Physical Chemistry , 1st ed.

(Darren Dugan) #1
TB

b

a
R





All of the liter units cancel, as well as the mole units. The atmosphere units
also cancel, leaving the unit of K (kelvins) in a denominator of the denomi-
nator, which makes it in the numerator. The final answer therefore has units
of K, which is what is expected for a temperature. Numerically, we evaluate
the fraction and find that
TB18.0 K
Experimentally, it is 25 K.
b.A similar procedure for methane, using a2.253 atmL^2 /mol^2 and b 
0.0428 L/mol, yields

641 K

The experimental value is 509 K.

The fact that the predicted Boyle temperatures are a bit off from the experi-
mental values should not be cause for alarm. Some approximations were made in
trying to find a correspondence between the virial equation of state and the van
der Waals equation of state. However, equation 1.23 does a good job of estimat-
ing the temperature at which a gas will act more like an ideal gas than at others.
We can also use these new equations of state, like the van der Waals equa-
tion of state, to derive how certain state variables vary as others are changed.
For example, recall that we used the ideal gas law to determine that


T

p
V,n

n
V

R

Suppose we use the van der Waals equation of state to determine how pressure
varies with respect to temperature, assuming volume and amount are constant.
First, we need to rewrite the van der Waals equation so that pressure is all by
itself on one side of the equation:

p


a
V

n
2

2
(Vnb)nRT

p

a
V

n
2

2

V

n


RT

nb




p
V

n


RT

nb




a
V

n
2

2


Next, we take the derivative of this expression with respect to temperature.
Note that the second term on the right does not have temperature as a vari-
able, so the derivative of it with respect to Tis zero. We get

2.253 a
m

tm
o



l

L

2

2


0.0428 
m

L

ol

0.08205 
m

L

o

a
l

t


m
K




0.03508a
m

tm
o



l

L

2

2


0.0237
m

L

ol

0.08205
m

L

o

a
l

t


m
K




16 CHAPTER 1 Gases and the Zeroth Law of Thermodynamics

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