The specific heat capacity of a solid at constant pressure and low temperatures is given by $$c_p=aT+bT^3$$ with $a$ and $b$ are constants.

How can I calculate a general expression for the entropy $S(T,p)$ and Gibbs free energy $G(T,p)$ as a function of $T$, assuming that $S$ and $G$ are zero at $T = 0~\mathrm{K}$?

I started with trying to calculate entropy by $$\mathrm{d}S = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T = \frac {c_p}{T} \mathrm{d}T$$ and $$S = \int_0^T \frac{c_p}{T}\mathrm{d}T = (aT + bT^3)\ln T$$

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    $\begingroup$ $c_p$ is itself a function of $T$, so $c_p/T = a + bT^2$ and $\int c_p\mathrm{d}T/T = aT + bT^3/3$ - you cannot simply take it out of the integral and integrate $\mathrm{d}T/T = \ln T$ $\endgroup$ Commented Jan 15, 2016 at 17:46
  • $\begingroup$ Understood, but is that approach correct, can I do it that way? For the free energy we have $G(T,p)=H-TS $, but I dont know how to determine H from either its total differential nor $H=U+pV $... $\endgroup$
    – user149868
    Commented Jan 15, 2016 at 18:22
  • $\begingroup$ The entropy part seems sensible to me, if there's a problem with it, I don't know. Gibbs free energy part, I suspect that you may need to use the equation $(\partial G/\partial T)_p = -S$, but I'm not entirely sure either. Sorry - I'm not that familiar with doing absolute thermodynamic quantities. $\endgroup$ Commented Jan 15, 2016 at 19:27
  • $\begingroup$ For $\mathrm{d}G=\left( \frac {\partial G}{\partial T}\right)_p \mathrm {d}T + \left(\frac {\partial G}{\partial p}\right)_T\mathrm {d}p = -S\mathrm{d}T+V\mathrm{d}p $ maybe the second part disappears because it's an isobaric process (because of the way the heat capacity is given) and because at $0$ K the free energy is zero ($\Delta G=G $) we would just have to integrate $S $ with respect to $T $? $\endgroup$
    – user149868
    Commented Jan 15, 2016 at 20:11
  • $\begingroup$ Your equation is messed up but I get what you mean. I was thinking along those lines too - as with the process for $S$, it seems very reasonable to me. $\endgroup$ Commented Jan 15, 2016 at 20:16

1 Answer 1


Since G and S are taken to be zero at T = 0, H is also 0 at T = 0. So you can calculate H by integrating the heat capacity from 0 to T: $$H=\int_0^TC_pdT=\frac{a}{2}T^2+\frac{b}{4}T^4$$ Then you substitute H into the equation $$\frac{\partial (G/T)}{\partial T}=-\frac{H}{T^2}=-\frac{a}{2}-\frac{b}{4}T^2$$ and integrate between 0 and T. This will give you G. You can then solve for S from G = H - TS. This gives you the same result as integrating directly to get S.

  • $\begingroup$ I calculated $G=-(aT)/2-(bT^3)/12$ and $S=a/2+(at)/2+(bT^2)/12-(bT^3)/4$, however I dont quite understand how you get to the expression $\frac{\partial (G/T)}{\partial T} $, how ddo you come up with $G/T $? $\endgroup$
    – user149868
    Commented Jan 16, 2016 at 8:35
  • $\begingroup$ Your solution for G is correct, but your solution for S is not. Probably an algebra error. Try again. The expression your are asking about comes from the equation for dG at constant pressure: $$dG=-SdT=-\frac{(H-G)}{T}dT$$. You can arrive at the final equation using the quotient rule for differentiation on the terms involving G. This equation is very useful because you can get the temperature dependence of G without involving S, and only involving H (which is much more easily obtained). $\endgroup$ Commented Jan 16, 2016 at 12:58
  • $\begingroup$ If I differentiate $-\frac{(H-G)}{T}dT $ with respect to $T$ I get $\frac{(H-G)}{T^2} $, but I still dont understand how you get from there to partially differentiating $G/T $ with respect to $T $ $\endgroup$
    – user149868
    Commented Jan 16, 2016 at 15:42
  • $\begingroup$ $\frac{dG}{dT}-\frac{G}{T}=-\frac{H}{T}$ so $\frac{T\frac{dG}{dT}-G}{T^2}=\frac{d(G/T)}{dT}=-\frac{H}{T^2}$ $\endgroup$ Commented Jan 16, 2016 at 15:58
  • $\begingroup$ Thats really neat. I calculated entropy to $S=-\frac {G-H}{T}=-a/2+bT^3/12-aT/2-bT^3/4$. Thanks man! $\endgroup$
    – user149868
    Commented Jan 16, 2016 at 18:54

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