Physical Chemistry , 1st ed.

Example 3.1

a.Determine the efficiency of a Carnot engine that takes in 855 J of heat, per-

forms 225 J of work, and gives off the remaining energy as heat.

b.Draw a diagram like Figure 3.1 showing the exact amounts of heat and

work going from place to place in the proper direction.

Solution

a.Using both definitions of efficiency, and recognizing the proper signs on

the heat and work:

e





2

8

2

5

5

5

J

J

0.263

e1 + (855

85



5J

225) J1 + (0.737) 0.263

b.The drawing is left to the student.

There is another way to define efficiency in terms of the temperatures of the

high- and low-temperature reservoirs. For the isothermal steps 1 and 3,

the change in the internal energy is zero because (U/V)T0. Therefore,

qwfor steps 1 and 3. From equation 2.7, for an ideal gas,

wnRTln 

V

V

f

i



For a reversible, isothermal process, the heats for steps 1 and 3 are

q 1 w 1 nRThighln 

V

V

A

B (3.7)

q 3 w 3 nRTlowln 

V

V

D

C

 (3.8)

The volume labels A, B, C, and D represent the initial and final points for each

step, as shown in Figure 3.2.Thighand Tloware the temperatures of the high-

temperature and low-temperature reservoirs, respectively. For the adiabatic

steps 2 and 4, we can use equation 2.47 to get

V

V

C

B



2/3



T

T

h

lo

ig

w

h



V

V

D

A



2/3



T

T

h

lo

ig

w

h



Equating the two volume expressions, which both equal Tlow/Thigh:

V

V

D

A



2/3



V

V

C

B



2/3

Raising both sides to the power of 3/2 and rearranging, we get



V

V

A

B



V

V

D

C



Substituting for VD/VCin equation 3.8, we get an expression for q 3 in terms of

volumes VAand VB:

q 3 nRTlowln 

V

V

A

B

nRTlowln 

V

V

A

B (3.9)

70 CHAPTER 3 The Second and Third Laws of Thermodynamics