The internal energy $U$ of a given mass of a real gas can be regarded as a function of temperature and volume i.e., $U(T,V)$. Under adiabatic free expansion, the change in the internal energy is zero because $\delta Q=0$. The workdone $\delta W=0$ too because the gas expands against vacuum. Hence, from first law of thermodynamics, $$dU=\delta Q-\delta W=0.$$ The change in the temperature of the gas is given by $$ dT=-\frac{\Big(\frac{\partial U}{\partial V}\Big)_TdV}{C_V}. $$ Since $C_V>0, dV>0$ (for expansion), the sign of $dT$ will be dictated by the sign of $\Big(\frac{\partial U}{\partial V}\Big)_T$.

This partial derivative is positive if $U$ is a monotonically increasing function of $V$ and negative if monotonically decreasing. Is there a way to determine the sign of this partial derivative from thermodynamic consideration without using any particular equation of state?

  • $\begingroup$ Comment on question v1: you should be careful when making arguments about free expansion using the differential forms $\delta Q$ and $\delta W$ because it's not a quasistatic process. It would be less notationally misleading to argue $\Delta U = Q - W = 0 - 0 = 0$. On the other hand, even for non-quasistatic processes, computing changes in state variables such as $T$ by integrating $dT$ along any path connecting the final and initial states is totally fine even though the process itself is only in equilibrium in the initial and final states. $\endgroup$ – joshphysics Jun 23 '17 at 18:13
  • $\begingroup$ @joshphysics I agree. $\endgroup$ – SRS Jun 23 '17 at 18:37
  • $\begingroup$ Do you count the virial as an equation of state, or do you just regard it as a Taylor series expansion? $\endgroup$ – Chet Miller Jun 23 '17 at 20:28
  • $\begingroup$ As you have already pointed out that $dU=0$, $(\frac{\partial U}{\partial V})_T$ is zero as well. $\endgroup$ – user115350 Jun 24 '17 at 0:10

Suppose we consider the Van der Waals gas as more accurate model for a real gas. It can be shown that its internal energy is

$$ U(T,V, N) = cNT - \frac{aN^2}{V}, \qquad c>0,\quad a>0. $$

Since for an adiabatic free expansion $Q = W = 0$, we have $\Delta U = 0$. It follows that for a gas of fixed particle number

\begin{align} 0 = U_f - U_i = cN(T_f-T_i) - aN^2\left(\frac{1}{V_f} - \frac{1}{V_i}\right), \end{align}

and therefore

$$ \Delta T = \frac{a}{c}N\left(\frac{1}{V_f} - \frac{1}{V_i}\right). $$

There is a nonzero change in temperature that depends on the increase in volume. This expression can be made more immediately informative if we define $x = V_f/V_i$ so that $V_f = xV_i$ and

$$ \Delta T = \frac{a}{c}\frac{N}{V_i}\left(\frac{1}{x} - 1\right). $$

If the gas expands then $x>0$, and the temperature decreases.

Edit. More General Considerations

Assuming the virial expansion (which amounts to quite a general equation of state, we have $$ \frac{PV}{NkT} = 1 + B(T)\frac{N}{V} + C(T)\left(\frac{N}{V}\right)^2 + \cdots $$ It follows that $$ \left(\frac{\partial U}{\partial V}\right)_{T,N} = T^2\left(\frac{\partial(P/T)}{\partial T}\right)_{T,N} = kT^2\frac{N}{V}\left(B'(T)\frac{N}{V} + C'(T)\left(\frac{N}{V}\right)^2 + \cdots\right) $$ For gases that are not too dense, we can truncate at the second virial coefficient $B(T)$ term to good approximation, so the sign of this derivative is the same as the sign of $B'(T)$. For many real gases, the second virial coefficient is a monotonically increasing function of temperature (at least for temperatures that are not too high, see diagram below), so $B'(T) > 0$ and thus to good approximation $$ \left(\frac{\partial U}{\partial V}\right)_{T,N} \gtrapprox 0. $$ enter image description here

  • $\begingroup$ Can we do without resorting to any particular equation of state? @joshphysics $\endgroup$ – SRS Jun 23 '17 at 19:38
  • $\begingroup$ Well simply the fact that the ideal gas and VdW gas have different behaviors in this respect (due to their different equations of state) is evidence that the result depends on the equation of state. In comparing those two models, the key factor is the inclusion of the effects of interactions. One might be able to argue that for a general class of gases about which certain reasonable hypotheses are made regarding molecular interactions, this result will still hold -- is that the sort of statement you're looking for? $\endgroup$ – joshphysics Jun 23 '17 at 19:59
  • $\begingroup$ In other words, perhaps the better question is "what are physically intuitive necessary and/or sufficient conditions under which a gas will have this temperature decrease property for adiabatic free expansion?" $\endgroup$ – joshphysics Jun 23 '17 at 20:01
  • $\begingroup$ @SRS See edit employing virial expansion for a reasonably general result. $\endgroup$ – joshphysics Jun 24 '17 at 2:29

By theoretically, in adiabatic free expansion the internal energy of a system is utilised to increase the temperature, as we look into atomic range the bond energy or molecular bond energy is utilised to increase the temperature

  • $\begingroup$ The internal energy does not change after the process. The internal energy during the process is not even defined as it is not a quasistatic process. $\endgroup$ – Yashas Jul 4 '17 at 3:40

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