84
2 Work, Heat, and Energy: The First Law of Thermodynamics
Exercise 2.27
a. Find ∆U for the process of the previous example.
b. Find ∆H, q, and w for the process in which the system of the previous example is first
heated at constant volume from 298.15 K to 373.15 K and then expanded isothermally from
a volume of 5.000 L to a volume of 10.000 L.
Enthalpy Changes for Irreversible Processes
Consider an irreversible process that begins with an equilibrium or metastable state
and ends with an equilibrium state. To calculate ∆H for such an irreversible process,
we find a reversible process with the same initial and final states, calculate ∆H for that
process, and assign that value to the irreversible process, using the fact that H is a state
function.
E X A M P L E 2.28
Find ∆H and q if 2.000 mol of supercooled liquid water at −15.00 ◦ C freezes irreversibly at
a constant pressure of 1.000 atm and a temperature of −15.00 ◦ C. Assume that C P,m of liquid
water is constant and equal to 75.48 J K −1 mol −1 , and that C P,m of ice is constant and equal
to 37.15 J K −1 mol −1 .
Solution
We assume that the supercooled water is metastable and treat it as though it were at equilibrium. We calculate ∆H along the reversible path shown in Figure 2.9. Step 1 is the reversible
heating of the supercooled liquid to 0.00 ◦ C, the equilibrium freezing temperature. Step 2 is
the reversible freezing of the system at 0.00 ◦ C, and step 3 is the reversible cooling of the
solid to −15.00 ◦ C. The enthalpy change for the irreversible process is equal to the enthalpy
change of this reversible process.
∆H 1
273.15 K
258.15 K
C P (l)dT C P (l)∆T
(2.000 mol)
75.48 J K −1 mol −1
(15.00 K) 2264 J
The enthalpy change of step 2 was calculated in an earlier example:
∆H 2 −12020 J
The enthalpy change of step 3 is similar to that of step 1.
∆H 3
258.15 K
273.15 K
C P (s)dT C P (s)∆T
(2.000 mol)
37.15 J K −1 mol −1
(15.00 K) 1114 J
2 Work, Heat, and Energy: The First Law of Thermodynamics
Exercise 2.27
a. Find ∆U for the process of the previous example.
b. Find ∆H, q, and w for the process in which the system of the previous example is first
heated at constant volume from 298.15 K to 373.15 K and then expanded isothermally from
a volume of 5.000 L to a volume of 10.000 L.
Enthalpy Changes for Irreversible Processes
Consider an irreversible process that begins with an equilibrium or metastable state
and ends with an equilibrium state. To calculate ∆H for such an irreversible process,
we find a reversible process with the same initial and final states, calculate ∆H for that
process, and assign that value to the irreversible process, using the fact that H is a state
function.
E X A M P L E 2.28
Find ∆H and q if 2.000 mol of supercooled liquid water at −15.00 ◦ C freezes irreversibly at
a constant pressure of 1.000 atm and a temperature of −15.00 ◦ C. Assume that C P,m of liquid
water is constant and equal to 75.48 J K −1 mol −1 , and that C P,m of ice is constant and equal
to 37.15 J K −1 mol −1 .
Solution
We assume that the supercooled water is metastable and treat it as though it were at equilibrium. We calculate ∆H along the reversible path shown in Figure 2.9. Step 1 is the reversible
heating of the supercooled liquid to 0.00 ◦ C, the equilibrium freezing temperature. Step 2 is
the reversible freezing of the system at 0.00 ◦ C, and step 3 is the reversible cooling of the
solid to −15.00 ◦ C. The enthalpy change for the irreversible process is equal to the enthalpy
change of this reversible process.
∆H 1
273.15 K
258.15 K
C P (l)dT C P (l)∆T
(2.000 mol)
75.48 J K −1 mol −1
(15.00 K) 2264 J
The enthalpy change of step 2 was calculated in an earlier example:
∆H 2 −12020 J
The enthalpy change of step 3 is similar to that of step 1.
∆H 3
258.15 K
273.15 K
C P (s)dT C P (s)∆T
(2.000 mol)
37.15 J K −1 mol −1
(15.00 K) 1114 J