Physical Chemistry , 1st ed.

(Darren Dugan) #1
Fortunately, most expressions for heat capacity are simple power series in T,
whose integrals are easy to evaluate on a term-by-term basis.
There is no Vor psubscript on the symbol for the heat capacity in equa-
tion 3.18. That’s because it depends on the conditions of the process. If it oc-
curs under conditions of constant volume, use CV. If it occurs under condi-
tions of constant pressure, use Cp. Usually the particular process involved
dictates the choice.
Now consider gas-phase processes. What if the temperature were constant
but the pressure or the volume changed? If the gas is ideal,Ufor the process
is exactly zero, so dqdwp dV. Substituting again for dq, then:

dS

d
T

q


p
T

dV


SdSp
T

dV

At this point, we can substitute for either por dVusing the ideal gas law. If we
substitute for pin terms ofV(that is,pnRT/V):

SnR
V

T

T

dVnR
V

dVnRd
V

V

SnRln 

V

V

f
i

 (3.19)

Similarly, for a change in pressure one gets:

SnRln 

p
p

f
i

 (3.20)

Because entropy is a state function, the change in entropy is dictated by the
conditions of the system, not how the system arrived at those conditions.
Therefore, any process can usually be broken down into smaller steps, the en-
tropy of each step can be evaluated using the growing number of expressions
for S, and the Sfor the overall process is the combination of all of the S’s
of the individual steps.

Example 3.3
Determine the overall change in entropy for the following process using
1.00 mole of He:
He (298.0 K, 1.50 atm) →He (100.0 K, 15.0 atm)
The heat capacity of He is 20.78 J/molK. Assume the helium acts ideally.

Solution
The overall reaction can be divided into two parts:
Step 1: He (298.0 K, 1.50 atm) →He (298.0 K, 15.0 atm)
(change in pressure step)
Step 2: He (298.0 K, 15.0 atm) →He (100.0 K, 15.0 atm)
(change in temperature step)
The change in entropy for step 1, the isothermal step, can be determined from
equation 3.20:

76 CHAPTER 3 The Second and Third Laws of Thermodynamics

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