3.3 The Calculation of Entropy Changes
127
Exercise 3.10
Find ∆S, ∆S surr , and ∆S univ if 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.
E X A M P L E 3.10
Calculate ∆S, ∆S surr , and ∆S univ for 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, ∆S is the same as in Example 3.6:
∆S 47.0 J K −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.
∆S surr
q surr
T surr
−
q
T surr
−
(2.000 mol)(18.01 g mol −1 )(4.184 J K −1 g −1 )(100.00 K)
374.15 K
−40.28 J K −1
∆S univ ∆S + ∆S surr 47.0 J K −1 − 40.28 J K −1 6.73 J K −1
Exercise 3.11
Find ∆S and ∆S surr for the irreversible heating of 2.000 mol of gaseous neon from 0.00 ◦ C to
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 considering reversible processes with the same initial and final states.
E X A M P L E 3.11
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
C P (l)
T
dT C P (l) ln
273.15 K
258.15 K
(2.000 mol)(75.48 J K mol −1 ) ln(1.0581) 8.526 J K −1
127
Exercise 3.10
Find ∆S, ∆S surr , and ∆S univ if 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.
E X A M P L E 3.10
Calculate ∆S, ∆S surr , and ∆S univ for 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, ∆S is the same as in Example 3.6:
∆S 47.0 J K −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.
∆S surr
q surr
T surr
−
q
T surr
−
(2.000 mol)(18.01 g mol −1 )(4.184 J K −1 g −1 )(100.00 K)
374.15 K
−40.28 J K −1
∆S univ ∆S + ∆S surr 47.0 J K −1 − 40.28 J K −1 6.73 J K −1
Exercise 3.11
Find ∆S and ∆S surr for the irreversible heating of 2.000 mol of gaseous neon from 0.00 ◦ C to
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 considering reversible processes with the same initial and final states.
E X A M P L E 3.11
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
C P (l)
T
dT C P (l) ln
273.15 K
258.15 K
(2.000 mol)(75.48 J K mol −1 ) ln(1.0581) 8.526 J K −1