then do all of Problems 1 below. After #1, 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 density of SO2 gas in g/L 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. f7 Page 414 Module 17 — Ideal Gases 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 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.) g SO2 gas = 64.1 g SO2 ● 1 mol 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 right-side up.) ? g = 0.00205 g gas ● 1 mL ● 22.4 L any gas at STP = 45.9 g 1 mol gas mol mol 1 mL gas at STP 10─3 L 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 ©2011 ChemReview.net v. f7 (STP Prompt) Page 415 Module 17 — Ideal Gases SOLVE: (These conversions below may be in any order.) 0.313 g gas g= ● 1 mL ● 22.4 L any gas STP = 28.0 g 1 mol gas mol mol 250. mL gas at STP 10─3 L ? 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 7. WANTED: 1 mol gas = 8.96 x 103 mL F2(g) STP 10─3 L ? 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 ©2011 ChemReview.net v. f7 Page 416 Module 17 — Ideal Gases 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 ***** Lesson 17C: Cancellation of Complex Units Solving gas law and heat problems requires 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. f7 Page 417 Module 17 — Ideal Gases 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. Remove the parentheses and simplify: 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: 1=C B B C In exponents: (B/C) ─1 = (B ● C─1) ─1 = B─1 ● C 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. Solve these in your notebook. Q2. Remove the parentheses and simplify: A/(B/C) ***** ©2011 ChemReview.net v. f7 Apply Rule 2a, then 2C. Page 418 Module 17 — Ideal Gases 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 Apply Rule 2B. ***** 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. A way to summarize Rule 2 is: To simplify a term with more than one fraction, separate the fractions. Q4. Simplify (A/B)/(C •(D/E)) = ***** (A/B)/(C •(D/E)) A= B C• D E = Rewrite for clarity↑ A● B C 1 D E = A● 1 ● B C ↑Rule 2a ↑Rule 2b 1 D E = 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. = ©2011 ChemReview.net v. f7 4. (meters/sec) sec = Page 419 Module 17 — Ideal Gases 5. meters/sec/sec = 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 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. 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. ©2011 ChemReview.net v. f7 Page 420 Module 17 — Ideal Gases 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 1-2c. Practice B: Apply the rules above to simplify these. 1. calories = calorie • gram gram • ºC 3. (mole) • atm ● L • (K) = mole ● K liters 2. atm ● L = (mole) • atm ● L mole ● K ANSWERS Practice A 1. 3. 2. (D/E)─1 = X/(Y/Z) = X • 1 = X • Z Y Y Z meters meters sec = meters • 1= meters sec = 1= D E E or (D─1 • E) D meters • sec = sec meters = meters • 1 = meters sec sec sec2 4. (meters/sec) sec 5. meters/sec/sec = Cannot be evaluated unless you know the numerator or denominator. Practice B 1. 2. = calories calorie • gram gram • ºC atm ● L (mole) • atm ● L mole ● K ©2011 ChemReview.net v. f7 = 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 Page 421 Module 17 — Ideal Gases 3. (mole) • atm ● L • (K) = (mol) • ( atm • L ) • (K) ● 1 mole ● K mol • K L liters = atm ***** Lesson 17D: The Ideal-Gas 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 piv-nert). 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. When using R to solve problems, R must have consistent units: 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 soda-can equality), the different numbers and units used for R do not change the constant quantity that R represents. 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 even when the gases have very different molar masses and chemical formulas. Non-Ideal 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 “non-ideal” 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. ©2011 ChemReview.net v. f7 Page 422 Module 17 — Ideal Gases 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 in 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 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. 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 ©2011 ChemReview.net v. f7 Page 423 Module 17 — Ideal Gases 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. Symbols move more quickly (and with fewer mistakes) than numbers with their units. 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 = (1 atm)(22.4 L) (1 mol)(273 K) = 0.0821 atm ● L mol ● K 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 a required set of units, use PV=nRT , one mole of gas, and the values for P, V, and T at STP in the required units. ***** 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. ©2011 ChemReview.net v. f7 Page 424 Module 17 — Ideal Gases 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. 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. 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. ©2011 ChemReview.net v. f7 Page 425 Module 17 — Ideal Gases 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. ***** 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. ©2011 ChemReview.net v. f7 Page 426 Module 17 — Ideal Gases 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.) ***** 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. ***** 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. ©2011 ChemReview.net v. f7 Page 427 Module 17 — Ideal Gases Practice A: 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? 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 unit that you will use for each equation symbol, a systematic approach is to write those units in the DATA table after each symbol, such as DATA: P in atm = V in L = T in 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. ***** 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 in 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 in kPa = 2.00 atm ● 101 kPa 1 atm = 202 kPa V in L = 125 mL ©2011 ChemReview.net v. f7 Page 428 Module 17 — Ideal Gases V must be in L to be consistent with the R units. Convert V, then finish the table and solve. ***** V in L = 125 mL = 0.125 L (by inspection: mL is 1000x L) n in 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. ***** 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: If an equation has a constant that includes units, and the constant must be used, • In the DATA table, list the constant first. • After each symbol for a variable, write a unit to convert DATA to. Choose if available a unit that matches the constants of the equation. • First SOLVE in the consistent unit, then convert to the WANTED unit if needed. Practice B: 1. If 2.00 moles of Ne gas has a pressure of 0.500 atm and and a temperature of 25 ºC, what will be its volume in liters? (USE: R = 8.31 kPa•L/mol•K ) ©2011 ChemReview.net v. f7 Page 429 Module 17 — Ideal Gases ANSWERS Practice A 1. PV = nRT DATA: P in torr = WANTED (the only R value that uses torr) R = 62.4 torr • L mol • K V in L to match R unit = 560. mL = 0.560 L (mL to L: see Lesson 12A) n = 0.0500 mol T must be in K : SOLVE: K = ºC +273 = 0 ºC + 273 = 273 K 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, re-write the equation with the symbols in the denominator separated into a reciprocal, then plug in numbers into this separated format (see Lesson 17C). However, you may do the math for the numbers and units in any way you choose, provided you do both numbers and units. Practice B 1. PV = nRT DATA: V in liters = WANTED (If a certain constant must be used, write it first) R = 8.31 kPa • L mol • K P in kPa = 0.500 atm ● 101 kPa 1 atm = 50.5 kPa ( ↑ convert P in data to kPa to be consistent with the units of R) V in liters = WANTED T must be in K : SOLVE: K = ºC +273 = 25 ºC + 273 = 298 K PV = nRT 1 = 98.1 L ? = V = nRT = nRT ● 1 = (2.00 mol)( 8.31 kPa • L )(298 K) ● P P mol • K 50.5 kPa ***** ©2011 ChemReview.net v. f7 Page 430 Module 17 — Ideal Gases 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) ***** 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) ©2011 ChemReview.net v. f7 Page 431 Module 17 — Ideal Gases 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 = 4.00 g mol Practice A: Use the above methods. Answers are at the end of this lesson. (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? 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 ©2011 ChemReview.net v. f7 Dgrams = grams Any V unit (1) (2) Page 432 Module 17 — Ideal Gases Substituting equation (1) into (2), Dgrams = moles ● MM , which can be rewritten as V Since PV = nRT can be re-written as P = Dgrams ● RT MM (3) P = moles ● RT , V Substituting with equation (3) gives moles = Dgrams V 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 gas-law 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. 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. ©2011 ChemReview.net v. f7 Page 433 Module 17 — Ideal Gases 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). 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. ©2011 ChemReview.net v. f7 Page 434 Module 17 — Ideal Gases 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 b. 202 kPa P = Dgrams • RT MM 2. Solve 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 A 1. Hints: 1. The data will include 1.76 g UF6 2. It says to use R. Write the only relationship that we know so far which uses R. ***** PV = nRT DATA: Make your data table to match the equation: n= ? V = ? WANTED Plus: P = 380 torr R = 62.4 torr • L mol • K T = 600. K 1.76 g UF6 352.0 g UF6 = 1 mol UF6 Strategy: (P uses torr) (g prompt) 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 ©2011 ChemReview.net v. f7 Page 435 Module 17 — Ideal Gases 2. WANT: ? g Ne DATA: PV = nRT (It says to use a relationship with R) P = 2.0 atm V = 8.96 L R = 0.0821 L • atm mol • K T = 298 K 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. ? = n = PV = PV ● 1 = 2.00 atm ● 8.96 L ● mol • K = 0.7324 mol Ne 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 Practice B 1. a. 122 g/mol MM 2. Solve b. 202 kPa P P = Dgrams • RT MM for b) MM = Dgrams • R • T P c. 13.5 g/mL D d. 30˚C t 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 (Strategy: P 1.90 g = 1 L (3 sf – doubt in one’s place when adding) 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)). ©2011 ChemReview.net v. f7 Page 436 Module 17 — Ideal Gases If an equation fits the symbols, try the equation. Solving for the wanted MM:) 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? 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. WANTED: ? g H2 gas L H2 gas at STP DATA: 2.016 g H2 = 1 mol H2 1 mol any gas = 22.4 L any gas at STP Strategy: (grams prompt) (STP Prompt) Analyze your units. Equalities lend themselves to conversions. You want grams over liters. You know grams to moles and moles to liters. 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. f7 Page 437 Module 17 — Ideal Gases Lesson 17G: Using the Combined Equation Many gas-law 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 are 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). You may also see it referred to as the two-point equation since it is based on initial and final conditions. 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 apply 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. f7 Page 438 Module 17 — Ideal Gases 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? ***** P1V1 = P2V2 T1 T2 DATA: P2 = 2.5 atm P1 = 1.0 atm V1 = 15.0 Liters V2 = ? T1 = 273 K (std T) 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 = ©2011 ChemReview.net v. f7 b. P1 = c. V2 = Page 439 Module 17 — Ideal Gases 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? 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 aerosol-spray 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 T2 T1 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 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 ©2011 ChemReview.net v. f7 = 0.500 atm ( or P1 = 455 kPa) Page 440 Module 17 — Ideal Gases 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 = (4.5 atm) (0.250 L) (273 K) = 2.0 L P2 T1 (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? Simplifying Conditions In a calculation involving a change from initial to final conditions for a constant moles of gas (gas in a sealed container), 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 (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 DATA: T2 Solve for the WANTED symbol first, then substitute the DATA. ***** ©2011 ChemReview.net v. f7 Page 441 Module 17 — Ideal Gases P1V1 SOLVE: 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 A 1. a. T2 If P1V1 = P2V2 , then T PVT 2= 221 T1 T2 P1 V1 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 T1 T2 DATA: P1 = 100. atm P2 = 2.00 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. f7 = (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 442 Module 17 — Ideal Gases 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 ºC = K ─ 273 = 434 K ─ 273 = 161 ºC Practice C 1. For trapped gas changing to new conditions, use P1V1 = P2V2 T1 T2 DATA: P2 = ? kPa P1 = 380. mm Hg = 50.5 kPa V1 = 0.500 L = V2 (a sealed metal container will have a constant volume) T1 = 30.ºC + 273 = 303 K 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 ● SOLVE: = 50.5 kPa 101 kPa 760 mm Hg P2 = P1 V1 T2 T1 V2 = (50.5 kPa) (273 K) = (303 K) 45.5 kPa ***** ©2011 ChemReview.net v. f7 Page 443 Module 17 — Ideal Gases 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. 9. For numbers, units, or symbols: 1/(1/X) = X ; A/(B/C) = (A · C)/B ***** ©2011 ChemReview.net v. f7 Page 444 Module 17 — Ideal Gases 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. 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? 2. How many gas molecules will there be, per milliliter, for all gases at STP? 3. 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? 4. If the density of a gas is 0.0147 g/mL at 20.ºC and 1.00 atm pressure, what is its molar mass? 5. 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? 6. 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: DATA: ? V at end 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: SOLVE: 2. WANTED: DATA: T2 P2 The WANTED and DATA symbols match ? = V2 = P1V1 T2 P 2 T1 P1V1 = P2V2 T1 T2 = (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 Strategy: P1 (“molecules” calls the Avogadro prompt) (STP Prompt) 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. ©2011 ChemReview.net v. f7 Page 445 Module 17 — Ideal Gases 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 3. WANTED: DATA: ? P -- problem uses kPa units, and wants P at end = P2 P = 112 kPa = P at start = P1 t = 25oC initial ; t = 100.oC final Strategy: K = 25oC + 273 = 298 K = T1 K = 100.oC + 273 = 373 K = T2 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. SOLVE: P1V1 = P2V2 ; T2 T1 4. 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 P = 1.00 atm Strategy: If the gas were at STP, we could use conversions, but it is not. Try adding symbols. The D and MM symbols fit the equation P = Dgrams • RT MM You need an R value to use this equation. Since P is in atm, choose R = 0.0821 atm • L / mol • K SOLVE: ? = Dg = MM = mol Dg • RT = 0.0147 g • 1 • 0.0821 atm L • 293 K = ??? P mL 1 atm mol K 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. This would have meant converting the “1 mL : in the density to 0.001 exact liters, in order to match the unit of volume in the R you chose. But the above, though messy, works. If the units cancel properly, it’s probably right.) 5. WANTED: ? ºC ©2011 ChemReview.net v. f7 t Page 446 Module 17 — Ideal Gases DATA: 70.4 g UF6 352.0 g UF6 = 1 mol UF6 4.48 L gas (grams prompt) V 202.6 kPa 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? ** *** = 0.200 mol UF6 n = ? mol UF6 = 70.4 g UF6 • 1 mol UF6 352.0 g UF6 PV = nRT 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. 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. f7 Page 447 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 re-written 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. f7 , where c is constant. Page 448 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. f7 Page 449 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? 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? ©2011 ChemReview.net v. f7 Page 450 Module 18 — Gas Labs, Gas Reactions 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 y-intercept, the value of y at x = 0 (which is the value of y where the graphed line crosses the y-axis). If the value of the y-intercept (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 y-intercept (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 y-axis variable = (constant slope of line)(x-axis 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. 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 y-axis and T on the x-axis, 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. ***** ©2011 ChemReview.net v. f7 Page 451 T Module 18 — Gas Labs, Gas Reactions 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 y-axis, and data for the other variable X is graphed on the x-axis, 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. 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 y-axis, and absolute temperature on the x-axis, what will be the shape of the graph? ©2011 ChemReview.net v. f7 Page 452 Module 18 — Gas Labs, Gas Reactions 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 = C 2 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 y-axis, versus ºC on the x-axis, 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 y-axis versus ºC on the x-axis, what will be the value of the slope? What will be the value of the y-intercept? 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. 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 one-half T , so P would be predicted to be about ½ of 450 = ~ 225 kPa. ©2011 ChemReview.net v. f7 Page 453 Module 18 — Gas Labs, Gas Reactions Practice B 1. a. P and T b. V and n P1 = P2 T2 T1 P = c and T d. 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 False ( x ≠ my ) D True ( y = mx ) True, since C = mD e. D = (1/m) C f. C C1 = C2 D2 D1 True; since C = mD, C/D = m = constant 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 y-intercept is zero (the line passes through the origin). Here, the y-intercept 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 y-intercept? b = + 32 **** ©2011 ChemReview.net v. f7 Page 454 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 mid-1600’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 one-third 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 re-written 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 long-term 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. f7 Page 455 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 _________? 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 x-values increase, y-values decrease. As x-values approach zero, y-values 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 y-axis and values calculated for 1/V are assigned to the x-axis, 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 y-intercept (b) is not zero, the y versus 1/x relationship will not be a direct proportion, and the data on the y- and x-axes 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. f7 Page 456 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 y-axis, and time on the x-axis, what will be the shape of the graph? f. If rate is plotted on the y-axis, what must be plotted on the x-axis to produce a graph which is a straight line through the origin? 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 re-write A 1/B as a. If A is tripled? AB = c B is reduced to 1/3 of its original value. b. If B goes from 400 to 200, A must go from 300 to 600 . ***** ©2011 ChemReview.net v. f7 Page 457 Module 18 — Gas Labs, Gas Reactions 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 . 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 whole-number 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. f7 Page 458 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. 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? ©2011 ChemReview.net v. f7 Page 459 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 volume-volume conversion. solve using standard 7-step 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 (WANTED # given formula) 1 mol gas = 22.4 L any gas at STP (STP prompt) 32.0 g O2 = 1 mol O2 (g prompt) 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. C3H8(g) + 5 O2(g) 4. Bridge. 5 mol O2 = 1 mol C3H8 3 CO2(g) + 4 H2O(g) (the WANTED to given mol ratio) ***** Steps 5-7: When a single unit is WANTED, convert “final unit WANTED = # and unit of given substance” to moles given, to moles WANTED, to units WANTED. ©2011 ChemReview.net v. f7 Page 460 Module 18 — Gas Labs, Gas Reactions ? unit WANTED = unit given ? g O2 = 3.50 L C3H8 gas STP ● >> mol given >> bridge-mol 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? 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 rice-moles 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 ©2011 ChemReview.net v. f7 Page 461 Module 18 — Gas Labs, Gas Reactions All supplied units to all moles to rice-moles 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 un-stuck, 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) ***** 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) ©2011 ChemReview.net v. f7 1 MgCl2(aq) + 1 H2(g) Page 462 Module 18 — Gas Labs, Gas Reactions Reaction 2 HCl 1 MgCl2 1 H2 excess -- -- ― 0.0800 mol + 0.0400 mol + 0.0400 mol excess + 0.0400 mol + 0.0400 mol 1 Mg 0.0400 mol Initial Change ― 0.0400 mol End 0 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. 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 volume-volume conversion, based on Avogadro’s hypothesis, using the coefficients of the balanced equation as whole-number gas volume ratios. 2. If all of the gas volumes in the WANTED and given are at STP. • Solve using standard 7-step 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 > rice-moles table > WANTED units ©2011 ChemReview.net v. f7 Page 463 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 A 1. a. H2 b. O2. 20. L H2 uses up only 10. L O2. c. O2 and H2O d. Zero H2, 10. L O2, 20. L H2O. 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 5-7:. ? 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. f7 Page 464 Module 18 — Gas Labs, Gas Reactions Practice C 1. 1. WANTED: 2. DATA: ? L CO2 gas at 740. torr and 30˚C (Want L, a single unit) 0.250 mol C6H12O6 (This must be the single-unit 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 > rice-moles 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. 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 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 ©2011 ChemReview.net v. f7 (g prompt) Page 465 Module 18 — Gas Labs, Gas Reactions 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 > rice-moles 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. ***** 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. f7 Page 466 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 and become part of the vapor above the liquid or solid. Vapor pressure 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. 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 below the surface of a liquid, and not just at its edges. 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 long-term memory. ©2011 ChemReview.net v. f7 Page 467 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 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 ©2011 ChemReview.net v. f7 4.50 moles He Page 468 Module 18 — Gas Labs, Gas Reactions ? Moles O2 = 16.0 g O2 • 1 mole O2 = 32.0 g O2 Equation (2) uses mole fraction. 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)? 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) ©2011 ChemReview.net v. f7 Page 469 Module 18 — Gas Labs, Gas Reactions 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). 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. f7 Page 470 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 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. f7 Page 471 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 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. ©2011 ChemReview.net v. f7 Page 472 Module 18 — Gas Labs, Gas Reactions SOLVE: P1V1 = P2V2 T2 T1 ? = V2 = P1V1 P2 = P1V1 = P2V2 ; = (740. torr) (0.370 L) = 0.360 L at 20.˚C and SP (760 torr) ***** 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 X2 X1 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 rice-moles 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. f7 Page 473 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. f7 Page 474 ...
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