Page 114 - Physical chemistry eng

5.4 CALCULATING CHANGES IN ENTROPY 91

This equation can be regarded as the mathematical statement of the second law.
What conclusions can be drawn from Equation (5.11)? Because the cyclic integral of
dq reversible >T is zero, this quantity must be the exact differential of a state function.
This state function is called the entropy, and given the symbol S

dq reversible
dS K (5.12)
T
For a macroscopic change,
dq reversible
¢S = (5.13)
T
L
Note that whereas dq reversible is not an exact differential, multiplying this quantity by
1>T makes the differential exact.

EXAMPLE PROBLEM 5.2

a. Show that the following differential expression is not an exact differential:
RT
dP + RdT
P
b. Show that RTdP + PRdT , obtained by multiplying the function in part (a) by P,
is an exact differential.
Solution

a. For the expression f(P,T)dP + g(P,T)dT to be an exact differential, the condi-
tion (0f(P,T)>0T) = (0g(P,T)>0P) T must be satisfied as discussed in
P
Section 3.1. Because
RT
0a b
P R 0R
P Q = and = 0
0T P 0P
P
the condition is not fulfilled.
b. Because (0(RT)>0T) = R and (0(RP)>0P) T = R , RTdP + RPdT is an
P
exact differential.

Keep in mind that it has only been shown that S is a state function. It has not
yet been demonstrated that S is a suitable function for measuring the natural
direction of change in a process that the system may undergo. We will do so in
Section 5.5.

5.4 Calculating Changes in Entropy

The most important thing to remember in doing entropy calculations using Equation (5.13)
is that ¢S must be calculated along a reversible path. In considering an irreversible
process, ¢S must be calculated for a reversible process that proceeds between the same
initial and final states corresponding to the irreversible process. Because S is a state func-
tion, ¢S is necessarily path independent, provided that the transformation is between the
same initial and final states in both processes.
We first consider two cases that require no calculation. For any reversible adiabatic
= 0, so that ¢S =
process, q reversible (dq reversible >T) = 0 . For any cyclic process,
1
¢S = (dq reversible >T) = 0 , because the change in any state function for a cyclic
A
process is zero.