1
$\begingroup$

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)

$\endgroup$
10
  • $\begingroup$ I know about ∆U. I want to know about ∆H. $\endgroup$ Jan 17 at 11:33
  • $\begingroup$ I didn't understand your point. ∆H = ∆U + nR∆T is just applicable to ideal gases, right? $\endgroup$ Jan 17 at 11:49
  • $\begingroup$ So how do I know about ∆H? (for real gas) $\endgroup$ Jan 17 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 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 at 12:24
1
$\begingroup$

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:

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

$\endgroup$
1
  • $\begingroup$ Thank you so much! $\endgroup$ Jan 18 at 8:07
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.