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For an adiabatic free expansion, $W = 0$ and $Q=0$. Therefore, by the first law of thermodynamics, $\Delta U = Q -W = 0$.

For a Van der Waals model of a real gas, $\Delta U = n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right)$. This means that $n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right) = 0$, or $n C_v \Delta T = a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right)$. Now, since the gas is expanding, $V_2 > V_1$. So, $a \, n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right) < 0$. Therefore, $n C_v \Delta T<0$. This means that the temperature of the gas decreases in an 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?

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dryairship
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2 Answers2

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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 $$\eqalign{ H=U+PV&\simeq {3\over 2}Nk_BT-{N^2a\over V} +\left({Nk_BT\over V-Nb}-{N^2a\over V^2}\right)V\cr &={3\over 2}Nk_BT-{2N^2a\over V} +Nk_BT\left(1-{Nb\over V}\right)^{-1}\cr &\simeq {3\over 2}Nk_BT-{2N^2a\over V} +Nk_BT\left(1+{Nb\over V}\right)\cr &={5\over 2}Nk_BT-{2N^2a\over V}+{N^2bk_BT\over V}\cr &={5\over 2}Nk_BT+{N^2\over V}\big(bk_BT-2a\big)\cr }$$ 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.

Christophe
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    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? – dryairship Mar 25 '17 at 11:56
  • I believe this answer is more in reference to a Joule-Thompson setup, specifically two regions of pressure separated by a porous plug. – Pecan Lim Oct 11 '23 at 06:49
  • Related answer: https://physics.stackexchange.com/questions/341037/does-the-temperature-of-a-real-gas-decrease-in-the-adiabatic-free-expansion?rq=1 – Christophe Oct 12 '23 at 07:56
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this is more a comment on your answer: in any free expansion the pressures in each side are not kept constant: the pressure decreases in one side and increases in the other side. The correct statement is: "It is only when the gas flows between two regions of different pressures (each kept constant) that the total entalpy $H_1 +H_2$ is conserved. For a real gas, it follows that the sign of $\Delta T$ depends on T".

Fabio Lima
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