Why is temperature constant after gas has expanded?

[Lnewqban]

.... There is also the possibility of some sort of longitudinal wave being produced to complicate matters!

With the proper timing and velocity through the valve, an oscillation of pressure values could occur between both tanks, before pressure equalizes.

For the first cycle at least, low pressure inside the left A cylinder could lead to lower temperature in there, while simultaneously, the opposite happens inside the right B cylinder.

Please, see:
https://en.wikipedia.org/wiki/Kadenacy_effect

https://en.wikipedia.org/wiki/Expansion_chamber

 

[Steve4Physics]

That's new to me but I can see how it makes complete sense in this situation. Thankyou.

@vcsharp2003, please note I edited Post #29 as I thought something I wrote wasn’t clear.

In a little more detail...

For an adiabatic compression or expansion of an ideal gas, the total kinetic energy of the particles changes if particles bounce off a moving wall. However that’s not the case in your Post #1 problem, so the total kinetic energy is constant in your problem.

(It's a useful exercise to think about how, say compressing a gas by pushing-in a piston, increases the kinetic energy of the gas particles. (You consider the behaviour of particles bouncing off a moving wall.).)

 

[Chestermiller]

Mass Balance: $$n_1+n_2=n_0$$where $n_0$ is the number of moles of gas initially in the left chamber, $n_1$ is the final number of moles of gas in the left chamber, and $n_2$ is the final number of moles in the right chamber. From the ideal gas law, this gives: $$\frac{PV}{RT_1}+\frac{PV}{RT_2}=\frac{P_0V}{RT_0}$$where P is the final pressure in both chambers, V is the volume of each of the two chambers, $P_0$ is the initial pressure in the left chamber, $T_1$ is the final temperature in the left chamber, $T_2$ is the final temperature in the right chamber, and $T_0$ is the initial temperature of the gas in the left chamber. So, $$\frac{T_0}{T_1}+\frac{T_0}{T_2}=\frac{P_0}{P}\tag{1}$$Energy Balance: $$\Delta U=n_1C_V(T-T_0)+n_2C_v(T_2-T_0)=0$$, or, combining with the ideal gas law:$$1-\frac{T_0}{T_1}+1-\frac{T_0}{T_2}=0$$or$$\frac{T_0}{T_1}+\frac{T_0}{T_2}=2\tag{2}$$So, combining Eqns. 1 and 2, we have $$P=\frac{P_0}{2}\tag{3}$$Thus, the final pressure in both chambers is half the initial pressure in the left chamber.

The gas finally remaining in the left chamber has experienced an adiabatic reversible expansion in gradually pushing the gas ahead of it into the stopcock, and into the right chamber. Thus, $$\frac{T_1}{T_0}=\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}}\tag{4}$$Next, combining Eqns. 2 and 4 yields:$$\frac{T_2}{T_0}=\frac{1}{2-2^{\frac{\gamma-1}{\gamma}}}>1$$

 

[Chestermiller]

Given these results, what is the entropy change for this version of the problem and how does it compare quantitatively with the entropy change for the other version?