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Problem Set 2 Solutions - Physical Chemistry I | CHE 360.00, Assignments of Physical Chemistry

Material Type: Assignment; Professor: Standard; Class: Physical Chemistry I; Subject: Chemistry ; University: Illinois State University; Term: Unknown 1989;

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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 tempe rature 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 CO2 present (0.96) times the total pressure,
P
CO2 = 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 CO2 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–1 K–1
( )
730 K
( )
= 62.64 g/L
( )
1L
1000 cm3
d = 0.06264 g/cm3.
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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

  • K -

( 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

  • aV mbPV m

2 − abRTV m

2 = 0.

Finally, we can divide by P and collect terms to give

V

m

3 −

RT

P

  • b

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

  • b =

0.08205L atm mol

− 1 K

− 1

(^660 K)

91 atm

  • 0.08445L mol

− 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

  • 0.095209 V m

− 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

  • K -

( 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

  • K -

( 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

  • b ,

or

V =

nRT

P

  • nb.

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).