0
$\begingroup$

For free expansion, $W = 0$

And for an adiabatic process, $\Delta Q = 0$

Therefore, by the equation of the first law of thermodynamics, $\Delta U = 0$

For the Van der Waal model of a real gas, $\Delta U = n C_v\Delta T - an^2 (\frac{1}{V_2}-\frac{1}{V_1})$

This means that

$n C_v\Delta T - an^2 (\frac{1}{V_2}-\frac{1}{V_1}) = 0$
or $n C_v\Delta T = an^2 (\frac{1}{V_2}-\frac{1}{V_1})$

Since the gas is expanding, $V_2 > V_1$

So, $an^2 (\frac{1}{V_2}-\frac{1}{V_1}) < 0$

So, $n C_v\Delta T<0$

This means that temperature of the gas decreases in adiabatic free expansion.

Will the temperature always decrease? I mean, what if the gas is above its inversion temperature or at it? Does inversion temperature play no role in adiabatic free expansion?

$\endgroup$
  • $\begingroup$ The equation you chose is not for a real gas, but for a Van der Waals model of a real gas. $\endgroup$ – Chet Miller Mar 24 '17 at 14:34
  • $\begingroup$ @ChesterMiller Yes. I actually meant a Van der Waal gas. I'll write that. $\endgroup$ – dryairship Mar 25 '17 at 10:23
0
$\begingroup$

Your calculation indeed shows that the temperature of the van der Waals gas always decreases in adiabatic free expansion. It is only when the gas flows between two regions of different pressures that the sign of the variation of temperature depends on the temperature. Enthalpy, rather than energy, is conserved in this case. Using the expression of pressure for the van der Waals gas $$P={Nk_BT\over V-Nb}-{N^2a\over V^2}$$ enthalpy reads $$\eqalign{ H=U+PV&\simeq {3\over 2}Nk_BT-{N^2a\over V} +\left({Nk_BT\over V-Nb}-{N^2a\over V^2}\right)V\cr &={3\over 2}Nk_BT-{2N^2a\over V} +Nk_BT\left(1-{Nb\over V}\right)^{-1}\cr &\simeq {3\over 2}Nk_BT-{2N^2a\over V} +Nk_BT\left(1+{Nb\over V}\right)^{-1}\cr &={5\over 2}Nk_BT-{2N^2a\over V}+{N^2bk_BT\over V}\cr &={5\over 2}Nk_BT+{N^2\over V}\big(bk_BT-2a\big)\cr }$$ which can be rewritten as $$H=C_pT+{N^2bk_B\over V}(T-T_i)$$ where the inversion temperature is $$T_i={2a\over bk_B}$$ For $T>T_i$, the expansion leads to an increase of the temperature.

$\endgroup$
  • $\begingroup$ You said that "It is only when the gas flows between two regions of different pressures that the sign of the variation of temperature depends on the temperature" Isn't the gas flowing from high pressure to zero pressure in free expansion? $\endgroup$ – dryairship Mar 25 '17 at 11:56

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.