# Temperature change of an ideal gas during an adiabatic free expansion

During an adiabatic process there is no heat exchange and thus, q = o. During a free expansion no work is done and so w = o.

Since q = U + w, thus we can say that during the overall process, change in internal energy is also 0.

And since in an ideal gas, internal energy energy only depends on the kinetic energy, and K.E is proportional to temperature, we can say that there is no overall change in temperature.

My doubt is whether there is a point during this process where the temperature varies and then returns back to the original position, or is the process isothermal?

If there is a change in temperature during the process, how do we reason for this? How do we reason whether the temperature decreases then returns back or whether it increases and then returns back?

My doubt is whether there is a point during this process where the temperature varies and then returns back to the original position, or is the process isothermal?

There is no single "temperature" of the gas during the free expansion. The same applies to pressure. During the free adiabatic expansion temperature and pressure gradients exist within the gas and these gradients are changing in time.

Although the initial and final equilibrium temperatures are the same, the process is not isothermal (it is not a constant temperature process). The ideal gas law only applies under equilibrium conditions. So for the adiabatic free expansion $$P_{f}V_{f}=P_{i}V_{i}$$ means $$T_{f}=T_{i}$$, only for the initial and final states. But it does not necessarily mean the temperature $$T$$ is constant during the expansion between the equilibrium states.

Let's say you had thermometers randomly located on both sides the chamber. If the gas is initially internally in thermal equilibrium, those thermometers on the gas side would all theoretically read the same (the initial equilibrium temperature of the gas). Those on the vacuum side would theoretically read zero since there is no gas possessing kinetic energy.

Now an opening is created and the gas starts rushing into the vacuum side. All the thermometers will start having different readings reflecting the temperature (and pressure) gradients created during the expansion. So for this scenario, the question for you is which thermometer reading shall be called "the" temperature of the gas during the expansion?.

On the other hand, if one waits long enough the readings of all the thermometers will eventually converge to the equal the initial temperatures measured on the original gas side. Although during the expansion the distribution of the speeds (and thus kinetic energy) of the gas molecules would not follow the Maxwell-Boltzmann distribution that is associated with a specific temperature, the total kinetic energy and thus the internal energy does not change.

Bottom line: The temperature (and pressure) of the gas only has meaning when the gas is in internal equilibrium, i.e., before the expansion occurs and after it ends and the gas is allowed to re-equilibriate (settle down).

Hope this helps.

• Hmmm if this is so then how would you explain joule Thompson effect? You wouldn't be able to define it mathematically as you won't be able to measure the thermodynamic variables – Buraian Sep 30 '20 at 10:12
• The Joule Thompson effect in which there is a temperature change applies to a real gas not an ideal gas. The subject of the OP is an ideal gas not a real gas. – Bob D Sep 30 '20 at 10:24
• Sure but the point I brought up applies for real gas as well. How do you define the state variables mathematically if it is non equilibrium?/ speak of joule thompson coefficent then? – Buraian Sep 30 '20 at 10:26
• An equation of state for a real or ideal gas applies to equilibrium conditions. How would you mathematically define the temperature of the gas, real or ideal, when not in equilibrium? – Bob D Sep 30 '20 at 10:44
• Yes that is the point I have doubt in. So, if that is the case it should be not possible to mathematically describle joule expansion co-efficent as it should be a non-equilibrium process ( the gas flows at very high velocity.. which I Think will cause non equilibrium) – Buraian Sep 30 '20 at 10:46