Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
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)
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.
