Physical Chemistry Third Edition

124 3 The Second and Third Laws of Thermodynamics: Entropy

Solution

∆S

(3.000 mol)(18.02 g mol−^1 )(79.7 cal g−^1 )(4.184 J cal−^1 )

273 .15 K

− 66 .0JK−^1

Since the process is reversible,∆Suniv0 and

∆Ssurr−∆S 66 .0JK−^1

Exercise 3.6

Calculate∆H,q, and∆Sfor the reversible vaporization of 50.0 g of ethanol at 1.000 atm. The

molar enthalpy change of vaporization is equal to 40.48kJmol−^1 and the boiling temperature

at 1.000 atm is 78.5◦C.

Entropy Changes for Reversible Temperature Changes

Another simple class of processes consists of temperature changes in closed systems

without phase change or chemical reaction. If the pressure is constant, the relations of

Eqs. (2.5-6) and (2.5-8) give

dqdHCPdT (closed system, constant pressure) (3.3-5)

so that Eq. (3.3-2) becomes

∆S

∫T 2

T 1

CP

T

dT (closed system, constant pressure process) (3.3-6)

whereT 1 is the initial temperature andT 2 is the final temperature.

EXAMPLE 3.6

Calculate∆Sand∆Ssurrfor reversibly heating 2.000 mol of liquid water from 0.00◦Cto

100.00◦C at a constant pressure of 1.00 atm.

Solution

The specific heat capacity of liquid water is nearly constant and equal to 1.00 cal K−^1 g−^1 or

4 .18JK−^1 g−^1.

∆S

∫ 373 .15 K

273 .15 K

CP

T

dTCP

∫ 373 .15 K

273 .15 K

1

T

dTCPln

(

373 .15 K

273 .15 K

)

(2.000 mol−^1 )(18.02 g mol−^1 )(4.18JK−^1 g−^1 )ln

(

373 .15 K

273 .15 K

)

 47 .0JK−^1

∆Ssurr−∆S− 47 .0JK−^1