I have been attempting to solve a problem involving a non-ideal gas. The gas is assumed to be non-ideal and hence modelled by the Van der Waals Equation, it is considered to be pure hydrogen gas and is contained in a cylinder that has a changing length and fixed radius, hence the volume can be changed and is not fixed.

I am trying to calculate the pressure and temperature at a given volume, I understand that the pressure and temperature will increase at their own rates $dP/dV$ and $dT/dV$ when the volume decreases, though the only formula I have is the non-ideal gas equation:

$$(p+((an^2)/V))(V-nb) = nRT.$$

Hence I cannot simply remove one of the variables to try to find values of another and I do not have equations for $dP/dV$ or $dT/dV$.

I began looking at trying ways to find values for $dP/dV$ and $dT/dV$. First, I tried using the non-ideal gas equation and deriving either $dP/dV$ or $dT/dV$ from this but I could not solve them without having all three variables present, (which is the initial problem).

Instead, I began looking at other peoples work and potentially using graphs of $P$ against $V$ or $T$ against $V$ to formulate a function and then work backwards to determine an accurate differential $dP/dV$ or $dT/dV$. Unfortunetly using this method did not seem to work as the gas that I was researching (Hydrogen) seemed to have no such graphs or data available.

Hence I am now stuck and asking for assistance with my problem. If anyone can offer help in finding a method to link the dynamic system together that would be greatly appreciated.

I believe that the key to finding the solution lies in finding a numerical answer for $dP/dV$ and or $dT/dV$.

  • $\begingroup$ If you know how the internal energy $U$ changes according to net work and heating, perhaps you can apply the general equation $dU=nc_V\,dT+(\alpha TK-P)dV$, which applies to all gases. Here $c_V$ is the constant-volume molar heat capacity, $\alpha$ is the thermal expansion coefficient, and $K$ is the bulk modulus. $\endgroup$ Commented Feb 19, 2022 at 20:56
  • $\begingroup$ Hi, welcome to physics StackExchange. Is the process adiabatic? Is heat being lost to the environment? Are pressures high enough so you have a mixture of gas and liquid H2? Is energy conserved? Is exergy conserved? I think you need Cengel's book. $\endgroup$
    – Arc
    Commented Feb 19, 2022 at 23:09
  • $\begingroup$ Substitute the given volume into the equation. You then have the pressure as a function of temperature. Simple as that. $\endgroup$ Commented Feb 20, 2022 at 4:19
  • $\begingroup$ @ChetMiller, no, there are a lot of details we are not being told about, so perhaps its not that simple. $\endgroup$
    – Arc
    Commented Feb 20, 2022 at 4:54
  • $\begingroup$ What is the exact statement of the problem? $\endgroup$ Commented Feb 20, 2022 at 9:38

1 Answer 1


If the process is adiabatic and reversible, then $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=0\tag{1}$$with $$\left(\frac{\partial S}{\partial T}\right)_V=\frac{nC_v}{T}\tag{2}$$and,from $dF=-SdT-PdV$ and the related Maxwell relation, $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{V-nb}\tag{3}$$From Eqns. 2 and 3 it follows that Cv is a function only of temperature, so that it must equal Cv in the ideal gas limit $C_v(T,V)=C_v^{IG}(T)$Therefore, from the above equations, we have: $$\frac{C_v^{IG}}{T}dT=-\frac{R}{V-nb}dV$$This can be integrated directly to give T as a function of V.

  • $\begingroup$ Up to my rusty thermodynamic skills, this is the answer. Nice to think again about Maxwell's relations. $\endgroup$
    – Arc
    Commented Feb 23, 2022 at 1:54
  • $\begingroup$ This seems to make sense, cheers aye. $\endgroup$
    – user322011
    Commented Feb 23, 2022 at 21:43

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