126
3 The Second and Third Laws of Thermodynamics: Entropy
Exercise 3.8
a. Calculate ∆S if 2.500 mol of neon gas is heated reversibly from 80.0 ◦ C to 250.0 ◦ C at a
constant volume of 100.0 L.
b. Calculate ∆S surr .
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.
E X A M P L E 3.9
Calculate ∆S, ∆S surr , and ∆S univ if 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 that P(transferred) P ext and 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
q rev
T
nR ln
V 2
V 1
(2.000 mol)(8.3145 J K −1 mol −1 ) ln
40.00 L
10.00 L
23.05 J K −1
In the ∆S calculation we have used q rev , not the actual value of q, since we must integrate
on a reversible path. In the calculation of ∆S surr we must use the actual value of the heat
transferred to the surroundings. We assume that P ext is equal to P(transferred). Since ∆U 0
for an isothermal process in an ideal gas,
q surr −q −∆U + w w −P ext ∆V −
101325 N m −2
0.030 m 3
−3040 J
∆S surr −
3040 J
298.15 K
−10.20 J K −1
∆S univ 23.05 J K −1 − 10.20 J K −1 12.85 J K −1
Exercise 3.9
Calculate ∆S surr if 2.500 mol of neon gas is heated irreversibly from 80.0 ◦ C to 250.0 ◦ C at a
constant volume of 100.0 L with the surroundings remaining at equilibrium at 265.0 ◦ C.
3 The Second and Third Laws of Thermodynamics: Entropy
Exercise 3.8
a. Calculate ∆S if 2.500 mol of neon gas is heated reversibly from 80.0 ◦ C to 250.0 ◦ C at a
constant volume of 100.0 L.
b. Calculate ∆S surr .
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.
E X A M P L E 3.9
Calculate ∆S, ∆S surr , and ∆S univ if 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 that P(transferred) P ext and 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
q rev
T
nR ln
V 2
V 1
(2.000 mol)(8.3145 J K −1 mol −1 ) ln
40.00 L
10.00 L
23.05 J K −1
In the ∆S calculation we have used q rev , not the actual value of q, since we must integrate
on a reversible path. In the calculation of ∆S surr we must use the actual value of the heat
transferred to the surroundings. We assume that P ext is equal to P(transferred). Since ∆U 0
for an isothermal process in an ideal gas,
q surr −q −∆U + w w −P ext ∆V −
101325 N m −2
0.030 m 3
−3040 J
∆S surr −
3040 J
298.15 K
−10.20 J K −1
∆S univ 23.05 J K −1 − 10.20 J K −1 12.85 J K −1
Exercise 3.9
Calculate ∆S surr if 2.500 mol of neon gas is heated irreversibly from 80.0 ◦ C to 250.0 ◦ C at a
constant volume of 100.0 L with the surroundings remaining at equilibrium at 265.0 ◦ C.