Physical Chemistry Third Edition

3.3 The Calculation of Entropy Changes 127

Exercise 3.10

Find∆S,∆Ssurr, and∆Sunivif 3.000 mol of argon (assumed ideal) expand isothermally into a

vacuum at 298.15 K, expanding from a volume of 15.00 L to a volume of 40.00 L.

EXAMPLE3.10

Calculate∆S,∆Ssurr, and∆Sunivfor the irreversible heating of 2.000 mol of liquid water

from 0.00◦C to 100.00◦C at a constant pressure of 1.000 atm. Assume that the surroundings

remain at equilibrium at 101◦C as the system warms up.

Solution

Since entropy is a state function,∆Sis the same as in Example 3.6:

∆S 47 .0JK−^1

Since the process was carried out at constant pressure,q∆H, and is the same as for the

reversible process in Example 3.6.

∆Ssurr

qsurr

Tsurr

−

q

Tsurr

−

(2.000 mol)(18.01 g mol−^1 )(4.184JK−^1 g−^1 )(100.00 K)

374 .15 K

− 40 .28JK−^1

∆Suniv∆S+∆Ssurr 47 .0JK−^1 − 40 .28JK−^1  6 .73JK−^1

Exercise 3.11

Find∆Sand∆Ssurrfor the irreversible heating of 2.000 mol of gaseous neon from 0.00◦Cto

100.00◦C at constant pressure, assuming that the surroundings remain at a constant temperature

of 100.00◦C.

Metastable supercooled or superheated systems can undergo irreversible phase

changes at constant pressure, and their entropy changes can be calculated by con-

sidering reversible processes with the same initial and final states.

EXAMPLE3.11

Calculate the entropy change of the system, the surroundings, and the universe for the process

of Example 2.28. Assume that the surroundings remain at equilibrium at− 15. 00 ◦C.

Solution

We use the same reversible path as in Example 2.28:

∆S 1 

∫ 273 .15K

258 .15K

CP(l)

T

dTCP(l) ln

(

273 .15 K

258 .15 K

)

(2.000 mol)(75.48 J K mol−^1 )ln(1.0581) 8 .526JK−^1