Page 61 - Physical chemistry eng

38 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

Segment 2 : 3
As previously noted, w = 0 , and
¢U 2:3 = q 2:3 = C (T - T )
2
3
V
-1
= 2.50mol * 20.79Jmol -1 K (79.9K - 2000K)
=-99.6 kJ
The numerical result is equal in magnitude, but opposite in sign to ¢U 1:2 because
T = T 1 . For the same reason, ¢H 2:3 =-¢H 1:2 .
3
Segment 3 : 1
For this segment, ¢U 3:1 = 0 and ¢H 3:1 = 0 as noted earlier, and w 3:1 =-q 3:1 .
Because this is a reversible isothermal compression,
V 1
w 3:1 =-nRTln =-2.50mol * 8.314Jmol -1 K -1 * 79.9K
V 3
1.00 * 10 -3 m 3
* ln -3 3
25.0 * 10 m
= 5.35 kJ
The results for the individual segments and for the cycle in the indicated direction are
given in the following table. If the cycle is traversed in the reverse fashion, the magni-
tudes of all quantities in the table remain the same, but all signs change.

Path q (kJ) w (kJ) ≤U (kJ) ≤H (kJ)
1 : 2 139.4 – 39.8 99.6 139.4
2 : 3 – 99.6 0 – 99.6 – 139.4
3 : 1 – 5.35 5.35 0 0
Cycle 34.5 – 34.5 0 0

EXAMPLE PROBLEM 2.6
-1 -1
In this example, 2.50 mol of an ideal gas with C V,m = 12.47 J mol K is expanded
adiabatically against a constant external pressure of 1.00 bar. The initial temperature
and pressure of the gas are 325 K and 2.50 bar, respectively. The final pressure is
1.25 bar. Calculate the final temperature, q, w, ¢U , and ¢H .
Solution
Because the process is adiabatic, q = 0 , and ¢U = w . Therefore,
¢U = nC v,m (T - T ) =-P external (V - V )
f
i
f
i
Using the ideal gas law,
T f T i
nC v,m (T - T ) =-nRP external ¢ - ≤
f
i
P f P i
nRP external nRP external
T ¢nC v,m + ≤ = T ¢nC v,m + ≤
i
f
P f P i
RP external
+
C v,m
P i
T = T § RP ¥
i
f
C v,m + external
P f
-1 -1
8.314 J mol K * 1.00 bar
-1 -1
12.47 J mol K +
= 325 K * § 2.50 bar ¥ = 268 K
-1
8.314 J mol K -1 * 1.00 bar
-1
-1
12.47 J mol K +
1.25 bar