Page 102 - Physical Chemistry

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we are considering a reversible engine, the gas also must remain at temperature T H Section 3.2
throughout the heat absorption from the reservoir. (Heat flow between two bodies Heat Engines
with a finite difference in temperature is an irreversible process.) Thus the first step
of the cycle is an isothermal process. Moreover, since U 0 for an isothermal
process in a perfect gas [Eq. (2.67)], it follows that, to maintain U as constant, the gas
must expand and do work on the surroundings equal to the heat absorbed in the first
step. The first step of the cycle is thus a reversible isothermal expansion, as shown by
the line from state 1 to state 2 in Fig 3.4a. Similarly, when the gas gives up heat at
T , we have a reversible isothermal compression at temperature T . The T isotherm
C
C
C
lies below the T isotherm and is the line from state 3 to state 4 in Fig. 3.4a. To have
H
a complete cycle, we must have steps that connect states 2 and 3 and states 4 and 1.
We assumed that heat is transferred only at T and T . Therefore the two isotherms
H
C
in Fig. 3.4a must be connected by two steps with no heat transfer, that is, by two
reversible adiabats.
This reversible cycle is called a Carnot cycle (Fig. 3.4b). The working substance
need not be a perfect gas. A Carnot cycle is defined as a reversible cycle that consists
of two isothermal steps at different temperatures and two adiabatic steps.
We now calculate the Carnot-cycle efficiency e rev on the ideal-gas temperature
scale T. We use a perfect gas as the working substance and restrict ourselves to P-V
work. The first law gives dU dq dw dq P dV for a reversible volume change.
For a perfect gas, P nRT/V and dU C (T) dT. The first law becomes
V
C dT dq nRT dV>V
V
for a perfect gas. Dividing by T and integrating over the Carnot cycle, we get

C 1T2 dT dq nR dV (3.9)
V
T T V
Each integral in (3.9) is the sum of four line integrals, one for each step of the Carnot
cycle in Fig. 3.4b. We have

C 1T2 dT T 2 C 1T2 dT T 3 C 1T2 dT T 4 C 1T2 dT T 1 C 1T2 dT Figure 3.4
V
V
V
V
V
T T T T T
T 1 T 2 T 3 T 4
(a) Isothermal steps of the
(3.10) reversible heat-engine cycle.
(b) The complete Carnot cycle.
Each integral on the right side of (3.10) has an integrand that is a function of T only, (Not to scale.)
and hence each such integral is an ordinary definite integral. Use of the identity
c
c
b
f(T ) dT f(T ) dT f(T ) dT (Sec. 1.8) shows that the sum of the first two in-
a
b
a
tegrals on the right side of (3.10) is 1C /T2 dT and the sum of the last two integrals
T 3
V
T 1
on the right side of (3.10) is 1C V /T2 dT. Hence the right side of (3.10) equals
T 1

1C /T2 dT 1C /T2 dT T 3 1C /T2 dT 0. Therefore (3.10) becomes
T 1
T 3
T 1
V
V
T 1 T 3 V T 1
dT
C 1T2
V 0 (3.11)
T
The cyclic integral in (3.11) must vanish because [C (T)/T] dT is the differential of a
V
state function, namely, a certain function of T whose derivative is C (T)/T. (Recall
V
Sec. 2.10.) Note, however, that the integral of PdV does not vanish for a cycle, since
PdV is not the differential of a state function.
The second integral on the right side of (3.9) must also vanish. This is because
dV/V is the differential of a state function (namely, ln V), and its line integral is there-
fore zero for a cyclic process.