Consider a realistic gas. If it is insulated and then allowed to expand into a vacuum, how does its temperature change?

We have to consider the sign of $\left(\dfrac{\partial T}{\partial V }\right)_E$.

Can this quantity take both signs?

  • $\begingroup$ Gas expansion into virtually infinite vacuum space means as the gas moves out, beyond certain point it does not have temperature any more, it moves preferentially in single direction and gets arbitrarily rarified. To maintain the meaning of temperature, it would have to expand into some other container of finite volume so that it can reach equilibrium again. In any case, you have to specify where in space is the temperature measured, the sign may depend on that. $\endgroup$ – Ján Lalinský May 12 at 14:03
  • $\begingroup$ You are right. Let the gas expand from a smaller chamber into a larger chamber. $\endgroup$ – John May 12 at 22:20
  • $\begingroup$ OK, then now we need to know whether the expansion is slow enough so that temperature is still meaningful, or so fast that the gas will acquire violent non-equilibrium states, shock waves etc. If the expansion is throttled enough, then gas in both chambers will be close to equilibrium to have a meaningful temperature. The initial chamber will cool down, the new chamber will heat up. Only later after much time their temperatures will meet the same value. So it is hard to see how $\partial T/\partial V_E$ is a meaningful expression here: there is no single temperature if we have two chambers. $\endgroup$ – Ján Lalinský May 12 at 22:42
  • $\begingroup$ On the other hand, if there is only one chamber that expands its volume (such as a combustion chamber with a piston that moves out), then the gas expands into larger volume and if slow enough, there is only one temperature of gas. Perhaps this is the case the question was meant to be about? However, it is not clear how constancy of energy can be achieved then, because by pulling the piston out, the gas does work on the piston, i.e. energy of the system is decreased. So I think you really should push back on the source of the question and ask how exactly is this process supposed to happen. $\endgroup$ – Ján Lalinský May 12 at 22:46

It follows from the thermodynamic equation: $$dE=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$that $$\left(\frac{\partial T}{\partial V}\right)_E=\frac{\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]}{C_v}$$Try this out for various real gas models, like the van Der Waals equation and see what you get.

  • $\begingroup$ I've seen you use the first equation in other posts as well as linking it to enthalpy, but I have yet to find a complete derivation of the second term. Can you site a reference? Thanks. $\endgroup$ – Bob D May 12 at 13:28
  • $\begingroup$ @BobD It is derived in exactly the same way, except using the Maxwell relationship based on dA rather than dG. I think the derivation is in most Thermo books. $\endgroup$ – Chet Miller May 12 at 13:32
  • $\begingroup$ Thanks Chet. Found a complete derivation of the Joule coefficient in the following link. orca.phys.uvic.ca/~tatum/thermod/thermod10.pdf $\endgroup$ – Bob D May 12 at 14:06
  • $\begingroup$ I did not know that this is called the Joule coefficient. Thanks! $\endgroup$ – John May 12 at 22:22

From a pure thermodynamic point of view, $\left( \frac{\partial{T}}{\partial{V}} \right)_E$ may take both signs. The reason is that $$ \left( \frac{\partial{T}}{\partial{V}} \right)_E = - \frac{\left( \frac{\partial{E}}{\partial{V}} \right)_T}{\left( \frac{\partial{E}}{\partial{T}} \right)_V} = - \frac{\left( T \left( \frac{\partial{S}}{\partial{V}} \right)_T -P \right)}{T \left( \frac{\partial{S}}{\partial{T}} \right)_V}= \frac{\left( P -T \left( \frac{\partial{P}}{\partial{T}} \right)_V \right)}{C_V}=\frac{\left( P -T \frac{\alpha}{\chi_T} \right)}{C_V} $$ where $\alpha = \frac{1}{V} \left( \frac{\partial{V}}{\partial{T}} \right)_P$ is the (constant pressure) thermal expansion coefficient, $\chi_T = - \frac{1}{V} \left( \frac{\partial{V}}{\partial{P}} \right)_T$ is the isothermal compressibility, and $C_V=\left( \frac{\partial{E}}{\partial{T}} \right)_V$ is the constant volume heat capacitance.

While $C_V$ and $\chi_T$ must be positive in a stable homogeneous thermodynamic system, thermal expansion coefficient $\alpha$ may have both signs.

However, every real system behaves increasingly better as a perfect gas when pressure decreases and temperature increases, i.e. close to the perfect gas limit, the sign of $\alpha$ is the opposite of the sign of $\left( \frac{\partial{T}}{\partial{V}} \right)_E$. We know that just a few substances have a negative thermal expansion coefficient, and usually, over a limited range of temperatures not too far from melting temperature. Therefore, in the high $T$, low $P$ limit, we expect $\alpha>0$ and $\left( \frac{\partial{T}}{\partial{V}} \right)_E < 0$.

The possibility of a positive sign, looks very unrealistic, from a statistical mechanics point of view, provided that the definition of gas would correspond to a system with low coordination number and large intermolecular distances.


As @Chet Miller pointed out, the equation for the Joule coefficient is

$$\Biggl( \frac{\delta T}{\delta V}\Biggr)_U=\frac{1}{C_V}\Biggl [P-T\Biggl(\frac{\delta P}{\delta T}\Biggr)_V\Biggr ]$$

A complete derivation of this equation can be found at


It is my understanding that the Joule coefficient for an adiabatic expansion in a vacuum can be positive, negative or negligible, depending on the extent to which the initial temperature of the gas is above, below, or close to its inversion temperature. In general, the closer to the inversion temperature, the less the effect, and the more the gas behaves like an ideal gas. Examples of inversion temperatures for some gases are (source NASA.gov):

$N_{2}$ =126 C, $O_{2}$ =155 C, $CO_{2}$ = 304 C, $He$= 5.2 C, $H_{2}$ = 33 C, $Ne$ = 44.

For an adiabatic free expansion of a gas in a vacuum, $\Delta U=0$. Therefore

$$\Delta KE+\Delta PE =0$$

For an ideal gas (negligible Joule coefficient) there are no intermolecular forces. For a real gas there are intermolecular forces.

If the initial temperature is above the inversion temperature, repulsive intermolecular forces dominate. Expansion of the gas will result in a decrease in potential energy and therefore increase in kinetic energy, and thus an increase in temperature.

If the initial temperature is below the inversion temperature, there are weak (Van der Waals) attractive intermolecular forces that dominate. Expansion of the gas increases potential energy and decreases kinetic energy. That results in a decrease in temperature.

Hope this helps


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.