Page 117 - Physical chemistry eng

94 CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics

EXAMPLE PROBLEM 5.4
One mole of CO gas is transformed from an initial state characterized by T = 320. K
i
and V = 80.0 L to a final state characterized by T = 650. K and V = 120.0 L. Using
f
i
f
k
Equation (5.22), calculate ¢S for this process. Use the ideal gas values for and .
b
For CO,
C V, m T T 2 T 3
-8
-5
= 31.08 - 0.01452 + 3.1415 * 10 - 1.4973 * 10
-1
Jmol K -1 K K 2 K 3
Solution
For an ideal gas,
1 0V 1 03nRT>P4 1
b = a b = a b = and
V 0T P V 0T P T
1 0V 1 03nRT>P4 1
k =- a b =- a b =
V 0P T V 0P T P
Consider the following reversible process in order to calculate ¢S . The gas is first
heated reversibly from 320. to 650. K at a constant volume of 80.0 L. Subsequently, the
gas is reversibly expanded at a constant temperature of 650. K from a volume of 80.0 L
to a volume of 120.0 L. The entropy change for this process is obtained using the inte-
k
b
gral form of Equation (5.22) with the values of and cited earlier. The result is
T f
C V V f
¢S = dT + nR ln
T V i
3
T i
¢S =
T T 2 T 3
-8
-5
650. (31.08 - 0.01452 + 3.1415 * 10 - 1.4973 * 10 )
K K 2 K 3 T
1 mol * d
T K
3
320.
K
120.0 L
-1
-1
+ 1 mol * 8.314 J K mol * ln
80.0 L
= 22.025 J K -1 - 4.792 J K -1 + 5.028 J K -1
- 1.207 J K -1 + 3.371 JK -1
= 24.4 J K -1

EXAMPLE PROBLEM 5.5
In this problem, 2.50 mol of CO gas is transformed from an initial state characterized
2
by T = 450. K and P = 1.35 bar to a final state characterized by T = 800. K and P =
i
f
i
f
3.45 bar. Using Equation (5.23), calculate ¢S for this process. Assume ideal gas
behavior and use the ideal gas value for . For CO ,
b
2
C P,m T T 2 T 3
-8
-2
-5
= 18.86 + 7.937 * 10 - 6.7834 * 10 + 2.4426 * 10
-1
J mol K -1 K K 2 K 3
Solution
Consider the following reversible process in order to calculate ¢S . The gas is first
heated reversibly from 450. to 800. K at a constant pressure of 1.35 bar. Subsequently,
the gas is reversibly compressed at a constant temperature of 800. K from a pressure of