Page 82 - Physical chemistry eng

3.6 THE VARIATION OF ENTHALPY WITH PRESSURE AT CONSTANT TEMPERATURE 59

For isothermal processes, dT = 0 , and Equation (3.42) can be rearranged to
0H 0U 0V
a b = ca b + P da b + V (3.43)
0P T 0V T 0P T
Using Equation (3.19) for (0U>0V) T ,

0H 0P 0V
a b = Ta b a b + V
0P T 0T V 0P T
0V
= V - Ta b = V(1 - Tb) (3.44)
0T P
The second formulation of Equation (3.44) is obtained through application of the cyclic
rule [Equation (3.7)]. This equation is applicable to all systems containing pure sub-
stances or mixtures at a fixed composition, provided that no phase changes or chemical
reactions take place. The quantity (0H>0P) T is evaluated for an ideal gas in Example
Problem 3.8.

EXAMPLE PROBLEM 3.8
Evaluate (0H>0P) T for an ideal gas.

Solution
(0P>0T) = (0[nRT>V]>0T) = nR>V and (0V>0P) = (d[nRT>P]>dP) =
V
T
V
T
-nRT>P 2 for an ideal gas. Therefore,
0H 0P 0V nR nRT nRT nRT
a b = Ta b a b + V = T a - b + V =- + V = 0
0P T 0T V 0P T V P 2 P nRT
This result could have been derived directly from the definition H = U + PV . For an
ideal gas, U = U(T) only and PV = nRT . Therefore, H = H(T) for an ideal gas
and (0H>0P) = 0 .
T

Because Example Problem 3.8 shows that H is only a function of T for an ideal gas,
T f T f
¢H = C (T)dT = n C P,m (T)dT (3.45)
P
3 3
T i T i
for an ideal gas. Because H is only a function of T, Equation (3.45) holds for an ideal
gas even if P is not constant. This result is also understandable in terms of the potential
function of Figure 1.10. Because ideal gas molecules do not attract or repel one
another, no energy is required to change their average distance of separation (increase
or decrease P).
Equation (3.44) is next applied to several types of systems. We have seen that
(0H>0P) = 0 for an ideal gas. For liquids and solids, 1 W Tb for T 6 1000 K as
T
can be seen from the data in Table 3.1. Therefore, for liquids and solids,
(0H>0P) L V to a good approximation, and dH can be written as
T
dH L C dT + V dP (3.46)
P
for systems that consist only of liquids or solids.

EXAMPLE PROBLEM 3.9
Calculate the change in enthalpy when 124 g of liquid methanol initially at 1.00 bar
and 298 K undergoes a change of state to 2.50 bar and 425 K. The density of liquid
–3
methanol under these conditions is 0.791 g cm , and C P, m for liquid methanol is
–1
–1
81.1 J K mol .