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Chemistry 360
Dr. Jean M. Standard
Problem Set 2 Solutions
1. The atmospheric surface pressure on Venus is 90 bar. The atmosphere at the surface is
composed of 96% carbon dioxide and 4% other gases. Given a surface temperature of 730
K, what is the mass of carbon dioxide present per cubic centimeter at the surface?
Assume ideal gas behavior.
The carbon dioxide partial pressure is related to the proportion of CO 2 present (0.96) times the total pressure,
P
CO 2
= ( 0.96) ( 90 bar) = 86.4 bar.
The mass per cubic centimeter is the density d ,
d =
m
V
where m is the mass and V is the volume. The mass m can be expressed in terms of the number of moles n and
the molecular weight M ,
m = nM.
The number of moles n can be determined from the ideal gas equation. Solving the ideal gas equation for n
yields
n =
PV
RT
This expression for the number of moles n can be substituted into the equation for the mass m ,
m = nM ,
or m =
PVM
RT
Finally, this equation for the mass m can be inserted into the relation for the density d ,
d =
m
V
or d =
PM
RT
Now, for the Venus atmosphere, the temperature is 730 K, the pressure of carbon dioxide is 86.4 bar, the
molecular weight of CO 2 is 44.0 g/mol, or 0.0440 kg/mol. Substituting into the expression for the density,
the result is
d =
PM
RT
(86.4 bar) (44.0 g/mol)
0.08314 L bar mol
( 730 K)
= (62.64 g/L)
1 L
1000 cm
3
d = 0.06264 g/cm
3 .
2. A bicycle tire was inflated to 9.3 bar total pressure in the cool of the morning when the
temperature was 50 ° F. Later, the bicycle was used for a race on hot roads in the
afternoon, when the temperature of the tires rose to 120 ° F. Assuming that the volume of
the tire stays the same and that the gases behave ideally, determine the pressure in the hot
tire.
The bicycle tire can be treated as a closed system, so the number of moles of air in the tire stays fixed. Since
the volume also stays constant, we can solve for the ratio of V/n ,
V
n
RT
1
P
1
where
T
1
and
P
1
are the temperature and pressure of the gas in the tire in the morning. However, since V/n
remains fixed, we also have
V
n
RT
2
P
2
where
T
2
and
P
2
are the temperature and pressure of the gas in the tire in the afternoon. Equating these two
results, we have
RT
1
P
1
RT
2
P
2
The pressure
P
2 is^ unknown^ and^ is^ what^ we^ want.^ Solving^ for
P
2 leads to
P
2
P
1
T
2
T
1
(9.3 bar) (322.04 K)
( 283.15K)
P
2 = 10.5bar.
4. Continued
Then, multiply the terms on the left out and move everything to the left side to yield
PV
m
3
2 − ab − RTV m
2 = 0.
Finally, we can divide by P and collect terms to give
V
m
3 −
RT
P
Vm
2
a
P
V
m
ab
P
Substituting numerical values and using the van der Waals coefficients from Engel & Reid Table 7.4, we have
RT
P
0.08205L atm mol
− 1 K
− 1
(^660 K)
91 atm
− 1
= 0.67954 L mol
− 1 ,
a
P
8.664 L
2 atm mol
− 2
91 atm
= 0.095209 L
2 mol
− 2 ,
and
ab
P
8.664 L
2 atm mol
− 2
0.08445L mol
− 1
91 atm
= 8.04038 × 10
− 3 L
3 mol
− 3 .
These quantities yield the cubic equation
V
m
3 − 0.67954 V m
2
− 8.04038 × 10
− 3 = 0.
This equation can be solved numerically, using a calculator for example, to obtain
V
m = 0.528 L/mol.
5. Use the virial equation to determine the pressure in atm of 1 mole of carbon dioxide gas
contained in a volume of 5.0 L at 300 K. Compare your result to the pressure that would
have been obtained from the ideal gas equation.
The virial equation is
Z =
PV
m
RT
B
V
m
C
V
m
2
+ K
Using this equation but only including terms up to the second virial coefficient B (since that is all that your
textbook includes), we can solve for the pressure,
P =
RT
V
m
B
V
m
5. Continued
Substituting numerical values of T = 300 K and
V
m = 5.0 L/mol, and using the second virial coefficient for
carbon dioxide from Engel & Reid Table 7.1 ( B = –126 cm
3 /mol = –0.126 L/mol), we have
P =
RT
V
m
B
V
m
0.08205 L atm mol
( 300 K)
5.0 L mol
0.126 L mol
5.0 L mol
P = 4.80 atm.
Comparing this with the pressure from the ideal gas equation, we have
P =
RT
V
m
0.08205 L atm mol
( 300 K)
5.0 L mol
P = 4.92 atm.
So we see that the pressure of a real gas is somewhat lower than that of an ideal gas, even at moderate
temperatures and pressures. This effect is due to the attractive interactions between real gas molecules that lower
the pressure exerted by the molecules on the walls of the container.
6. By definition, the compressibility factor (or the compression factor) Z for an ideal gas
equals 1. By approximately what percentage does this change for diatomic hydrogen upon
inclusion of the second virial coefficient term? Recalculate this result for water vapor.
Give the conditions under which you made this estimate.
The virial equation is
Z =
PV
m
RT
B
V
m
C
V
m
2
+ K
Including only the second virial coefficient B , the expression becomes
Z =
PV
m
RT
B
V
m
For diatomic hydrogen, the second virial coefficient B is 15 cm
3 /mol. The molar volume
V
m can be estimated
using the ideal gas equation,
V
m
RT
P
7. Derive expressions for the cubic expansion coefficient α and the isothermal
compressibility κ for a real gas that obeys an equation of state given by
P V
m ( −^ b ) =^ RT^.
Compare your results to the ideal gas case. See Engel & Reid Section 3.1 or Problem Set
1 for the definitions of the cubic expansion coefficient and isothermal compressibility.
(Note that Engel & Reid refer to the cubic expansion coefficient as the thermal expansion
coefficient, β , but this is the same thing as α defined in Problem Set 1.)
The cubic expansion coefficient α is defined as
α =
V
∂ V
∂ T
P
and the isothermal compressibility κ is defined as
κ = −
V
∂ V
∂ P
T
To evaluate the partial derivatives, we have to solve the equation of state for V ,
V
m
RT
P
or
V =
nRT
P
Evaluating the cubic expansion coefficient,
α =
V
∂ V
∂ T
P
V
nR
P
P
nRT + nbP
nR
P
α =
R
RT + bP
This can be put in a useful form by pulling out a factor of 1/T and dividing numerator and denominator by R ,
α =
T
1 + bP / RT
This shows that the cubic expansion coefficient for this equation of state can be written as the ideal gas value
( 1/T ) times a correction factor (in parentheses).
7. Continued
Evaluating the isothermal compressibility,
κ = −
V
∂ V
∂ P
T
V
nRT
P
2
P
nRT + nbP
nRT
P
2
κ =
RT
P (^) ( RT + bP )
This can also be put in a useful form by dividing numerator and denominator by RT ,
κ =
P
1 + bP / RT
This shows that the isothermal compressibility for this equation of state can be written as the ideal gas value
( 1/P ) times a correction factor (in parentheses).
