Physical Chemistry Third Edition

126 3 The Second and Third Laws of Thermodynamics: Entropy

Exercise 3.8

a.Calculate∆Sif 2.500 mol of neon gas is heated reversibly from 80.0◦C to 250.0◦Cata

constant volume of 100.0 L.

b.Calculate∆Ssurr.

Entropy Changes for Irreversible Processes

If a system undergoes any irreversible process with equilibrium initial and final states,

we can calculate the entropy change of the system using a reversible process that

has the same initial and final states as the irreversible process. A separate calculation

is necessary for the entropy change of the surroundings, since the final state of the

surroundings will not necessarily be the same for the reversible process as for the

irreversible process.

EXAMPLE 3.9

Calculate∆S,∆Ssurr, and∆Sunivif 2.000 mol of argon (assume ideal) expands isothermally

and irreversibly at 298.15 K from a volume of 10.00 L to a volume of 40.00 L at a constant

external pressure of 1.000 atm. Assume thatP(transferred)Pextand that the surroundings

remain at equilibrium at 298.15 K.

Solution

The initial and final states of the system are the same as for a reversible isothermal expansion,

so that

∆S

qrev

T

nRln

(

V 2

V 1

)

(2.000 mol)(8.3145 J K−^1 mol−^1 )ln

(

40 .00 L

10 .00 L

)

 23 .05JK−^1

In the∆Scalculation we have usedqrev, not the actual value ofq, since we must integrate

on a reversible path. In the calculation of∆Ssurrwe must use the actual value of the heat

transferred to the surroundings. We assume thatPextis equal toP(transferred). Since∆U 0

for an isothermal process in an ideal gas,

qsurr−q−∆U+ww−Pext∆V−

(

101325 N m−^2

)(

0 .030 m^3

)

−3040 J

∆Ssurr−

3040 J

298 .15 K

− 10 .20JK−^1

∆Suniv 23 .05JK−^1 − 10 .20JK−^1  12 .85JK−^1

Exercise 3.9

Calculate∆Ssurrif 2.500 mol of neon gas is heated irreversibly from 80.0◦C to 250.0◦Cata

constant volume of 100.0 L with the surroundings remaining at equilibrium at 265.0◦C.