I have been studying isothermal expansion process, the heat change particulary for a real gas. On the internet is many solved problems about this, but it's always about changing volume from $v_i$ to $v_f$. But what if I want to study how the presure changes? Because volume and pressure are tied, but for real gas I cannot use the law stating $\;$$p_iv_i=$ const. explicitly.

The state equations for one mole of the real gas are:

$p = \frac {RT}{v-b} - \frac {a}{v^2}$ $\;$ and $\;$ $u = cRT - \frac{a}{v}$, $\;$ where $a$ and $b$ are constants.

Then for the volume change we have:

$\int_{v_i}^{v_f}\frac {RT}{v-b} - \frac {a}{v^2}dV = [RTln(v-b)+a/v]^{v_f}_{v_i}$

  • $\begingroup$ You seem to be assuming that the expansion is isothermal and reversible. Is that correct? $\endgroup$ Commented Jan 19, 2019 at 17:06
  • $\begingroup$ Yes, I do assume it. And I made the mistake with that interval energii part so I am going to edit it. $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 18:37
  • $\begingroup$ My question is maybe broader,... what to do, if my differential form is expressed in the way I dont want it. I was thinking about expressing dV as total differential with the respect of the pressure and temperature. $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 18:38
  • $\begingroup$ The mathematics is such that you can't do that analytically. The equation in cubic in v. Why don't you follow @Philip Wood's suggestion? $\endgroup$ Commented Jan 19, 2019 at 19:03
  • $\begingroup$ I want to calculate the heat change, I need to integrate. So do you mean calculate the heat via standard volume change then express initial and final volume from the state equation and plug it inside? Unfortunately its not possible exactly to express the volume in terms of pressure and temperature for real gas. $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 19:30

1 Answer 1


"For isothermal process we have that $dQ=pdv,$ because $du$ is zero in this case."

This is the case for an ideal gas, for which $U$ is a function of $T$ only, but it is not generally true for a real gas.

As for your question about pressure change in a V der W gas, why can't you just substitute the volumes at start and finish into the V der W equation to obtain the pressure values at start and finish, then subtract one pressure from the other?

  • $\begingroup$ Thank you for the comment about the internal energy but I think I didnt understand the second part $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 18:41
  • $\begingroup$ I think the problem is that I misinterpreted the question. I thought you wanted to find the pressure change – in terms of $what?$ $\endgroup$ Commented Jan 19, 2019 at 18:48
  • $\begingroup$ Oh I meant the heat when I change the pressure $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 19:18
  • $\begingroup$ In all solved problems I found the heat was expressed via volume and for an ideal gas it was easy to get the expression with pressure. But no such example with the real gas. I think this model problem would help me to understand the differential forms in thermodynamics. There must be a trick in differentation. $\endgroup$
    – Andromeda
    Commented Jan 19, 2019 at 19:22
  • $\begingroup$ So, just to be clear, you want to find the pressure change for a given heat input, $\Delta Q?$ Or do you want to find the heat input or output for a given change of pressure? Or neither?$ $\endgroup$ Commented Jan 19, 2019 at 19:23

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