3.3 The Calculation of Entropy Changes
123
Entropy Changes for Processes That Begin and End
at the Same Temperature
If a process is not isothermal but has a final temperature that is equal to its initial
temperature, we can calculate ∆S for the process by integrating dq rev /T on a reversible
isothermal path. The actual process does not have to be reversible or isothermal, but the
initial and final states must be equilibrium or metastable states at the same temperature.
E X A M P L E 3.4
Calculate the entropy change for the following process: A sample containing 2.000 mol of
helium gas originally at 298.15 K and 1.000 bar is cooled to its normal boiling temperature of
4.00 K, condensed to a liquid, and then cooled further to 2.00 K, where it undergoes another
phase transition to a second liquid form, called liquid helium II. This liquid phase is suddenly
vaporized by a beam of laser light, and the helium is brought to a temperature of 298.15 K
and a pressure of 0.500 bar.
Solution
Since entropy is a state function the entropy change is the same as for an isothermal reversible
expansion from 1.000 bar to 0.500 bar:
∆S nR ln
V 2
V 1
nR ln
P 1
P 2
where we have used Boyle’s law, PV constant at constant temperature.
∆S (2.000 mol)(8.3145 J K −1 mol −1 ) ln
1.000 bar
0.500 bar
11.5 J K −1
Entropy Changes for Reversible Phase Changes
Two phases of a single substance can be at equilibrium with each other at a fixed
temperature that depends on the pressure. For example, liquid and gaseous water can
be at equilibrium with each other at 100.00 ◦ C if the pressure is 1.000 atm (760.0 torr),
and can be at equilibrium with each other at 25.00 ◦ C if the pressure is 23.756 torr. If an
equilibrium phase change is carried out at constant pressure and temperature Eq. (3.3-2)
applies. Since the pressure is constant, q is equal to ∆H, and
∆S
q rev
T
∆H
T
(reversible phase change
at constant pressure)
(3.3-4)
E X A M P L E 3.5
Find the entropy change of the system and of the surroundings if 3.000 mol of water freezes
reversibly at 1.000 atm. The freezing temperature is 0.00 ◦ C at this pressure, and the specific
enthalpy change of fusion is equal to 79.7 cal g −1 at this temperature.
123
Entropy Changes for Processes That Begin and End
at the Same Temperature
If a process is not isothermal but has a final temperature that is equal to its initial
temperature, we can calculate ∆S for the process by integrating dq rev /T on a reversible
isothermal path. The actual process does not have to be reversible or isothermal, but the
initial and final states must be equilibrium or metastable states at the same temperature.
E X A M P L E 3.4
Calculate the entropy change for the following process: A sample containing 2.000 mol of
helium gas originally at 298.15 K and 1.000 bar is cooled to its normal boiling temperature of
4.00 K, condensed to a liquid, and then cooled further to 2.00 K, where it undergoes another
phase transition to a second liquid form, called liquid helium II. This liquid phase is suddenly
vaporized by a beam of laser light, and the helium is brought to a temperature of 298.15 K
and a pressure of 0.500 bar.
Solution
Since entropy is a state function the entropy change is the same as for an isothermal reversible
expansion from 1.000 bar to 0.500 bar:
∆S nR ln
V 2
V 1
nR ln
P 1
P 2
where we have used Boyle’s law, PV constant at constant temperature.
∆S (2.000 mol)(8.3145 J K −1 mol −1 ) ln
1.000 bar
0.500 bar
11.5 J K −1
Entropy Changes for Reversible Phase Changes
Two phases of a single substance can be at equilibrium with each other at a fixed
temperature that depends on the pressure. For example, liquid and gaseous water can
be at equilibrium with each other at 100.00 ◦ C if the pressure is 1.000 atm (760.0 torr),
and can be at equilibrium with each other at 25.00 ◦ C if the pressure is 23.756 torr. If an
equilibrium phase change is carried out at constant pressure and temperature Eq. (3.3-2)
applies. Since the pressure is constant, q is equal to ∆H, and
∆S
q rev
T
∆H
T
(reversible phase change
at constant pressure)
(3.3-4)
E X A M P L E 3.5
Find the entropy change of the system and of the surroundings if 3.000 mol of water freezes
reversibly at 1.000 atm. The freezing temperature is 0.00 ◦ C at this pressure, and the specific
enthalpy change of fusion is equal to 79.7 cal g −1 at this temperature.