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August 18, 2010 11:36 9in x 6in b985-ch03 Elementary Physical Chemistry
The Second Law of Thermodynamics 25
Similarly, for the adiabatic transition from the lower (T 1,V ) to the higher
1
isotherm (T 2 ,V 1 )weobtain
ln(T 2 /T 1 )+(γ − 1) ln(V 1 /V ) = 0 (3.12b)
1
which is equivalent to
ln(T 1 /T 2 )+(γ − 1) ln(V 1 /V ) = 0 (3.12c)
1
Obviously, comparing (3.12a) with (3.12c) shows that V /V 2 = V /V 1
1
2
or V /V = V 2 /V 1 . Substituting this result in Eq. (3.10b), shows that
2
1
q 1 = −RT 1 ln V 2 /V 1 (3.12d)
Finally, substituting this expression in Eq. (3.8) and using (3.10a) yields
ξ =1 − T 1 /T 2 (3.13)
Note that since ξ = −w/q 2 =(q 1 + q 2)/q 2 =1 + q 1 /q 2 it follws
that q 1 /q 2 = −T 1 /T 2 and thus, q 1 /T 1 + q 2 /T 2 = 0. Replacing this by the
differential form dq rev/T (emphasizing the reversible nature of the process)
yields, upon integration along a closed contour, dq rev /T =0, which is
another indication that dq rev/T is an exact differential.
3.3.2. Irreversible Process
What is the efficiency when the process is irreversible?
Let us first assume that the efficiency is greater for the irreversible
than for the reversible case. Denoting the irreversible efficiency as ξ and
∗
∗
the reversible as ξ, and since ξ >ξ,we musthave
∗ ∗
−w /q > −w/q 2 (3.14)
2
Consider now two Carnot cycles, the starred and the non-starred one,
coupled together (Fig. 3.3). Let us, for simplicity, adjust the engines so
that −w ∗ = −w. Dividing both sides of (3.14) by −w ,1/q ∗ 2 < 1/q 2
∗
∗
and therefore, q 2 <q . Hence, since the w’s are the same, q >q 1.This
∗
1
2
means that heat, in the amount of q − q 1 , is transferred from the lower
∗
1
temperature reservoir to the higher temperature one, in violation of the
Clausius Principle. So the efficiency of an irreversible change cannot be
greater than that of a reversible change.
Can the efficiency of an irreversible change be less than that of a
reversible change? Yes, there is no prohibition! Since ξ =1 − T 1 /T 2 for