Page 80 - Physical chemistry eng

3.5 HOW ARE C P AND C V RELATED? 57

EXAMPLE PROBLEM 3.7

A 143.0 g sample of C(s) in the form of graphite is heated from 300. to 600. K at a
constant pressure. Over this temperature range, C P, m has been determined to be
C P,m T T 2 T 3
-7
-1 -1 =-12.19 + 0.1126 - 1.947 * 10 -4 2 + 1.919 * 10 3
J K mol K K K
T 4
-11
-7.800 * 10 4
K
Calculate ¢H and q . How large is the relative error in ¢H if we neglect the
P
temperature-dependent terms in C P, m and assume that C P, m maintains its value at 300. K
throughout the temperature interval?

Solution
T f
m
¢H = C P,m (T)dT
M
3
T i
2
T - 4 T
600. -12.19 + 0.1126 - 1.947 * 10 + 1.919
143.0 g J K 2 T
= § 3 4 K ¥d
12.00 g mol -1 mol -7 T -11 T K
3
300. * 10 3 - 7.800 * 10 4
K K
T T 2 -5 T 3 600.
-12.19 + 0.0563 - 6.49 * 10 + 4.798
143.0 K K 2 K 3
= * D T J = 46.9 kJ
12.00 -8 T 4 -11 T 5
* 10 4 - 1.56 * 10 5
K K 300.
From Equation (3.28), ¢H = q P .
-1
-1
If we had assumed C P, m = 8.617 J mol K , which is the calculated value at
-1
300. K, ¢H = 143.0 g>12.00 g mol -1 * 8.617 JK mol -1 * [600. K - 300. K] =
30.8 kJ . The relative error is 100 * (30.8 kJ - 46.9 kJ)>46.9 kJ =-34.3% . In this
case, it is not reasonable to assume that C P, m is independent of temperature.

3.5 How Are C and C Related?
V
P
To this point, two separate heat capacities, C and C , have been defined. How are
V
P
these quantities related? To answer this question, the differential form of the first law is
written as
0U
dq = C dT + a b dV + P external dV (3.35)
V
0V T
Consider a process that proceeds at constant pressure for which P = P external . In this
case, Equation (3.35) becomes

0U
dq = C dT + a b dV + PdV (3.36)
V
P
0V T
Because dq = C dT ,
P
P
0U 0V 0V 0U 0V
C = C + a b a b + Pa b = C + B a b + PR a b
V
V
P
0V T 0T P 0T P 0V T 0T P
0P 0V
= C + Ta b a b (3.37)
V
0T V 0T P