2.4 Calculation of Amounts of Heat and Energy Changes
67
b. Calculate q and w for the reversible process corresponding to the following path. Step 1: The
system is heated from 298.15 K to 373.15 K at a constant volume of 2.000 L; step 2: It is then
expanded isothermally to a volume of 20.000 L.
c. Comment on the difference between the q and w values for parts a and b. What is the value
of ∆U for the process of part b?
Reversible Adiabatic Processes
An adiabatic process is one in which no heat is transferred to or from a closed system,
so that dq is equal to zero for every infinitesimal step of the process:
dU dq + dw dw (adiabatic process)
(2.4-16)
Consider a reversible adiabatic process of an ideal gas.
dU C V dT (closed ideal gas, reversible process)
(2.4-17)
dw −PdV −
nRT
V
dV (closed ideal gas, reversible process)
(2.4-18)
Because dq 0 for an adiabatic process, we equate dU and dw:
C V dT −
nRT
V
dV (closed ideal gas, reversible adiabatic process)
(2.4-19)
This is a differential equation that can be solved to give T as a function of V if the
dependence of C V on T and V is known.
We first assume that C V is constant. We can solve Eq. (2.4-19) by separation of
variables. We divide by T to separate the variables (remove any V dependence from
the left-hand side and any T dependence from the right-hand side):
C V
T
dT −
nR
V
dV
(2.4-20)
Because each integrand contains only one variable we can integrate Eq. (2.4-20) from
the initial state, denoted by V 1 and T 1 , to the final state, denoted by V 2 and T 2 . A definite
integration gives
C V ln
T 2
T 1
−nR ln
V 2
V 1
We divide by C V and take the exponential (antilogarithm) of both sides of this
equation:
T 2
T 1
V 1
V 2
nR/C V
V 1
V 2
R/C V,m (reversible adiabatic process,
ideal gas, C V constant)
(2.4-21a)
67
b. Calculate q and w for the reversible process corresponding to the following path. Step 1: The
system is heated from 298.15 K to 373.15 K at a constant volume of 2.000 L; step 2: It is then
expanded isothermally to a volume of 20.000 L.
c. Comment on the difference between the q and w values for parts a and b. What is the value
of ∆U for the process of part b?
Reversible Adiabatic Processes
An adiabatic process is one in which no heat is transferred to or from a closed system,
so that dq is equal to zero for every infinitesimal step of the process:
dU dq + dw dw (adiabatic process)
(2.4-16)
Consider a reversible adiabatic process of an ideal gas.
dU C V dT (closed ideal gas, reversible process)
(2.4-17)
dw −PdV −
nRT
V
dV (closed ideal gas, reversible process)
(2.4-18)
Because dq 0 for an adiabatic process, we equate dU and dw:
C V dT −
nRT
V
dV (closed ideal gas, reversible adiabatic process)
(2.4-19)
This is a differential equation that can be solved to give T as a function of V if the
dependence of C V on T and V is known.
We first assume that C V is constant. We can solve Eq. (2.4-19) by separation of
variables. We divide by T to separate the variables (remove any V dependence from
the left-hand side and any T dependence from the right-hand side):
C V
T
dT −
nR
V
dV
(2.4-20)
Because each integrand contains only one variable we can integrate Eq. (2.4-20) from
the initial state, denoted by V 1 and T 1 , to the final state, denoted by V 2 and T 2 . A definite
integration gives
C V ln
T 2
T 1
−nR ln
V 2
V 1
We divide by C V and take the exponential (antilogarithm) of both sides of this
equation:
T 2
T 1
V 1
V 2
nR/C V
V 1
V 2
R/C V,m (reversible adiabatic process,
ideal gas, C V constant)
(2.4-21a)