Page 83 - Physical Chemistry

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Chapter 2 Since H is a state function, we can use the integral in (2.79) to find H for
The First Law of Thermodynamics any process whose initial and final states have the same pressure, whether or not
the entire process occurs at constant pressure.
3. Constant-volume heating with no phase change. Since V is constant, w 0.
Integration of C dq /dT and use of U q w q give
V
V
V
2
¢U C dT q V const. (2.80)
V V
1
As with (2.79), Eq. (2.80) holds whether or not the heating is reversible. H is
found from H U (PV) U V P.
4. Perfect-gas change of state. Since U and H of a perfect gas depend on T only, we
integrate dU C dT and dH C dT [(2.68) and (2.70)] to give
P
V
T 2 T 2
¢U C 1T2 dT, ¢H C 1T2 dT perf. gas (2.81)
P
V
T 1 T 1
If C (T) or C (T) is known, we can use C C nR and integrate to find U
V
P
P
V
and H. The equations of (2.81) apply to any perfect-gas change of state includ-
ing irreversible changes and changes in which P and V change. The values of
2
q and w depend on the path. If the process is reversible, then w PdV
1
2
nR (T/V) dV, and we can find w if we know how T varies as a function of V.
1
Having found w, we use U q w to find q.
5. Reversible isothermal process in a perfect gas. Since U and H of the perfect gas
2
are functions of T only, we have U 0 and H 0. Also, w PdV
1
nRT ln (V /V ) [Eq. (2.74)] and q w, since q w U 0.
1
2
6. Reversible adiabatic process in a perfect gas. The process is adiabatic, so q
0. We find U and H from Eq. (2.81). The first law gives w U. If C is es-
V
g
sentially constant, the final state of the gas can be found from P V P V g 2
1
2
1
[Eq. (2.77)], where g C /C .
P
V
7. Adiabatic expansion of a perfect gas into vacuum. Here (Sec. 2.7) q 0, w
0, U q w 0, and H U (PV) U nR T 0.
Equations (2.79) and (2.80) tell us how a temperature change at constant P or at
constant V affects H and U. At this point, we are not yet able to find the effects of a
change in P or V on H and U. This will be dealt with in Chapter 4.
A word about units. Heat-capacity and latent-heat data are sometimes tabulated in
calories, so q is sometimes calculated in calories. Pressures are often given in atmos-
3
pheres, so P-V work is often calculated in cm atm. The SI unit for q, w, U, and H
3
is the joule. Hence we frequently want to convert between joules, calories, and cm atm.
We do this by using the values of R in (1.19) to (1.21). See Example 2.2 in Sec. 2.2.
A useful strategy to find a quantity such as U or q for a process is to write the
expression for the corresponding infinitesimal quantity and then integrate this expres-
sion from the initial state to the final state. For example, to find U in an ideal-gas
2
change of state, we write dU C dT and U C (T) dT; to find q in a constant-
1
V
V
2
pressure process, we write dq C dT and q C dT. The infinitesimal change
P
P
1
P
P
in a state function under the condition of constant P or T or V can often be found from
the appropriate partial derivative. For example, if we want dU in a constant-volume
process, we use ( U/ T) C to write dU C dT for V constant, and U
V
V
V
2
C dT, where the integration is at constant V.
V
1
When evaluating an integral from state 1 to 2, you can take quantities that are con-
stant outside the integral, but anything that varies during the process must remain inside
2
2
the integral. Thus, for a constant-pressure process, PdV P dV P(V V ),
2
1
1
1
2
2
and for an isothermal process, (nRT/V) dV nRT (1/V) dV nRT ln (V /V ).
1
1
1
2