I am trying to understand at the molecular level why a real gas with attractive forces between particles experiences an increase in internal energy when expanding isothermally. Here's my thinking:

For a gas, the path-independent differential form of change in internal energy is

\begin{equation} dU = \frac{\partial U}{\partial V}{ \delta V} + \frac{\partial U}{\partial T}{ \delta T} \end{equation}

For an isothermal expansion, \begin{equation} { \delta T} = 0 \end{equation}

so \begin{equation} dU = \frac{\partial U}{\partial V}{ \delta V} \end{equation}

(For an ideal gas, internal energy is proportional to temperature, so an isothermal expansion doesn't change the internal energy.)

The way I'm looking at the real gas situation is that:

  1. When the real gas expands, say by free expansion into an evacuated chamber, the Δ U = Δ P.E. + Δ K.E.

  2. The increase in volume causes the average distance between particles to increase and therefore increases the potential energy.

  3. The increase in potential energy comes as the expense of a decrease in kinetic energy, because Δ U= 0 = Δ P.E. + Δ K.E., in the same way that throwing a ball upwards into the air causes it to lose K.E.

  4. However, in the case of the gas, this is a very short-lived decrease in K.E., because the decrease in K.E. of the particles lasts no longer until they reach a wall.

  5. Because the walls of the evacuated flask are maintaining an isothermal environment, they transfer heat energy to the particles inside, either by contact conduction, or radiation, thereby restoring K.E. to initial value.

  6. U final = final P.E. (higher than initial conditions) + initial (restored) K.E., and is therefore increased.

(In the case of a volume expansion of a real gas by doing work against a piston, I believe that there would be an even larger transient drop in K.E., immediately counteracted by the introduction of correspondingly more heat because of the isothermal constraint. So the end-of-pathway conditions are still the same, increased P.E. and restored K.E.)

Is this reasoning correct? The version of Atkins I have asserts that for a real gas \begin{equation} \frac{\partial U}{\partial V} > 0 \end{equation}

but doesn't provide a molecular explanation.

  • $\begingroup$ Searching for the same thing. $\endgroup$ – Kashmiri Dec 7 '20 at 12:38

The internal energy change with temperature can be found from one of the Maxwell equations and is

$$ \left (\frac{\partial U}{\partial V} \right)_T = T\left (\frac{\partial p}{\partial T} \right)_V -p$$

and a real gas can be represented with the empirical van-der-Waals equation $\displaystyle (p+a/V^2)(V-b)+RT$ where a represents the attractive potential and b the molecular volume.

Performing the calculation gives

$$ \left (\frac{\partial U}{\partial V} \right)_T = \frac{a}{V^2}$$

which shows what you note in your question.

As the gas expands work has to be done to separate the molecules because they have an attractive potential between them, thus the internal energy has to increase because the temperature is held constant. If the gas was thermally isolated during expansion it would cool and this is observed in the Joule-Thomson effect.

  • $\begingroup$ Won't the gas do work as it expands contributing a factor towards reduction in the internal energy? $\endgroup$ – Kashmiri Dec 7 '20 at 12:36

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.