# Change in Enthalpy in Free Expansion of Ideal and Real gas against Vacuum

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)

• I know about ∆U. I want to know about ∆H. Jan 17 at 11:33
• I didn't understand your point. ∆H = ∆U + nR∆T is just applicable to ideal gases, right? Jan 17 at 11:49
• So how do I know about ∆H? (for real gas) Jan 17 at 11:50
• 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 Jan 17 at 12:18
• @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 Jan 17 at 12:24

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:

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

• Thank you so much! Jan 18 at 8:07

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.