The Gibbs free energy is defined as

$$ G = U-TS+pV$$

It is also known that the chemical potential is the per-particle Gibbs free energy so:

$$ G=\mu N $$

Now, putting the two together we get

$$ U = TS - pV + \mu N$$

I can also take the differential of the above:

$$ dU= d(TS) - d(pV) + d(\mu N) $$

Expanding the products and rearranging terms:

$$ dU = TdS-pdV + \mu dN + S dT - Vdp + N d\mu$$

This, however, seems at odds with the second law of thermodynamics which should write as

$$ dU=TdS-pdV+\mu dN $$

My question is, how do I resolve this contradiction? From here I would conclude that

$$ SdT - Vdp + N d\mu =0 $$

Is this true? If so, how do I prove it? If not, where does the contradiction come from?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.