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I'm confused about the third law of Thermodynamics.

The entropy of a perfect crystal at absolute zero is exactly equal to zero.

A consequence of third law is that

The heat capacity must go to zero at absolute zero

$$\displaystyle \lim _{T\rightarrow 0}C(T,X)=0 \tag{1}$$

The law is also known in this way

In 1912 Nernst stated the law thus: "It is impossible for any procedure to lead to the isotherm $T = 0 $ in a finite number of steps."

I do not understand how $(1)$ agrees with the Nernst first statement.

We have $$Q=C \Delta T \implies \Delta T=Q/ C$$

So $C \to 0$ means that a small amount of heat exchanged causes a very large variation of temperature.

So why "it becomes difficult to lower a body's temperatue near $0K$"? Shouldn't it be easier instead ($C \to 0 \implies \Delta T \to \infty$)?

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    $\begingroup$ Nernst statement suggest that it's impossible to actually reach the absolute zero $\endgroup$ – physnolimits Sep 14 '16 at 22:57
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    $\begingroup$ Your argument about $\Delta T = \Delta Q/C$ becoming large when $C$ is small suffers from a zero-over-zero problem, since both the heat content and the heat capacity depend on the temperature; it's necessary to tread carefully. $\endgroup$ – rob Sep 14 '16 at 23:48
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    $\begingroup$ As you pointed out near $0K$ the heat capacity is very small, therefore the slightest heat absorption will lead to a relatively large increase of the body temperature, that is all. $\endgroup$ – hyportnex Sep 15 '16 at 1:42
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Look at Clausius's statement of the second law of thermodynamics. The short version is that heat only ever flows from high temperature areas to low temperature ones. So, in order bring a body's temperature to $0\operatorname{K}$ by the transfer of heat energy you need to have a $0\operatorname{K}$ heat bath to dump the heat into. Note that a heat bath needs to be able to absorb heat without changing temperature, so the existence of a $0\operatorname{K}$ heat bath would violate the claim in the question that $C\rightarrow 0$ for all bodies as $T\rightarrow 0 \operatorname{K}$.

There are other ways to reach $0\operatorname{K}$ that are equally nonphysical. Adiabatically expanding an ideal gas to infinite volume, for example, will get you to the $0\operatorname{K}$ isotherm. You just need to have an infinite volume to expand into, and an infinite amount of time to do it quasi-statically so you can extract the heat as work.

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You can't see Nernst's theorem by looking at the heat transport, or you get just the problem you are asking about. The key to Nernst's theorem is that you can't just force a system to give up heat-- if you could, you'd be right, it would be much easier to get to T=0 than any other T because it doesn't take much heat extracted to get that last drop in T. The problem is, it's hard to extract the heat. I've looked at derivations of Nernst's theorem, and I can't say I have seen any insightful ones yet, so I can't be much more help. But it's clear that one does not get to imagine that you have a direct handle on either dQ or dS, or else you could just step S down to zero. It seems that what you can actually do in a single step always amounts to either a vertical or horizontal step between two curves that converge on zero, so that's why it takes an infinite number of steps-- you're constrained about what you are allowed to "step", and it's never either dQ or dS that you can make whatever you like, it's always something like the pressure or the field strength or some externally controllable parameter like that. Then you just get whatever dQ and dS you get, and it gets harder and harder to get significant dQ or dS as T and S go to zero.

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