Physical Chemistry Third Edition

2.6 Calculation of Enthalpy Changes of Processes without Chemical Reactions 83

only onnandT,∆Hfor the first step vanishes. For the second step

∆H 2 

∫T 2

T 1

CPdTnCP, m∆T

( 1 .000 mol)

(

5

2

)(

8 .3145 J K−^1 mol−^1

)

( 75 .00 K)1559 J

∆H∆H 1 +∆H 2  0 +1559 J1559 J

Although∆His the same for any process with the same initial and final states as

the overall process in Example 2.26,qandware dependent on the path of a particular

process. If the pressure is not constant during the entire process,qis not necessarily

equal to∆H.

EXAMPLE2.27

Findqandwfor the process used in the calculation of Example 2.26.

Solution

We first findV 2 andP 2 , the volume and pressure at the end of step 2. From the ideal gas law,

V 2 (10.00 L)

298 .15 K

373 .15 K

 7 .990 L

P 2 

( 1 .000 mol)

(

8 .3145 J K−^1 mol−^1

)

( 298 .15 K)

0 .007990 m^3

 3. 103 × 105 Pa

q 1 ( 1 .000 mol)

(

8 .3145 J K−^1 mol−^1

)

( 298 .15 K)ln

(

7 .990 L

5 .000 L

)

1162 J

Because the pressure was constant during step 2,

q 2 ∆H 2 1559 J

qq 1 +q 2 1162 J+1559 J2721 J

This differs considerably from the value of∆H, 1559 J.

We can calculatewfor the process. For the first step

w 1 ∆U 1 −q 1 −q 1 −1162 J

For the second step, we letV 3 be the final volume

w 2 −

∫V 3

V 2

PdV−P

∫V 3

V 2

dV−P∆V

−

(

3. 103 × 105 Pa

)(

0 .01000 m^3 − 0 .007990 m^3

)

− 623 .6J

ww 1 +w 2 −1162 J−624 J−1786 J