Page 64 - Physical Chemistry

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Section 2.2
3
gas constant R 8.314 J/(mol K) and R 82.06 cm atm/(mol K) [Eqs. (1.19) P-V Work
and (1.20)]:
8.314 J mol 1 K 1
3
w rev 1500 cm atm 152 J
3
1
82.06 cm atm mol K 1
An alternative procedure is to note that no work is done during the constant-
volume part of process (a); all the work is done during the second step of the
process, in which P is held constant at P . Therefore
2
2 V 2 V 2 V 2
w rev P dV P dV P 2 dV P V `
2
2

1 V 1 V 1 V 1
3
P 1V V 2 11.00 atm211500 cm 2 152 J
2
2
1
3
Similarly, we find for process (b) that w 4500 cm atm 456 J (see
the exercise in this example).
Processes (a) and (b) are expansions. Hence the system does positive work
on its surroundings, and the work w done on the system is negative in these
processes.
For the reverse of process (a), all the work is done during the first step, during
3
3
which P is constant at 1.00 atm and V starts at 2000 cm and ends at 500 cm . Hence
w 500 cm 3 3 (1.00 atm) dV (1.00 atm)(500 cm 2000 cm ) 152 J.
3
3
2000 cm
Exercise
Find w rev for process (b) of Fig. 2.3 using the P , V , P , V values given for
2
2
1
1
3
process (a). (Answer: 4500 cm atm 456 J.)
Irreversible P-V Work
The work w in a mechanically irreversible volume change sometimes cannot be cal-
culated with thermodynamics.
For example, suppose the external pressure on the piston in Fig. 2.2 is suddenly reduced
by a finite amount and is held fixed thereafter. The inner pressure on the piston is then
greater than the outer pressure by a finite amount, and the piston is accelerated outward.
This initial acceleration of the piston away from the system will destroy the uniform pres-
sure in the enclosed gas. The system’s pressure will be lower near the piston than farther
away from it. Moreover, the piston’s acceleration produces turbulence in the gas. Thus we
cannot give a thermodynamic description of the state of the system.
We have dw F dx. For P-V work, F is the force at the system–surroundings bound-
x
x
ary, which is where the displacement dx is occurring. This boundary is the inner face of
the piston, so dw irrev P surf dV, where P surf is the pressure the system exerts on the inner
face of the piston. (By Newton’s third law, P surf is also the pressure the piston’s inner face
exerts on the system.) Because we cannot use thermodynamics to calculate P surf during the
turbulent, irreversible expansion, we cannot find dw irrev from thermodynamics.
The law of conservation of energy can be used to show that, for a frictionless piston
(Prob. 2.22),
(2.29)
dw irrev P ext dV dK pist
where P is the external pressure on the outer face of the piston and dK pist is the infinitesimal
ext
2
change in piston kinetic energy. The integrated form of (2.29) is w irrev P dV K .
1
ext
pist