I am having some trouble with calculation of entropy and internal energy for ideal gas. All sources which provide such derivations which I managed to find either assume a priori that $U$ is some constant times $T$ or that $C_V$ is constant. I am now wondering if this calculation is even possible without a priori assuming one of these things, or in other words: is equation of state sufficient physical input to derive all thermodynamical quantities? So far I managed to show the following (basing only on EOS and principles of thermodynamics):

  1. $U$ depends only on T, $U=U(T)$. Similarly, $C_V$ only depends on $T$. Therefore $\mathrm d U=C_V(T) \mathrm d T$.

  2. U=TS-pV.

  3. $\mathrm d S=\frac{nR}{V} \mathrm d V+\frac{C_V}{T} \mathrm d T$.

  4. Integration of point 3. and using 2. I get that

$U=nRT( log(\frac{V}{V_0})+\int_{T_0}^T\frac{C_V(T')}{nRT'} \mathrm d T' -1)$,

where $T_0$ and $V_0$ are values in some reference state. But this formula seems to contradict 1.! Clearly, this expression cannot be expressed by temperature alone. What is wrong in my reasoning? Can I calculate $U$ and $S$ at all if I don't assume something more than EOS?

For reference of people who might have the same problem in the future I add proofs of points (1-3):

Ad 1. $\mathrm d S = \frac{1}{T}(\mathrm d T +p \mathrm d V)=\frac{1}{T}(p + (\frac{\partial U}{\partial V})_T) \mathrm d V + \frac{1}{T}(\frac{\partial U}{\partial T})_V \mathrm d T$.

This is an exact differential so mixed partial derivatives are equal. Using this and equation of state one gets the hypothesis.

Ad 2. This holds from first law because $U$ is extensive quanitity (formally one uses Euler's theorem about homogeneous functions). For a positive parameter $t$ we have that $U(tS,tV)=tU(S,V)$. Differentiating w.r.t. $t$ and then setting it to $1$ yields the hypothesis.

Ad 3. One just uses the first law, definition of $C_V$ and the fact that derivative of $U$ w.r.t. $V$ vanishes.

  • 1
    $\begingroup$ Eqn. 2 is incorrect. Otherwise the gibbs free energy G would be zero at all states. Eqn. 3 is also incorrect; there should be an n in front of the Cv. $\endgroup$ Apr 13, 2016 at 20:23
  • $\begingroup$ One can use the equipartition theorem. $\endgroup$
    – sbp
    Apr 13, 2016 at 20:27
  • $\begingroup$ For the n in front of $C_V$ I don't quite agree. I take $C_V$ to be extensive quanitity, $C_V=nc_V$. For incorrectness of equation two: It is after all just a simple consequence of extensivness of $U$ and a mathemathical theorem! For equipartition theorem: This is purely thermodynamical calculation, I don't want to appeal to statistical mechanics. $\endgroup$
    – Blazej
    Apr 13, 2016 at 20:28
  • $\begingroup$ As Chester Miller said eq.2 is incorrect, you cannot derive from $(*) dU=TdS-pdV$ your $(**) U=TS-pV$. The two are not the same. $\endgroup$
    – hyportnex
    Apr 13, 2016 at 20:34
  • $\begingroup$ I am not appealing to the first law alone, I use the fact that internal energy is extensive and Euler's theorem. Full proof is now in the post. Equivalent, differential statement of this fact is the so called Gibbs-Duhem relation $S \mathrm d T - V \mathrm d p =0$. $\endgroup$
    – Blazej
    Apr 13, 2016 at 20:36

1 Answer 1


The condition that the internal energy of an ideal gas is a function only of temperature follows empirically from the observed behavior of real gases in the limit of low pressures and high specific volumes. (It may also follow from statistical thermodynamics). In addition, the heat capacities of most real gases are observed to be a function of temperature at low pressures so that, at constant (high) specific volume, real gases give $dU=nc_vdT$. If integrated at constant (high specific) volume, this gives $\Delta U=n\int_{T_1}^{T_2}{c_v(T')dT'}$. So, for real gases in the ideal gas limit, $\Delta U$ is not equal to $nc_v(T_2-T_1)$. And for real gases, if we assume that, in the ideal gas limit, $U=nc_vT$ we will get the wrong answer for $\Delta U$ unless $c_v$ is constant.

For a real gas, $$dU=C_vdT+\left[T\left(\frac{\partial P}{dT}\right)_V-P\right]dV$$ The term in brackets is equal to zero for an ideal gas. The development that led to this equation is another way of proving that, for an ideal gas, the internal energy and the constant volume heat capacity are functions only of temperature.


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