5
The Chapter Ideal Gas Laws
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PHYSICAL PROPERTIES: STATES OF MATTER!
Physical properties of a substance include
appearance, density, melting point, boiling point,
refractive index, etc.
The most common physical property of a pure
substance is its physical state, that is, whether it is a
solid, a liquid or a gas.
Water vapor in
equilibrium with liquid
water in the clouds.
Solid water (ice).
Liquid water.
CHEMICAL AND PHYSICAL CHANGES!
The most common form of a physical change is a simple
change in state, in which a substance goes from a solid,
to a liquid, and finally to a gas as it is warmed.
PHYSICAL PROPERTIES: STATES OF MATTER!
Solids have definite shapes and definite volumes.
Liquids have a definite volume, but they take on the
shape of whatever container they occupy.
Gasses have neither definite shape nor definite volume
and they must by contained on all sides, taking on the
shape and volume of the container.
THE KINETIC MOLECULAR THEORY!
According to the Kinetic Molecular Theory, all
substances have an inherent attraction to each other
(intermolecular forces of various forms).
These attractive forces tend to make the particles of a
substance stick together in a lump.
Imagine a collection of
small magnets, held
together by their
attractive forces.
THE KINETIC MOLECULAR THEORY!
The Kinetic Molecular Theory tells us that the particles
that make up a substance are always in motion. The
more energy that is put into a substance, the faster (and
more energetic) that motion becomes. In a solid, the
particles simply sit there in more or less one place and
vibrate because they dont have enough energy to
break the attractive forces between them.
In a solid, the particles
dont have enough
kinetic energy to
overcome the
intermolecular forces.
THE KINETIC MOLECULAR THEORY!
In a liquid, the particles have enough energy to bend
the attractive forces, but not to break them. The result
is that the particles tend to slide over and around each
other, but still remain associated.
In a liquid, the particles
dont have enough
kinetic energy to totally
overcome the
intermolecular forces.
THE KINETIC MOLECULAR THEORY!
In a gas, the particles have enough energy to totally
break all of the attractive forces between them and they
fly around randomly, colliding with each other and with
the walls of their container.
In aThus, depending on
gas, the particles have
sufficient kinetic energy to
the amount of energy
totally overcome
that is put into the
intermolecular forces.
system, a substance
can exist as a solid, a
liquid or a gas.
PROPERTIES OF GASSES!
Molecules or atoms in the gas phase have
certain defined properties which include:
A gas has an indefinite shape and can take on the
shape of its container.
Gasses can expand and compress in response to
increases or decreases in pressure.
Gasses diffuse uniformly throughout their
containers and all samples within the container
will have the same composition.
GAS PRESSURE!
Gas pressure is the result of constantly moving
atoms or molecules striking the inside surface of
its container. The pressure that a gas exerts
depends on how often and how hard these atoms
or molecules strike the walls of the container.
Atmospheric pressure is the result
of air molecules (as a mixture)
striking the various surfaces of the
environment. The pressure of the
atmosphere is about 14.7 pounds of
force on every square inch of
surface.
GAS PRESSURE!
Atmospheric pressure is measured using
a barometer; an assortment of typical
barometers are shown below.
GAS PRESSURE!
The principle of measuring atmospheric
pressure is most readily shown using a
simple At ercury barometer. 760
m sea level and 0 oC, h =
mm of Hg, which is defined
A column of mercury in a tube as
sealed onatmospheric pressure. in
one end is supported
an open container of mercury. The
level of mercury in the tube falls until
the pressure of the atmosphere on
the surface of the container
equals the weight of the column
of mercury in the tube.
The height of the resulting column
of mercury is defined as the
atmospheric pressure in mm Hg.
GAS PRESSURE!
The pressure of a gas is measured
using a variety of units:
1 atm (atmosphere) is the
atmospheric pressure at sea
level and is equal to 760 mm
Hg, at 0 oC (standard
pressure).
mm Hg are also referred to as
torr; 760 mm Hg = 760 torr
Atmospheric pressure is also
reported in pounds per square
inch (psi); 760 mm Hg = 14.7
psi.
THE PRESSURE-VOLUME RELATIONSHIP!
As noted previously, gasses can expand and
compress in response to increases or
decreases in pressure. The relationship is
defined by Boyles Law which states that, at
constant temperature, the volume occupied
by a fixed quantity of gas is inversely
proportional to the applied pressure.
The piston shown on the right is sealed
and contains a fixed quantity of a gas. If
an object is placed on the top of the
piston, the gas will compress and the
pressure, measured by the gauge, will
increase.
THE PRESSURE-VOLUME RELATIONSHIP!
Thus, an object with a very small mass,
exhibits a small change in pressure.
While a more substantial mass
significantly compresses the gas
and increases the pressure.
The simple conclusion here is that
the volume of the gas is inversely
proportional to the pressure applied
to the piston.
THE PRESSURE-VOLUME RELATIONSHIP!
The quantitative nature
of the pressure-volume
relationship is shown in
the following animation.
As the pressure is
increased, the volume
decreases. But, the
volume can never drop
to zero, so the
relationship must be
non-linear.
THE PRESSURE-VOLUME RELATIONSHIP!
The relationship between pressure and volume is a
reciprocal relationship; as pressure increases,
volume decreases.
Mathematically,
Boyles Law can be
described as:
1
V
P
V= k
1
P
THE PRESSURE-VOLUME RELATIONSHIP!
This equation can be
re-written:
1
or
PV = k
V= k
P
This equation says that
for any given sample of
gas, PV (Pressure x
Volume) is a constant.
Thus, for our two
examples:
P1V1 = k
P1V1 = P2V2
P2V2 = k
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 0.82 atm. If the
pressure changes to 1.32 atm, what will the final
volume be?
Recalling that P1V1 = P2V2, you simply
need to solve the equation for V2 in
terms of P1, V1 and P2.
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 0.82 atm. If the
pressure changes to 1.32 atm, what will the final
volume be?
P1V1 = P2V2
V2 =
V2 =
P1V1
P2
(0.82 atm)(12.5 L)
1.32 atm
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 0.82 atm. If the
pressure changes to 1.32 atm, what will the final
volume be?
P1V1 = P2V2
V2 =
V2 =
P1V1
P2
(0.82 atm)(12.5 L)
V2 = 7.8 L
1.32 atm
IN-CLASS PROBLEM!
The pressure of 150.0 mL sample of CO2 is 35.3 mm
Hg. If the volume is decreased to 50.0 mL (with the
temperature being held constant) what will the final
pressure be?
Recalling that P1V1 = P2V2, you simply
need to solve the equation for P2 in
terms of P1, V1 and V2.
IN-CLASS PROBLEM!
The pressure of 150.0 mL sample of CO2 is 35.3 mm
Hg. If the volume is decreased to 50.0 mL (with the
temperature being held constant) what will the final
pressure be?
P1V1 = P2V2
P1V1
P2 =
V2
P2 =
(35.3 mm Hg)(150.0 mL)
50.0 mL
IN-CLASS PROBLEM!
The pressure of 150.0 mL sample of CO2 is 35.3 mm
Hg. If the volume is decreased to 50.0 mL (with the
temperature being held constant) what will the final
pressure be?
P1V1 = P2V2
P1V1
P2 =
V2
P2 =
(35.3 mm Hg)(150.0 mL)
P2 = 106 mm
50.0 mL
IN-CLASS PROBLEM!
A sample of helium gas has a pressure of 860 mm
Hg. This gas is transferred to a different container
having a volume of 25.0 L; in this new container, the
pressure is determined to be 770 mm Hg. What was
the initial volume of the gas?
Recalling that P1V1 = P2V2, you simply
need to solve the equation for V1 (the
initial volume) in terms of P1, P2 and V2.
IN-CLASS PROBLEM!
A sample of helium gas has a pressure of 860 mm
Hg. This gas is transferred to a different container
having a volume of 25.0 L; in this new container, the
pressure is determined to be 770 mm Hg. What was
the initial volume of the gas?
P1V1 = P2V2
V1 =
V1 =
P2V2
P1
(770.0 mm Hg)(25.0 L)
860.0 mm Hg
IN-CLASS PROBLEM!
A sample of helium gas has a pressure of 860.0 mm
Hg. This gas is transferred to a different container
having a volume of 25.0 L; in this new container, the
pressure is determined to be 770.0 mm Hg. What
was the initial volume of the gas?
P1V1 = P2V2
V1 =
V1 =
P2V2
P1
(770.0 mm Hg)(25.0 L)
V1 = 22.4 L
860.0 mm Hg
A gas system has initial pressure and volume of 5.05
atm and 5050 mL. If the volume changes to 5.62 L,
what will the resultant pressure be in atm?
Enter your Result:
P1 = 5.05 atm
V1 = 5050 mL
P2 = ? (in atm)
V2 = 5.62 L 5620 mL
Rearrange Boyle's Law to solve for the missing quantity
A gas system has initial pressure and volume of 5.05
atm and 5050 mL. If the volume changes to 5.62 L,
what will the resultant pressure be in atm?
Enter your Result:
P1 = 5.05 atm
V1 = 5050 mL
P2 = ? (in atm)
V2 = 5.62 L 5620 mL
Rearrange Boyle's Law to solve for the missing quantity
(P1V1) = (P2V2) P2 = P1(V1/V2)
Substitute and solve:
(5.05) (5050/5620) = 4.538 = 4.54 atm
A gas system has initial pressure and volume of 5.05
atm and 5050 mL. If the volume changes to 5.62 L,
what will the resultant pressure be in atm?
Enter your Result:
That is correct!
4.54
A gas system has initial pressure and volume of 2350
mm Hg and 8520 mL. If the pressure changes to 5.60
atm, what will the resultant volume be in L?
Enter your Result:
P1 = 2350 mm Hg
V1 = 8520 mL 8.52 L
P2 = 5.60 atm 4260 mm Hg
V2 = ? (in L)
Rearrange Boyle's Law to solve for the missing quantity
A gas system has initial pressure and volume of 2350
mm Hg and 8520 mL. If the pressure changes to 5.60
atm, what will the resultant volume be in L?
Enter your Result:
P1 = 2350 mm Hg
V1 = 8520 mL 8.52 L
P2 = 5.60 atm 4260 mm Hg
V2 = ? (in L)
Rearrange Boyle's Law to solve for the missing quantity
(P1V1) = (P2V2) V2 = P1(V1/P2)
Substitute and solve:
(2350) (8.52/4260) = 3575 = 3.58 L
A gas system has initial pressure and volume of 2350
mm Hg and 8520 mL. If the pressure changes to 5.60
atm, what will the resultant volume be in L?
Enter your Result:
That is correct!
3.58
THE TEMPERATURE-VOLUME RELATIONSHIP!
The balloon in the animation will be
The qualitative nature-196 oC.
cooled in liquid N2 at of
the temperature-volume
relationship is shown in
the following animation.
As the temperature is
decreased, the volume
decreases, and as the
temperature is increased,
the volume also increases.
The relationship between
the two is linear.
THE TEMPERATURE-VOLUME RELATIONSHIP!
The balloon deflates (the
volume decreases) as the
temperature is decreased.
As the balloon warms (as
the temperature increases)
the volume also increases.
Again, the relationship
between temperature and
volume is linear.
THE TEMPERATURE-VOLUME RELATIONSHIP!
Volume, L
20
15
In order to confirm the proposed linear
relationship between temperature and
volume, the volume of different molar
amounts of the same gas sample were
determined at various temperatures and
the results presented on a plot.
10
5
0
-200
-100
0
100
Temperature, degrees C
200
THE TEMPERATURE-VOLUME RELATIONSHIP!
Volume, L
20
The extrapolated xintercept for all of the
data intersect at the
same point: -273.15 oC
0.75 mole
0.50 mole
15
0.25 mole
10
The relationship is
linear, as predicted.
5
0
-200
-100
0
100
Temperature, degrees C
200
THE TEMPERATURE-VOLUME RELATIONSHIP!
This value, -273.15 oC,
was defined as 0 on the
Kelvin (K) scale; this is
absolute zero.
Volume, L
20
0.75 mole
0.50 mole
15
0.25 mole
10
5
0
0
73
173
273
Temperature, K
373
473
THE TEMPERATURE-VOLUME RELATIONSHIP!
The relationship between temperature and volume
is a linear relationship; as increases increases,
volume increases.
Mathematically, the relationship can be described
as:
V
k=
V T or V = kT
T
This is a mathematical statement of Charless Law
which can be generalized as:
V1
T1
=
V2
T2
Which relates temperature and volume for a given
sample of gas at two sets of experimental conditions.
THE TEMPERATURE-VOLUME RELATIONSHIP!
The Kelvin temperature scale must be used in all
gas-law problems whenever temperature is a
variable.
Convert the following temperatures into the
equivalent Kelvin temperatures.
1000.0 oC
-150.0 oC
100.0 oC
1273.2 K
123.2 K
373.2 K
The conversion is:
T (Kelvin) = t (Celsius) + 273.15
THE TEMPERATURE-VOLUME RELATIONSHIP!
The Kelvin temperature
100oC
scale must be used in all
gas-law problems whenever
temperature is a variable.
373.15 K
Boiling Point
of water.
"
Centigrade and Kelvin differ
by 273.15 degrees, but the
size of a degree is identical.
"
0 oC
273.15 K
Freezing
Point of water.
"
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its volume is 62.4 mL . What
is the final temperature of the gas?
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its volume is 62.4 mL . What
is the final temperature of the gas?
We know V1, T1 and V2. We need to solve
Charless Law in terms of T1.
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its volume is 62.4 mL . What
is the final temperature of the gas?
V1
V2
=
T1
T2
T2
T2 =
V2 T1
=
V1
(62.4 mL) (299.55 K)
50.0 mL
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its volume is 62.4 mL . What
is the final temperature of the gas?
V1
V2
=
T1
T2
T2
T2 =
V2 T1
=
V1
(62.4 mL) (299.55 K)
50.0 mL
T2 = 373.84 K
T2 = 374 K
or
101 oC
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 1 atm. If the
pressure changes to 800.0 mm Hg, what will the final
volume be?
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 1 atm. If the
pressure changes to 800.0 mm Hg, what will the final
volume be?
Recalling that P1V1 = P2V2, you simply
need to solve the equation for V2 in
terms of P1, V1 and P2.
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 1 atm. If the
pressure changes to 800.0 mm Hg, what will the final
volume be?
Remember, 1 atm and 760 mm
(written this way only) are
exact numbers and are not
included when determining
significant figures.
IN-CLASS PROBLEM!
The pressure of 12.5 L of a gas is 1 atm. If the
pressure changes to 800.0 mm Hg, what will the final
volume be?
P1V1 = P2V2
V2 =
V2 =
P1V1
P2
(1 atm)(12.5 L)
1.053 atm
V2 = 11.9 L
THE TEMPERATURE-VOLUME RELATIONSHIP!
The relationship between temperature and volume
is a linear relationship; as increases increases,
volume increases.
Mathematically, the relationship can be described
as:
V
k=
V T or V = kT
T
This is a mathematical statement of Charless Law
which can be generalized as:
V1
T1
=
V2
T2
Which relates temperature and volume for a given
sample of gas at two sets of experimental conditions.
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its temperature is 374 K .
What is the final volume of the gas?
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its temperature is 374 K .
What is the final volume of the gas?
We know V1, T1 and T2. We need to solve
Charless Law in terms of V2.
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its temperature is 374 K .
What is the final volume of the gas?
V1
V2
=
T1
T2
V2
V2 =
V1 T2
=
T1
(50.0 mL) (374 K)
299.6 K
IN-CLASS PROBLEM!
A 50.0 mL sample of gas at 26.4 oC, is heated at
constant pressure until its temperature is 374 K .
What is the final volume of the gas?
V1
V2
=
T1
T2
V2
V2 =
V1 T2
=
T1
(50.0 mL) (374 K)
299.6 K
V2 = 62.4 mL
A gas system has initial volume and temperature of
11000 mL and 106 oC If the volume changes to 7.66 L,
what will the resultant temperature be in K?
Enter your Result:
V1 = 11000 mL
T1 = 106 oC 379.0 K
V2 = 7.66 L 7660 mL
T2 = ? (in K)
Rearrange Charles Law to solve for the missing quantity
A gas system has initial volume and temperature of
11000 mL and 106 oC If the volume changes to 7.66 L,
what will the resultant temperature be in K?
Enter your Result:
V1 = 11000 mL
T1 = 106 oC 379.0 K
V2 = 7.66 L 7660 mL
T2 = ? (in K)
Rearrange Charles Law to solve for the missing quantity
(V1/T1) = (V2/T2) (T2 = V2T1/V1)
Substitute and solve:
(7660) (379.0/11000) = 264 K
A gas system has initial volume and temperature of
11000 mL and 106 oC If the volume changes to 7.66 L,
what will the resultant temperature be in K?
Enter your Result:
That is correct!
264
A gas system has an initial temperature of 436.0 K
with the volume unknown. When the temperature
changes to 164.0 K the volume is found to be 5040
mL. What was the initial volume in L?
Enter your Result:
V1 = ? (in L)
T1 = 436.0 K
V2 = 5040 mL 5.04 L
T2 = 164.0 K
Rearrange Charles Law to solve for the missing quantity
A gas system has an initial temperature of 436.0 K
with the volume unknown. When the temperature
changes to 164.0 K the volume is found to be 5040
mL. What was the initial volume in L?
Enter your Result:
V1 = ? (in L)
T1 = 436.0 K
V2 = 5040 mL 5.04 L
T2 = 164.0 K
Rearrange Charles Law to solve for the missing quantity
(V1/T1) = (V2/T2) V1 = (V2T1/T2)
Substitute and solve:
(5.04) (436.0/164.0) = 13.4 L
A gas system has an initial temperature of 436.0 K
with the volume unknown. When the temperature
changes to 164.0 K the volume is found to be 5040
mL. What was the initial volume in L?
Enter your Result: 13.4
That is correct!
THE MOLE-VOLUME RELATIONSHIP!
Volume, L
A second observation you can
make from the data that was
shown previously is the
20
dependence of volume on on
the number of moles of gas
that is present.
15
0.75 mole
0.50 mole
0.25 mole
10
5
0
0
73
173
273
Temperature, K
373
473
THE MOLE-VOLUME RELATIONSHIP!
The relationship
between number of
moles of a gas (n)
and its volume is
linear; this is called
Avogadros Law.
THE IDEAL GAS LAWS!
As previously described, the relationship between
volume and pressure (Boyles Law) is,
1
P
the relationship between volume and temperature
(Charless Law) is,
V
VT
and the relationship between volume and the
number (n) of moles (Avogadros Law) is:
Vn
THE IDEAL GAS LAWS!
These three relationships can be combined to give
a single equation of the form:
V
nT
P
The combined proportionality constant for this
relationship is called the universal gas constant,
and is given the symbol R.
V=R
nT
P
PV = nRT
THE IDEAL GAS LAWS!
These three relationships can be combined to give
a single equation of the form:
V
nT
P
The combined proportionality constant for this
relationship is called the universal gas constant,
and is given the symbol R.
V=R
nT
P
PV = nRT
THE IDEAL GAS LAWS!
Experimentally it has been determined that one
mole of any gas at exactly 273.15 K and 1 atm
pressure has a volume of 22.414 L. Knowing this,
the value of R can be calculated:
nT
V=R
P
PV
R=
nT
(1 atm)(22.414 L)
=
(1 mol)(273.1 K)
L atm
= 0.082057
mol K
THE IDEAL GAS LAWS!
Because R is a constant for any gas, changes in
pressure, volume, temperature or number of moles
o any sample of a gas
This fis the combined Gas Law can be calculated by simply
eequating thetwo state
xpression for a equations describing the two states,
system.
as shown below.
PV = nRT
P1V1
R=
n1T1
therefore,
and
P1V1
n1T1
=
P2V2
R=
n2T2
P2V2
n2T2
THE IDEAL GAS LAWS!
Gas law problems generally belong to two main
groups; single state problems and two state
problems. In a single state problem a gas is
described under a single set of conditions of pressure,
volume and temperature. These problems are solved
using the equation:
PV = nRT
In a two state problem a sample of gas undergoes a
change in pressure, volume or temperature. These
problems are solved using the equation:
P1V1
=
n1T1
P2V2
n2T2
IN-CLASS PROBLEM!
A 0.0500 L sample of a gas has a pressure of 745
mm Hg at 26.4 oC. The temperature is now raised to
404.4 K and the volume is allowed to expand until a
final pressure of 1.06 atm is reached. What is the
final volume of the gas?
P1V1
n1T1
(0.980 atm) (0.0500 L)
(299.55 K)
V2
=
=
P2V2
n2T2
(1.06 atm) (V2)
(404.4 K)
(0.980 atm) (0.0500 L)
=
(299.55 K)
404.4 K
1.06 atm
IN-CLASS PROBLEM!
A 0.0500 L sample of a gas has a pressure of 745
mm Hg at 26.4 oC. The temperature is now raised to
404.4 K and the volume is allowed to expand until a
final pressure of 1.06 atm is reached. What is the
final volume of the gas?
P1V1
n1T1
(0.980 atm) (0.0500 L)
(299.55 K)
V2
=
=
P2V2
n2T2
(1.06 atm) (V2)
(404.4 K)
(0.980 atm) (0.0500 L)
=
(299.55 K)
404.4 K
1.06 atm
IN-CLASS PROBLEM!
A 0.0500 L sample of a gas has a pressure of 745
mm Hg at 26.4 oC. The temperature is now raised to
404.4 K and the volume is allowed to expand until a
final pressure of 1.06 atm is reached. What is the
final volume of the gas?
P1V1
n1T1
(0.980 atm) (0.0500 L)
(299.55 K)
V2 = 0.0624 L
=
=
P2V2
n2T2
(1.06 atm) (V2)
(404.4 K)
IN-CLASS PROBLEM!
When 128.9 grams of cyclopropane (C3H6) are placed
into an 8.00 L cylinder at 298 K, the pressure is
observed to be 1.24 atm. A piston in the cylinder is
now adjusted so that the volume is now 12.00 L and
the pressure is 0.88 atm. What is the final temperature
of the gas?
C3H6: 3(12.011) + 6(1.0079) = 42.080 g/mol
IN-CLASS PROBLEM!
When 128.9 grams of cyclopropane (C3H6) are placed
into an 8.00 L cylinder at 298 K, the pressure is
observed to be 1.24 atm. A piston in the cylinder is
now adjusted so that the volume is now 12.00 L and
the pressure is 0.88 atm. What is the final temperature
of the gas?
128.9 g cyclohexane
42.08 g/mol
= 3.06 mol
But do we need to know the number of moles?
Nope! In the problem, n does not change.
IN-CLASS PROBLEM!
When 128.9 grams of cyclopropane (C3H6) are placed
into an 8.00 L cylinder at 298 K, the pressure is
observed to be 1.24 atm. A piston in the cylinder is
now adjusted so that the volume is now 12.00 L and
the pressure is 0.88 atm. What is the final temperature
of the gas?
P1V1
P2V2
You are given the initial pressure, volume
=
The tfirst question nd whether 2pressure one
is1 the final this is a and
n T2
and emperaturen1T
a
state the a two of moles does
volume; or numberstate problem. not
change. (0.88 atm) (12.00 L)
(1.24 atm) (8.00 L)
(298 K)
T2 =
=
(T2)
(0.88 atm) (12.00 L) (298 K)
(1.24 atm) (8.00 L)
IN-CLASS PROBLEM!
When 128.9 grams of cyclopropane (C3H6) are placed
into an 8.00 L cylinder at 298 K, the pressure is
observed to be 1.24 atm. A piston in the cylinder is
now adjusted so that the volume is now 12.00 L and
the pressure is 0.88 atm. What is the final temperature
of the gas?
P1V1
P2V2
=
n1T1
n2T2
(1.24 atm) (8.00 L)
(298 K)
T2 =
=
(0.88 atm) (12.00 L)
(T2)
(0.88 atm) (12.00 L) (298 K)
(1.24 atm) (8.00 L)
IN-CLASS PROBLEM!
When 128.9 grams of cyclopropane (C3H6) are placed
into an 8.00 L cylinder at 298 K, the pressure is
observed to be 1.24 atm. A piston in the cylinder is
now adjusted so that the volume is now 12.00 L and
the pressure is 0.88 atm. What is the final temperature
of the gas?
P1V1
P2V2
=
n1T1
n2T2
(1.24 atm) (8.00 L)
(298 K)
=
(0.88 atm) (12.00 L)
T2 = 320 K
(T2)
Combined Gas Laws
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
Work the problem and enter the answer in the space provided. Do not include
units!
A closed gas system initially has pressure and volume of 0.405
atm and 6.35 L with the temperature unknown. If the same
closed system has values of 1030 mm Hg, 2.02 L and 357 K,
what was the initial temperature in K?
Enter your Result:
P1 = 0.405 atm; V1 = 6.35 L; T1 = ? (in K);
P2 = 1030 mm Hg 1.35 atm; V2 = 2.02 L; T2 = 357 K;
Rearrange the combined gas equation toLawsfor the
Combined Gas solve
missing quantity:
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
Work the problem and enter the answer in the space provided. Do not include
units!
A closed gas system initially has pressure and volume of 0.405
atm and 6.35 L with the temperature unknown. If the same
closed system has values of 1030 mm Hg, 2.02 L and 357 K,
what was the initial temperature in K?
Enter your Result:
P1 = 0.405 atm; V1 = 6.35 L; T1 = ? (in K);
P2 = 1030 mm Hg 1.35 atm; V2 = 2.02 L; T2 = 357 K;
Rearrange the combined gas equation toLawsfor the
Combined Gas solve
missing quantity:
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
(P1V1/T1) = (P2V2/T2) T1 = (P1V1T2/P2V2)
Work the problem and enter the answer in the space provided. Do not include
Substitute and solve:
units!
(0.405 6.35 357)/(1.35 2.02) = 337 K
A closed gas system initially has pressure and volume of 0.405
atm and 6.35 L with the temperature unknown. If the same
closed system has values of 1030 mm Hg, 2.02 L and 357 K,
what was the initial temperature in K?
Enter your Result:
That is correct!
337
Combined Gas Laws
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
Work the problem and enter the answer in the space provided. Do not include
units!
A closed gas system initially has volume and temperature of
9150 mL and 364.0 oC with the pressure unknown. If the same
closed system has values of 1.18 atm, 2180 mL and -61.00 oC,
what was the initial pressure in atm?
Enter your Result:
P1 = ? (in atm); V1 = 9150 mL; T1 = 364.0 oC 637K;
P2 = 1.18 atm; V2 = 2180 mL; T2 = -61.00 oC 212K;
Rearrange the combined gas equation toLawsfor the
Combined Gas solve
missing quantity:
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
Work the problem and enter the answer in the space provided. Do not include
units!
A closed gas system initially has volume and temperature of
9150 mL and 364.0 oC with the pressure unknown. If the same
closed system has values of 1.18 atm, 2180 mL and -61.00 oC,
what was the initial pressure in atm?
Enter your Result:
P1 = ? (in atm); V1 = 9150 mL; T1 = 364.0 oC 637K;
P2 = 1.18 atm; V2 = 2180 mL; T2 = -61.00 oC 212K;
Rearrange the combined gas equation toLawsfor the
Combined Gas solve
missing quantity:
The combined gas law states that for a closed system, where the number of
moles of gas are constant: (P1V1/T1) = (P2V2/T2).
(P1V1/T1) = (P2V2/T2) P1 = (P2V2T1/T2V1)
Work the problem and enter the answer in the space provided. Do not include
Substitute and solve:
units!
(1.18 2180 637)/(212 2180) = 0.844 atm
A closed gas system initially has volume and temperature of
9150 mL and 364.0 oC with the pressure unknown. If the same
closed system has values of 1.18 atm, 2180 mL and -61.00 oC,
what was the initial pressure in atm?
Enter your Result:
That is correct!
0.844
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
4 Fe (s) + 3 CO2 (g)
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
4 Fe (s) + 3 CO2 (g)
Looking at the equation, 2 moles of Fe2O3 will
produce 3 moles of CO2. We need to convert
mass into moles (n), adjust for the
stoichiometry, and use PV=nRT to solve V.
IN-CLASS for PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
In this problem, we are producing a gas
under certain a set of conditions. For
this type of problem, we need to use the
combined Gas Law expression for a one
state system; PV=nRT.
4 Fe (s) + 3 CO2 (g)
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
96.6 grams
159.767 grams/mol
4 Fe (s) + 3 CO2 (g)
= 0.605 moles Fe2O3
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
3 moles CO2
2 moles Fe2O3
4 Fe (s) + 3 CO2 (g)
x 0.605 moles Fe2O3
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
3 moles CO2
2 moles Fe2O3
4 Fe (s) + 3 CO2 (g)
x 0.605 moles Fe2O3 = 0.907 mol CO2
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
PV = nRT
V=
nRT
P
4 Fe (s) + 3 CO2 (g)
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
4 Fe (s) + 3 CO2 (g)
PV = nRT
V=
(0.907 mol)(0.0821 L atm mol-1 K-1)(453 K)
1 atm
IN-CLASS PROBLEM!
When Fe2O3 is heated in the presence of carbon,
CO2 gas is produced, according to the equation
shown below. A sample of 96.9 grams of Fe2O3 is
heated in the presence of excess carbon and the
CO2 produced is collected and measured at 1 atm
and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
4 Fe (s) + 3 CO2 (g)
PV = nRT
V=
(0.907 mol)(0.0821 L atm mol-1 K-1)(453 K)
V = 33.7 L
1 atm
The Ideal Gas Law
The ideal gas equation states : PV=nRT. This equation is used in "single state"
gas law problems, usually involving a chemical reaction where a gas is
produced. In this problem set, we will simply be given conditions and we will
solve for an unknown value. When you press "New Problem", a question will
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential notation).
A gas system has volume , moles and temperature of
5.122 L, 0.396 moles and 291.9 K, respectively. What is
the pressure in mm Hg?
Enter your Result:
P = ?(in mm Hg)
V = 5.122 L
The Ideal Gas Law
n = 0.396 moles
The ideal gas equation states291.9nRT. This equation is used in "single state"
T = : PV= K
gas law problems, usually involving a chemical reaction where a gas is
Rearrange the ideal gas equation to solve for the missing we will
produced. In this problem set, we will simply be given conditions and
solve for an unknown value. When you press "New Problem", a question will
quantity:
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential notation).
A gas system has volume , moles and temperature of
5.122 L, 0.396 moles and 291.9 K, respectively. What is
the pressure in mm Hg?
Enter your Result:
P = ?(in mm Hg)
V = 5.122 L
The Ideal Gas Law
n = 0.396 moles
The ideal gas equation states291.9nRT. This equation is used in "single state"
T = : PV= K
gas law problems, usually involving a chemical reaction where a gas is
Rearrange the ideal gas equation to solve for the missing we will
produced. In this problem set, we will simply be given conditions and
solve for an unknown value. When you press "New Problem", a question will
quantity:
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential = nRT P = nRT/V
PV notation).
Substitute and solve:
(0.396 0.0821 291.9)/5.122 = 1.86 atm
A gas system has volume , moles and temperature of
5.122 L, 0.396 moles and 291.9 K, respectively. What is
the pressure in mm Hg?
(1.86 atm) x (760 mm Hg/atm) = 1408 mm Hg
Enter your Result: 1410
That is correct!
The Ideal Gas Law
The ideal gas equation states : PV=nRT. This equation is used in "single state"
gas law problems, usually involving a chemical reaction where a gas is
produced. In this problem set, we will simply be given conditions and we will
solve for an unknown value. When you press "New Problem", a question will
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential notation).
A gas system has pressure , volume and moles of
1.559 atm, 8357 mL and 0.706 moles, respectively.
What is the temperature in K?
Enter your Result:
P = 1.559 atm
V = 8357 mL 8.357L
The Ideal Gas Law
n = 0.706 moles
The ideal gas equation states ?PV=nRT. This equation is used in "single state"
T = : (in K)
gas law problems, usually involving a chemical reaction where a gas is
Rearrange the ideal gas equation to solve for the missing we will
produced. In this problem set, we will simply be given conditions and
solve for an unknown value. When you press "New Problem", a question will
quantity:
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential notation).
A gas system has pressure , volume and moles of
1.559 atm, 8357 mL and 0.706 moles, respectively.
What is the temperature in K?
Enter your Result:
P = 1.559 atm
V = 8357 mL 8.357L
The Ideal Gas Law
n = 0.706 moles
The ideal gas equation states ?PV=nRT. This equation is used in "single state"
T = : (in K)
gas law problems, usually involving a chemical reaction where a gas is
Rearrange the ideal gas equation to solve for the missing we will
produced. In this problem set, we will simply be given conditions and
solve for an unknown value. When you press "New Problem", a question will
quantity:
appear; determine the value of the answer and enter it as numbers only (no
units, no exponential = nRT T = PV/nR
PV notation).
Substitute and solve:
(1.559 8.357)/(0.0821 0.706) = 225 K
A gas system has pressure , volume and moles of
1.559 atm, 8357 mL and 0.706 moles, respectively.
What is the temperature in K?
Enter your Result:
That is correct!
225
IN-CLASS PROBLEM!
A gas system has an initial pressure of 586 mm
Hg with the volume unknown. When the
pressure changes to 5.17 atm the volume is
found to be 1.12 L. What was the initial volume
in mL?
a.
b.
c.
d.
e.
7510 mL
9.88 mL
28.1 mL
0.281 mL
none of the above
IN-CLASS PROBLEM!
A gas system has an initial pressure of 586 mm
Hg with the volume unknown. When the
pressure changes to 5.17 atm the volume is
found to be 1.12 L. What was the initial volume
in mL?
a.
b.
c.
d.
e.
586/760 = 0.771 atm
7510 mL
9.88 mL
28.1 mL
0.281 mL
none of the above
V1
(1,120 mL ) ( 5.17 atm)
=
0.771 atm
IN-CLASS PROBLEM!
A gas system has an initial temperature of 401.0
K with the volume unknown. When the
temperature changes to -202.2 oC the volume is
found to be 2560 mL. What was the initial
volume?
a.
b.
c.
d.
e.
0.0690 L
69.0 L
14.5 L
2,450 mL
none of the above
IN-CLASS PROBLEM!
A gas system has an initial temperature of 401.0
K with the volume unknown. When the
temperature changes to -202.2 oC the volume is
found to be 2560 mL. What was the initial
volume?
a.
b.
c.
d.
e.
0.0690 L
69.0 L
14.5 L
2,450 mL
none of the above
(-202.2) + 273 = 70.8 K
V1 V2
=
T1 T2
V1
( 2.56 L ) ( 401.0 K )
=
70.8 K
IN-CLASS PROBLEM!
A gas system has initial volume of 2.62 L and
contains 0.154 moles. If gas is added so that
the volume changes to 3.30 L, under conditions
of constant P and T, what will the resultant
number of moles be?
a.
b.
c.
d.
e.
0.122 moles
0.194 moles
8.18 moles
5.15 moles
none of the above
IN-CLASS PROBLEM!
A gas system has initial volume of 2.62 L and
contains 0.154 moles. If gas is added so that
the volume changes to 3.30 L, under conditions
of constant P and T, what will the resultant
number of moles be?
a.
b.
c.
d.
e.
0.122 moles
0.194 moles
8.18 moles
5.15 moles
none of the above
V1 V2
=
n1 n2
n2
( 3.30 L ) ( 0.154 moles )
=
2.62 L
THE IDEAL GAS LAWS!
As previously described, the relationship between
volume and pressure (Boyles Law) is,
1
P
the relationship between volume and temperature
(Charless Law) is,
V
VT
and the relationship between volume and the
number (n) of moles (Avogadros Law) is:
Vn
THE IDEAL GAS LAWS!
These three relationships can be combined to give
a single equation of the form:
V
nT
P
The combined proportionality constant for this
relationship is called the universal gas constant,
and is given the symbol R.
V=R
nT
P
PV = nRT
THE IDEAL GAS LAWS!
Experimentally it has been determined that one
mole of any gas at exactly 273.15 K and 1 atm
pressure has a volume of 22.414 L. Knowing this,
the value of R can be calculated:
nT
V=R
P
PV
R=
nT
(1 atm)(22.414 L)
=
(1 mol)(273.1 K)
L atm
= 0.082057
mol K
THE IDEAL GAS LAWS!
Gas law problems generally belong to two main
groups; single state problems and two state
problems. In a single state problem a gas is
described under a single set of conditions of pressure,
volume and temperature. These problems are solved
using the equation:
PV = nRT
In a two state problem a sample of gas undergoes a
change in pressure, volume or temperature. These
problems are solved using the equation:
P1V1
=
n1T1
P2V2
n2T2
IN-CLASS PROBLEM!
A gas system containing 0.534 moles has a
pressure of 1300.0 mm Hg, and a temperature of
82.0 oC. What is the volume of the gas?
a.
b.
c.
d.
e.
9.10 L
19 L
9,100 mL
Both a) and c) are correct.
None of the above are correct.
IN-CLASS PROBLEM!
A gas system containing 0.534 moles has a
pressure of 1300.0 mm Hg, and a temperature of
82.0 oC. What is the volume of the gas?
a.
b.
c.
d.
e.
V1 =
9.10 L
1300/760 = 1.71 atm
PV = nRT
82.0 + 273 = 355 K
19 L
9,100 mL
nRT
Both a) and c) are correct.
V=
P
None of the above are correct.
( 0.534 moles ) ( 0.0821L atm mol-1 K -1 ) ( 355 K )
1.71 atm
IN-CLASS PROBLEM!
A closed gas system initially has pressure and
temperature of 0.372 atm and 85.00 oC with the
volume unknown. If the same closed system has
values of 0.573 atm, 3.28 L and 139.0 oC, what
was the initial volume?
a.
b.
c.
d.
e.
4.39 L
1.19 L
4,400 mL
Both a) and c) are correct.
None of the above are correct.
IN-CLASS PROBLEM!
A closed gas system initially has pressure and
temperature of 0.372 atm and 85.00 oC with the
volume unknown. If the same closed system has
values of 0.573 atm, 3.28 L and 139.0 oC, what
was the initial volume?
P1V1 P2V2
=
a. 4.39 L
T1
T2
b. 1.19 L
c. 4,400 mL
P2V2T1
V1 =
d. Both a) and c) are correct.
T2 P1
e. None of the above are correct.
( 0.573 atm) ( 3.28 L ) ( 358 K )
V1 =
( 0.372 atm) ( 412 K )
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. If we react 1.5 moles of zinc and
collect the hydrogen gas in a 2.0 L vessel at 315 K,
what will be the pressure of the hydrogen gas in the
vessel?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. If we react 1.5 moles of zinc and
collect the hydrogen gas in a 2.0 L vessel at 315 K,
what will be the pressure of the hydrogen gas in the
vessel?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
Looking at the equation, 1.5 moles of Zn will
produce 1.5 moles of H2. We therefore have
n, V, T and we know R.
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. If we react 1.5 moles of zinc and
collect the hydrogen gas in a 2.0 L vessel at 315 K,
what will be the pressure of the hydrogen gas in the
vessel?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
PV = nRT
P=
(1.5 mol)(0.0821 L atm mol-1 K-1)(315 K)
2.0 L
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. If we react 1.5 moles of zinc and
collect the hydrogen gas in a 2.0 L vessel at 315 K,
what will be the pressure of the hydrogen gas in the
vessel?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
PV = nRT
(1.5 mol)(0.0821 L atm mol-1 K-1)(315 K)
P=
2.0 L
P = 19 atm
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. An unknown quantity of zinc in a
sample is observed to produce 7.50 L of hydrogen
gas at a temperature of 404 K and a pressure of 1.75
atm. How many moles of zinc were in the sample?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. An unknown quantity of zinc in a
sample is observed to produce 7.50 L of hydrogen
gas at a temperature of 404 K and a pressure of 1.75
atm. How many moles of zinc were in the sample?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
Looking at the equation, 1 mole of
Zn will produce 1 mole of H2. We
know P, V, T and R.
IN-CLASS PROBLEM!
The reaction of zinc and hydrochloric acid
generates hydrogen gas, according to the equation
shown below. An unknown quantity of zinc in a
sample is observed to produce 7.50 L of hydrogen
gas at a temperature of 404 K and a pressure of 1.75
atm. How many moles of zinc were in the sample?
Zn(s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
PV = nRT
(1.75 atm) (7.50 L)
n=
(0.0821 L atm mol-1 K-1)(404 K)
n = 0.396 mol
IN-CLASS PROBLEM!
The reaction of sodium and water generates
hydrogen gas, according to the equation shown
below. An unknown quantity of sodium metal is
observed to produce 7.50 L of hydrogen gas at a
temperature of 404 K and a pressure of 1.75 atm.
How many moles of sodium were in the sample?
2 Na(s) + 2 H2O
2 NaOH (aq) + H2 (g)
IN-CLASS PROBLEM!
The reaction of sodium and water generates
hydrogen gas, according to the equation shown
below. An unknown quantity of sodium metal is
observed to produce 7.50 L of hydrogen gas at a
temperature of 404 K and a pressure of 1.75 atm.
How many moles of sodium were in the sample?
2 Na(s) + 2 H2O
PV = nRT
2 NaOH (aq) + H2 (g)
(1.75 atm) (7.50 L)
n=
(0.0821 L atm mol-1 K-1)(404 K)
or
0.792 mol Na
IN-CLASS PROBLEM!
An 8.00 L cylinder at 298 K is filled with helium gas
and the pressure is observed to be 1.24 atm. A piston
in the cylinder is now adjusted so that the volume is
now 12.00 L and the pressure is 0.88 atm. What is the
final temperature of the gas?
P1V1
n1T1
(1.24 atm) (8.00 L)
(298 K)
T2 =
=
=
P2V2
n2T2
(0.88 atm) (12.00 L)
(T2)
(0.88 atm) (12.00 L) (298 K)
(1.24 atm) (8.00 L)
IN-CLASS PROBLEM!
An 8.00 L cylinder at 298 K is filled with helium gas
and the pressure is observed to be 1.24 atm. A piston
in the cylinder is now adjusted so that the volume is
now 12.00 L and the pressure is 0.88 atm. What is the
final temperature of the gas?
P1V1
n1T1
(1.24 atm) (8.00 L)
(298 K)
T2 =
=
=
P2V2
n2T2
(0.88 atm) (12.00 L)
(T2)
(0.88 atm) (12.00 L) (298 K)
(1.24 atm) (8.00 L)
IN-CLASS PROBLEM!
An 8.00 L cylinder at 298 K is filled with helium gas
and the pressure is observed to be 1.24 atm. A piston
in the cylinder is now adjusted so that the volume is
now 12.00 L and the pressure is 0.88 atm. What is the
final temperature of the gas?
P1V1
n1T1
(1.24 atm) (8.00 L)
(298 K)
=
=
P2V2
n2T2
(0.88 atm) (12.00 L)
T2 = 320 K
(T2)
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
25o C, according the equation given below, in a
flask with a volume of 2.50 L, what will be the
pressure due to the O2 that has been produced?
What will be the pressure due to the water vapor?
2 H2O2 (l)
2 H2O (g) + O2 (g)
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
25o C, according the equation given below, in a
flask with a volume of 2.50 L, what will be the
pressure due to the O2 that has been produced?
What will be the pressure due to the water vapor?
2 H2O2 (l)
2 H2O (g) + O2 (g)
Looking at the equation, 2 moles of H2O2 will
produce 2 moles of H2O and 1 mole of O2. If
we calculate moles of the products, we can
calculate P, from V, T and R.
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
25o C, according the equation given below, in a
flask with a volume of 2.50 L, what will be the
pressure due to the O2 that has been produced?
What will be the pressure due to the water vapor?
2 H2O2 (l)
2 H2O (g) + O2 (g)
0.032 mol H2O and 0.016 mol O2
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
25o C, according the equation given below, in a
flask with a volume of 2.50 L, what will be the
pressure due to the O2 that has been produced?
What will be the pressure due to the water vapor?
PO
2
Looking atm equation, 2 moles of H2O2 will
= 0.16 at the
produce 2 moles of H2O and 1 mole of O2. If
we calculate moles of the products, we can
calculate P, from V, T and R using PV=nRT.
IN-CLASS PROBLEM!
Avogadros Law tells us that the relationship
If 1.1 grams of hydrogen peroxide decompose at
25o C, according between numbergiven below, in (a ) and its
the equation of moles of a gas n
volume s linear. If we double
flask with a volume of 2i.50 L, what will be the number of
the
moles, but keep the volume constant at 2.50 L,
pressure due to the O2 that has been produced?
the pressure must double!
What will be the pressure due to the water vapor?
To calculate the pressure due to the water vapor we
could repeat the calculation, but recalling
Avogadros Law:
Because there are twice as
many moles of H2O produced,
PH2O must be twice PO2:
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
25o C, according the equation given below, in a
flask with a volume of 2.50 L, what will be the
pressure due to the O2 that has been produced?
What will be the pressure due to the water vapor?
To calculate the pressure due to the water vapor we
could repeat the calculation, but recalling
Avogadros Law:
Because there are twice as
many moles of H2O produced,
PH2O must be twice PO2:
PO = 0.16 atm
2
PH O = 0.32 atm
2
IN-CLASS PROBLEM!
If 1.1 grams of hydrogen peroxide decompose at
What is the total pressure in the flask? It is
25o C, according the equation given below, in a
simply the sum
flask with a volume of the.50 L,artial pressures.the
of 2 two p what will be
This is called Daltons Law of Partial
pressure due to the O2ressures. been produced?
P that has
What will be the pressure due to the water vapor?
2 H2O2 (l)
PO = 0.16 atm
2
2 H2O (g) + O2 (g)
PH O = 0.32 atm
2
Ptotal = PO2 + PH2O = 0.48 atm
DALTONS LAW: PARTIAL PRESSURES!
Daltons Law simply states that the total pressure
exerted by a gas mixture is equal to the sum of the
pressures of each individual gas.
Ptotal = P1 + P2 + P3 +
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
CH4 (g) + 2 O2 (g)
2 H2O (g) + CO2 (g)
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
CH4 (g) + 2 O2 (g)
2 H2O (g) + CO2 (g)
Looking at the equation, 1 mole of CH4 will
produce 2 moles of H2O and 1 mole of CO2. If
we calculate moles of the products, we can
calculate P, from V, T and R.
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
CH4 (g) + 2 O2 (g)
2 H2O (g) + CO2 (g)
0.020 mol H2O and 0.010 mol CO2
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
PCO
2
= 0.31 atm
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
Avogadros Law tells us that the relationship
reacts completely with oxygen and the gaseousits
between number of moles of a gas (n) and
products arevolume is linear. Ifa 1.0 L flask 373 K. of
collected at in we double the number
m the but keep the ue to the CO2 at 1.0 L, the
What will be oles,pressure dvolume constantand water
pressure must
vapor that has been produced? double!
To calculate the pressure due to the water vapor we
could repeat the calculation, but recalling
Avogadros Law:
Because there are twice as
many moles of H2O produced,
PH 2O must be twice PCO2:
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
To calculate the pressure due to the water vapor we
could repeat the calculation, but recalling
Avogadros Law:
Because there are twice as
many moles of H2O produced,
PH 2O must be twice PCO2:
PCO = 0.31 atm
2
PH
2O
= 0.62 atm
IN-CLASS PROBLEM!
A sample consisting of 0.010 moles of methane
reacts completely with oxygen and the gaseous
products are collected at in a 1.0 L flask 373 K.
What will be the pressure due to the CO2 and water
vapor that has been produced?
CH4 (g) + 2 O2 (g)
PCO = 0.31 atm
2
2 H2O (g) + CO2 (g)
PH
2O
= 0.62 atm
Ptotal = PCO2 + PH 2O = 0.93 atm
What is the total pressure in the flask? It is
simply the sum of the two partial pressures.
GAS LAWS AND DENSITY!
The mass of any gas can be simply calculated from
the moles (n) using the relationship:
Mass of the gas
Molar mass of
the gas
Solving PV = nRT for the number of moles:
or
If we multiply both sides by the
molar mass, M, and divide by the
volume, V, we get:
GAS LAWS AND DENSITY!
The mass of any gas can be simply calculated from
the moles (n) using the relationship:
Mass of the gas
Molar mass of
the gas
Solving PV = nRT for the number of moles:
or
If we multiply both sides by the
molar mass, M, and divide by the
volume, V, we get:
GAS LAWS AND DENSITY!
The mass of any gas can be simply calculated from
the moles (n) using the relationship:
Mass of the gas
Molar mass of
the gas
or
P
d=
M
RT
Remembering that mass
divided by volume is density
IN-CLASS PROBLEM!
Carbon tetrafluoride (CF4) is a gas at 1.00 atm and
50.0o C. Calculate the density of CF4 under these
conditions.
Calculate the molar mass (M) of CF4 and
insert it into the equation, d=(P/RT)M. We
now have M, P, T and we know R.
1.00 atm
d=
(88.00g mol-1)
(0.0821 L atm mol-1 K-1)(323.15 K)
IN-CLASS PROBLEM!
Carbon tetrafluoride (CF4) is a gas at 1.00 atm and
50.0o C. Calculate the density of CF4 under these
conditions.
1.00 atm
d=
(88.00g mol-1)
(0.0821 L atm mol-1 K-1)(323.15 K)
d = 3.32 g L-1
IDEAL GAS LAWS AND THE KMT!
The behavior of gasses that is predicted by the
Ideal Gas Laws are based on the Kinetic Molecular
Theory. The basic assumptions are:
IDEAL GAS LAWS AND THE KMT!
Because we assume that the molecules in a gas are
in motion, they must all have kinetic energy, such
that:
KE = (1/2)(mass)(speed)2
!
Thus, at the same temperature, all gasses must
have the same average kinetic energy.
Following from this, as the
temperature increases, KE
increases and so does the
average speed of gas
molecules.
IDEAL GAS LAWS AND THE KMT!
The actual average speed of gas molecules can be
predicted using Maxwells equation, that predicts
that the root mean square speed is directly
proportional to temperature and inversely
proportional to mass.
root mean squared speed =
3RT
M
Thus, average speed increases with temperature
(T) and decreases with mass (M).
IDEAL GAS LAWS AND THE KMT!
The actual average speed of gas molecules can be
predicted using Maxwells equation, that predicts
that the root mean square speed is directly
proportional to temperature and inversely
proportional to mass.
Oxygen at 25o C
Number of Molecules
Oxygen at 1000o C
0
200
400
600
800
1000
1200
Molecular Speed; m/sec
1400
1600
1800
IDEAL GAS LAWS AND THE KMT!
The actual average speed of gas molecules can be
predicted using Maxwells equation, that predicts
that the root mean square speed is directly
proportional to temperature and inversely
proportional to mass.
Number of Molecules
Oxygen at 25o C; 32 g/mol
Water at 25o C; 18 g/mol
Helium at 25o C; 4 g/mol
0
400
800
1200
1600
2000
2400
Molecular Speed; m/sec
2800
3200
3600
GAS DIFFUSION & EFFUSION!
Gas diffusion is defined as the gradual mixing of
molecules of different gasses. Gas effusion is the
movement of molecules, from one chamber into a
vacuum, through a small opening.
GAS DIFFUSION & EFFUSION!
Gas diffusion is defined as the gradual mixing of
molecules of different gasses. Gas effusion is the
movement of molecules, from one chamber into a
vacuum, through a small opening.
GAS DIFFUSION & EFFUSION!
Gas molecules diffuse and effuse at a rate that is
proportional to time, and inversely proportional to
the mass of the gas.
rate for gas A
mass of gas B
=
rate for gas B
mass of gas A
Grahams
Law.
GAS DIFFUSION & EFFUSION!
Gas molecules diffuse and effuse at a rate that is
proportional to time, and inversely proportional to
the mass of the gas.
In this demonstration, HCl
and NH3 diffuse from
opposite ends of the tube to
react and form NH4Cl(s).
HCl is heavier, thus NH4Cl
forms closer to the HCl end.
Cotton soaked in HCl.
Cotton soaked in NH3.
AVOGADROS LAW AND THE KMT!
Recall that pressure is proportional to n when V
and T are constant.
nRT
P =
V
CHARLESS LAW AND THE KMT!
Recall that pressure is proportional to T when V
and n are constant.
nRT
P =
V
BOYLESS LAW AND THE KMT!
Recall that pressure is inversely proportional to P
when T and n are constant.
nRT
P =
V
IDEAL GAS LAWS AND THE KMT!
The behavior of gasses that is predicted by the
Ideal Gas Laws are based on the Kinetic Molecular
Theory. The basic assumptions are:
DEVIATIONS FROM THE IDEAL GAS LAWS!
Because real gas molecules have volumes and
intermolecular forces do exist, even in gasses, the
ideal gas laws are only an approximation of the true
behavior of gasses.
n 2a
P + 2 (V nb) = nRT
V
This term accounts for
intermolecular forces.
This term accounts for
molecular volume.
You can account for the volume of gas molecules and
for intermolecular forces using the van der Waals
equation.
DEVIATIONS FROM THE IDEAL GAS LAWS!
Because real gas molecules have volumes and
intermolecular forces do exist, even in gasses, the
ideal gas laws are only an approximation of the true
behavior of gasses.
n 2a
P + 2 (V nb) = nRT
V
As an example, for Cl2 gas, a = 6.49, b = 0.0562
For 8.0 moles of Cl2 in a 4.0 L tank at 27 oC:
P(ideal) = nRT/V = 49.3 atm
P(van der Waals) = 29.5 atm