Here's the standard way of deriving the entropy of the ideal gas (see e.g. here): $$ dQ=dU+PdV=C_VdT+\dfrac{NkT}{V}dV $$ $$ dS = \dfrac{dQ}{T} $$ Integration of the latter gives the correct result. Note that nobody ever says here $``\text{consider a particular process, say isochoric}"$. The equation is written for an arbitrary reversible process.

Here's how the analogous thing is done for the photon gas (see e.g. here). $``\text{Consider an isochoric process. Then, } dU=TdS"$, and, so, $$ dS = \dfrac{dU}{T} $$ and so on.

Now, my question about these derivations is very simple.

Why can't we consider an isochoric process in the first case, in order to get $dS=dQ/T=dU/T$? Then, in analogy with the second case, we should throw away the last term in the first equation.

Or do the opposite, and follow the steps of the first derivation in the second case, is order to get $dQ = dU + PdV$? Then, we should add another term in the numerator of the last equation.

What am I missing??


1 Answer 1


Your book gives $U=aVT^4$ and $P=a\frac{T^4}{3}$. So, $$TdS=dU+PdV=4aT^3VdT+aT^4dV+aT^4dV/3=4aT^3VdT+\frac{4}{3}aT^4dV$$or $$dS=4aT^2VdT+\frac{4}{3}aT^3dV=d\left(\frac{4}{3}aVT^3\right)$$So, $$S=\frac{4}{3}aVT^3$$The question is: why does integrating dU/T at constant T also work for giving the entropy. Well, since P is independent of V, we have from the Maxwell relation $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=\frac{4}{3}aT^3$$This is a function only of temperature. So S is directly proportional to V. So both U and S are directly proportional to V, which enables us to factor out the V (which is equivalent to saying that we can integrate dU/T at constant V to get S).

  • $\begingroup$ Thank you very much for the complete answer. I actually first reached the same equation for the entropy as you did. However, instead of realizing that it's a total derivative, I mistakenly 'integrated' both terms separately. While, as you showed, whether we write the second term or not, the answer is the same. $\endgroup$
    – mavzolej
    Jan 30, 2019 at 12:15
  • $\begingroup$ Could you please also show the $``\text{isochoric}"$ derivation for the case of ideal gas? $\endgroup$
    – mavzolej
    Jan 30, 2019 at 16:32
  • $\begingroup$ I don't understand this question. $\endgroup$ Jan 30, 2019 at 16:37
  • $\begingroup$ Ohhh my bad. So is it correct that we can consider the isochoric process for photons only because the state equation is $V$-independent? And for the ideal gas, we have to derive it in full generality? $\endgroup$
    – mavzolej
    Jan 30, 2019 at 17:09
  • 1
    $\begingroup$ I think so. But I was still only comfortable by doing the photon gas in full generality. $\endgroup$ Jan 30, 2019 at 18:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.