# Regarding Joule's gas expansion experiment

Hi,
Joule's original gas expansion experiment is often presented like in the following link:
http://demonstrations.wolfram.com/JouleExperimentOnFreeExpansion/
The apparent lack of temperature change in this experiment is often used in textbooks to demonstrate that the energy of an ideal gas is only temp. dependent. It is, however, often mentioned that a non-negligible change in temp. for real gases occurs because of intermolecular interactions.
My question: wouldn't temp. change (decrease) also for the ideal case? After all the expanding gas will be doing work against the pressure that is being created in the right chamber? Am I missing something?

Sincerely,
Mike

Chestermiller
Mentor
Yes. Ordinarily you would expect the expanding gas to cool as it expanded. But, in this irreversible expansion, there is another physical mechanism at work. This is often referred to as "viscous heating," although there is minimal actual heat flow. A better term would be "viscous conversion of usable mechanical energy to internal energy. " This is the effect of viscous stresses that develop within the gas, which are determined not by the amount of deformation, but by the rate of deformation. For an ideal gas, the "heating" effect caused by viscous dissipation exactly cancels the cooling effect of the expansion, and the net result is no change in temperature.

In the experiment, there is no work done by the system (the total gas) on its surroundings because the combined chamber is rigid. And there is no heat transfer because the combined chamber is insulated. So, the overall change in internal energy is zero. For an ideal gas, the internal energy is a function only of temperature, so no change in internal energy means no change in temperature. For a real gas, the internal energy is also a function of specific volume, which decreases for the gas. So even though the internal energy is constant, a change in temperature occurs to offset the effect of the change in specific volume.

Chestermiller, thank you for your reply! I understand now were my logic broke. I was fixated on the the gas that is expanding in the left chamber and ignored the gas in the right chamber, which is being compressed as the system equilibrates, and thus the overall work in the system is zero. This should had been obvious since the system doesn't change volume. There is a similar experiment (Joule-Thomson) which was used to measure these small temperature changes in the compression/expansion of real gases. Thank you for your patience in explaining these basics!

Sincerely
Mike

Chestermiller
The question: wouldn't temp. change (decrease) also for the ideal case? After all the expanding gas will be doing work against the pressure that is being created in the right chamber? Am I missing something?

My answer: yes the LHS-gas will do some work on the RHS-gas, but the RHS-gas will receive this work, with w+w=0. Together its only one system with some internal work done that has no effect on the internal energy, and (if n constant) T constant.