Does the temperature of a real gas decrease in the adiabatic free expansion?

Question

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?

Answer 1

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 $$ \left(\frac{\partial U}{\partial V}\right)_{T,N} \gtrapprox 0. $$

Answer 2

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