By free expansion, I am referring to gas allowed to expand freely against vacuum in a Joule Expansion.

If gas is ideal then change in Internal Energy '∆U' and change in Enthalpy '∆H' is zero. (By ∆H = ∆U + nR∆T).

But I was wondering whether '∆H' would still be zero for real gas?

(I know that ∆U will be zero for real gas. I just want to ask about ∆H)

  • $\begingroup$ I know about ∆U. I want to know about ∆H. $\endgroup$ Jan 17, 2021 at 11:33
  • $\begingroup$ I didn't understand your point. ∆H = ∆U + nR∆T is just applicable to ideal gases, right? $\endgroup$ Jan 17, 2021 at 11:49
  • $\begingroup$ So how do I know about ∆H? (for real gas) $\endgroup$ Jan 17, 2021 at 11:50
  • $\begingroup$ See Joule-Thomson effect for real gas vacuum expansion and van der Waals equation for pV=f(T,p) dependence. As $\Delta H=\Delta U + f(T,p)$. Also look at en.wikipedia.org/wiki/Joule_expansion $\endgroup$
    – Poutnik
    Jan 17, 2021 at 12:18
  • 1
    $\begingroup$ @Poutnik Joule Thomson effect is a bit different from this as it has constant H but not constant U. Joule expansion is exactly what I am talking about, but its wiki page has nothing on H only U. I had asked this question after reading that only $\endgroup$ Jan 17, 2021 at 12:24

2 Answers 2


In a Joule expansion, which is an irreversible adiabatic expansion against a vacuum, $q = 0 \text{ and } p_{ext} = 0$. Thus, since the only type of work in a Joule expansion is $pV$-work:

$$\Delta \text{U} = q + w = w = -p_{ext} \Delta V = 0 \text{, always.}$$

And since $\Delta \text{H} = \Delta \text{U}+ \Delta pV$:

$$\Delta \text{H} = \Delta pV \text {, always.}$$

For ideal gases:

$pV$ = constant (at a given $T$), so $\Delta \text{H} = \Delta pV = 0.$

For real gases, we have two cases:

I. At the inversion temperature:

$pV = \text{constant (at a given } T \text{), so } \Delta \text{H} = \Delta pV = 0$.

II. Not at the inversion temperature:

$pV \ne \text{constant, so } \Delta \text{H} = \Delta pV \ne 0$.


In the Joule experiment, $\Delta H$ is not zero for a real gas. $\Delta H=\Delta U+\Delta (PV)$, and even if $\Delta U$ is zero, there is no physical reason to expect that $\Delta (PV)$ would be zero in this process.


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