Physical Chemistry Third Edition

66 2 Work, Heat, and Energy: The First Law of Thermodynamics

400

300

200

100

0

T/K

V/L

01015205

Step 2: Constant-volume heating

Step 1: Isothermal expansion

(2 L, 298.15 K) (20 L, 298.15 K)

(20 L, 373.15 K)

Figure 2.6 The Curve Representing the Path for the∆ULine Integral.

A change in internal energy for a nonisothermal process can be calculated by carrying

out a line integral ofdU.

EXAMPLE2.16

Calculate∆Ufor a process that takes 1.000 mol of argon fromT 298 .15 K andV

2 .000 L toT 373 .15 K andV 20 .000 L. Does the result depend on whether the process

is reversible?

Solution

SinceUis a state function, we can choose any path with the same initial state and final

state to calculate∆U. We choose to integrate along the reversible path shown in Figure 2.6,

consisting of an isothermal expansion and a constant-volume change in temperature. The first

step of the path is that of the previous example, so∆U 1 , the change in energy for that part,

is equal to 67 J mol−^1. For the second step

∆U 2 

∫T 2

T 1

CVdTCV(T 2 −T 1 )nCV, m(T 2 −T 1 )

(1.000 mol)

(

3

2

)

R(T 2 −T 1 )

(

12 .472JK−^1

)

( 75 .0K)935 J

∆U∆U 1 +∆U 2 67 J+935 J1002 J

BecauseUis a state function, the result does not depend on whether the actual process is

reversible, so long as the initial and final states are metastable or equilibrium states.

Exercise 2.12

a.Calculateqandwfor the reversible process that follows the path used in the solution of the

previous example.