Sunday, February 22, 2009

Equivalence of Kelvin-Planck and Clausius Statements

The Clausius statement implies the Kelvin-Planck statement: -

Suppose that we violated the Kelvin-Planck statement and constructed a magic heat engine. Then we could take its work output and use it to power a (non-magical) fridge, which draws heat from a cooler thermal reservoir and discharges this heat into the thermal reservoir that the magic heat engine runs off, as shown in the diagram to the right. Then this combined system of the fridge and the magic heat engine is a magic fridge, since it moves heat from a lower temperature to a higher temperature while having no other effect on its environment - in particular, without having an external power source.

So if we had a magic heat engine, we could have a magic fridge. Therefore, the statement that you can't have a magic fridge implies that you can't have a magic heat engine.

The Kelvin-Planck statement implies the Clausius inequality: -

We now wish to prove that the statement that you can't have a magic heat engine implies the Clausius inequality - that for any closed, stationary system operating in a cycle

\oint \delta Q / T \leq 0

where Q is as usual the heat transferred to the system and T is the temperature at the boundary of the system.

We shall prove this with the assistance of the piece of imaginary equipment shown in the diagram below. We choose any closed, stationary system (hereafter referred to as "the chosen system"). We imagine that the chosen system is attached to a reversible system which is connected to a thermal reservoir, as shown on the diagram.

By convention, the diagram shows heat flowing in to the chosen system and work coming out of it. We emphasize, however, that this is not a restriction on what the chosen system is allowed to do: it merely means that work input or heat output will have negative sign.

It is important to note that the inequality we're going to derive relates only to the boundary temperature and heat flow of the chosen system, and is therefore true whether or not the chosen system is connected to a reversible system which is connected to a thermal reservoir. The reversible system and the thermal reservoir are, so to speak, intellectual scaffolding, which will be discarded after we have produced our result.

So, consider the diagram. The combined system of the chosen system and the reversible system to which it's attached is, like everything else, subject to the First Law of Thermodynamics, which we give for the combined system in its differential form: δQr - δW = dE.

Now, as the reversible system is reversible, then if we are using a Thermodynamic Temperature Scale (eg degrees kelvin) then by the definition of a thermodynamic temperature scale we have δQr / δQ = Tr / T.

We can rearrange this as δQr = Tr δQ / T and substitute this into the First Law equation above, giving us: Tr δQ / T - δW = dE

For convenience, we shall assume that the reversible system performs an whole number of cycles in the same time that the chosen system takes to execute one. Now, let's integrate the equation above over one cycle of the chosen system.

\oint T_r \delta Q / T - \oint \delta W = \oint dE

Of course, \oint \delta W is simply W. To evaluate \oint dE, recall that over a cycle, the combined system can neither gain nor lose energy - or it wouldn't be a cycle - and so \oint dE = 0. This gives us

\oint T_r \delta Q / T - W = 0

which we shall rearrange to give

\oint T_r \delta Q / T = W

Now, here's the clever part. There is no such thing as a magic heat engine. That's the Kelvin-Planck statement of the Second Law, and the whole thing we're basing our argument on.

Now, look at the diagram given above, and observe that the combined system is connected to only one thermal reservoir. If the net work output W was greater than zero, the combined system would be a magic heat engine. Therefore, the net work output W of the combined system is less than or equal to zero, and so

\oint T_r \delta Q / T \leq 0

Finally, we note that Tr is a constant (being the temperature of a thermal reservoir) and positive, since we are using a thermodynamic temperature scale. So we can divide both sides of the inequality through by Tr, giving us

\oint \delta Q / T \leq 0


  1. this is awesome thanks!

  2. a very helpful page. thanks!

  3. thank you very much!

  4. can't u prove the 2nd part as easily as the 1st
    1st one is awesome

  5. very easy to understand thank u :)