## Isotherms and adiabats, part I

*November 7, 2010 at 4:40 pm* *
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In most thermodynamics textbooks that I’ve seen, it’s shown that isotherms (processes of constant temperature) and adiabats (processes with no heat exchange) must take a particular form when the working substance is an ideal gas. That form is as shown on the dry erase board in the CRAZY THERMODYNAMICS WOMAN picture. But did you know that the isotherms and adiabats look like that for ANY working substance; not just ideal gases?

Um, yes. Yes I did know that (Лъжа… Аз съм крокодил, и хич не ми пука!).

Yes, it’s true! How do you go about showing it? You start by assuming the following two statements, both inductive generalizations based on a vast amount of experience:

Statement J (after Joule, who did the first experiments providing evidence for this statement). Work done on an adiabatically isolated system increases the temperature of that system, and the required amount of work to get a given temperature change is the same, no matter how that work is carried out (for the proof below, we won’t need this second part of statement J).

Statement T (after Thompson). It is impossible to take heat from an isothermal reservoir and convert it completely into work, without affecting the thermodynamic state of any other body.

Corollary of T. Distinct thermodynamic states have different internal energies.

Proof. Suppose that states A and B are different, but that U(A) = U(B). By the first law, U(B)-U(A) = w + q, where w is work done on the system and q is heat transferred to the system. Hence, 0 = w + q or -w = q. The work done BY the system is equal to the heat transferred to the system. If we assume that the process does not appreciably affect the temperature of the environment (which is always the case if we consider the environment to be big enough), then heat from a constant T reservoir has been converted entirely into work, without affecting the state of any system, contradicting statement T. QED.

If statement T were false, it would be possible to extract heat from some not particularly hot place in the ground and power things with it for a very, very long time. My backyard would become a great energy supply! Unfortunately, to use heat from a reservoir, you have to dump a fraction of that heat into a cooler reservoir, and the maximum amount of work that you can get is determined by the ratio of the temperatures of the hot and cold reservoirs (find a thermodynamics book, and read the sections on Sadi Carnot and reversible heat engines). Now for the part that isn’t as easy to find.

Claim 1. An adiabat cannot cross an isotherm more than once. We can symbolize this statement as ~C.

Proof. Suppose that Claim 1 is false (that is, assume ~~C = C). Then there will be distinct points A and B, both lying on an adiabat, and such that the temperature at A is the same as the temperature at B. Consider a process that starts at A, follows the adiabat, and ends at B. Because A and B are distinct states, the internal energy of the system at A is different from that at B (by the corollary to statement T above). By the first law of thermodynamics, U(B)-U(A) = w + q, and so the work done by the system is

-w = q – [U(B)-U(A)] = U(A)-U(B),

the last equality following because q = 0 along an adiabat. Now either U(A)-U(B) < 0 or U(A)-U(B) > 0. Consider the first case. In this case, -w is also negative, meaning that work was done ON the system in going from A to B. But by statement J, the temperature at B would have to be higher than that at A, which it isn’t, because A and B lie on an isotherm! So it is impossible that U(A)-U(B) < 0. Consider the second case, U(A)-U(B) > 0. Then -w is also positive, and work was done BY the system in going from A to B. Now note that, because q = 0, no other system has to be thermodynamically affected. We can just use -w to lift a weight or something. Because q =0 and -w > 0, this weight-lifting feat is equivalent to converting heat from a constant T reservoir completely into work, which is impossible by statement T. In summary, we have

C -> (~T v ~J)

T & J

——————–

~C

And so the proof is complete.

Entry filed under: Thermodynamics.

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