Physical Chemistry , 1st ed.

therefore contains heats and temperatures from related parts of the universe

under consideration. Note that equation 3.11 includes allof the heats of the

Carnot cycle. The fact that these heats, divided by the absolute temperatures of

the two reservoirs involved, add up to exactly zero is interesting. Recall that the

cycle starts and stops at the same system conditions. But changes in state func-

tions are dictated solely by the conditions of the system, not by the path that

got the system to those conditions. If a system starts and stops at the same con-

ditions, overall changes in state functions are exactly zero. Equation 3.11 sug-

gests that for reversible changes, a relationship between heat and absolute tem-

perature is a state function.

3.4 Entropy and the Second Law of Thermodynamics

We define entropy, S, as an additional thermodynamic state function. The in-

finitesimal change in entropy,dS, is defined as

dS

dq

T

rev (3.12)

where “rev” on the infinitesimal for heat,dq, specifies that it must be the heat

for a reversible process. The temperature,T, must be in kelvins. Integrating

equation 3.12, we get

S

dq

T

rev (3.13)

where Sis now the change in entropy for a process. As indicated in the pre-

vious section, for the Carnot cycle (or any other closed cycle) Smust be zero.

For an isothermal, reversible process,the temperature can be taken out of the

integral and the integral can be evaluated easily:

S

T

1

dqrev

q

T

rev (3.14)

Equation 3.14 demonstrates that entropy has units of J/K. These may seem like

unusual units, but they are the correct ones. Also, keep in mind that the

amount of heat for a process depends on the amount of material, in grams or

moles, and so sometimes the unit for entropy becomes J/molK. Example 3.2

shows how to include amount in the unit.

Example 3.2

What is the change in entropy when 1.00 g of benzene, C 6 H 6 , boils reversibly

at its boiling point of 80.1°C and a constant pressure of 1.00 atm? The heat

of vaporization of benzene is 395 J/g.

Solution

Since the process occurs at constant pressure,vapHfor the process equals

the heat,q, for the process. Since vaporization is an endothermic (that is,

energy-in) process, the value for the heat is positive. Finally, 80.1°C equals

353.2 K. Using equation 3.14:

S

3



5

3

3

9

.2

5

K

J

1.12 

K

J



72 CHAPTER 3 The Second and Third Laws of Thermodynamics