I Movable Wall in an Adiabatic System: Same Final Temperature?

[Filipeml]

TL;DR Summary

A monatomic and a diatomic gas, initially at the same temperature and pressure, are separated by a movable, heat-conducting wall in an adiabatic container. How do they reach equilibrium?

Consider an adiabatic container divided into two chambers by a movable, heat-conducting wall. One side contains a monatomic ideal gas, while the other contains a diatomic ideal gas. Initially, both gases are at the same temperature and pressure. Over time, the wall moves until a new thermodynamic equilibrium is reached.

• How can the final position of the wall be determined based on the thermodynamic properties of the gases?
• Will the final temperatures of the two gases be the same? If not, what physical principle justifies this difference?
• If the gases were real instead of ideal, how might the system’s behavior deviate from the predictions of the ideal gas model?

 

[anuttarasammyak]

Are some initial p, V, N of the two gases given ?

 

[anuttarasammyak]

. Initially, both gases are at the same temperature and pressure.

Thanks. So I do not think the wall moves.

 

[Chestermiller]

I'm assuming that the wall is forced to move slowly by a rod passing through the wall of the container.

Let n be the total number of moles of the gases in the two chambers and let V be the total volume of the chamber. Let $\xi$ characterize to the fractional volume of the gases in the two chambers ($-1<\xi<+1$), such that $$V_1=0.5(1+\xi)V$$and $$V_2=0.5(1-\xi)V$$so that, at all times, $$V_1+V_2=V$$Let $\xi_0$ be the fractional volumes at the initial state. For an ideal gas, show that, at all times, $$n_1=0.5(1+\xi_0)n$$and $$n_2=0.5(1-\xi_0)n$$

 

[Chestermiller]

I am amazed that no one in PF has made an attempt to solve this very interesting problem.

 

[Filipeml]

Estou assumindo que a parede é forçada a se mover lentamente por uma haste que passa pela parede do contêiner.

Seja n o número total de mols dos gases nas duas câmaras e seja V o volume total da câmara. Seja $\xi$ caracterizado pelo volume fracionário dos gases nas duas câmaras ($-1<\xi<+1$), tal que $$V_1=0,5(1+\xi)V$$e $$V_2=0,5(1-\xi)V$$de modo que, em todos os momentos, $$V_1+V_2=V$$Sejam $\xi_0$ os volumes fracionários no estado inicial. Para um gás ideal, mostre que, em todos os momentos, $$n_1=0,5(1+\xi_0)n$$e $$n_2=0,5(1-\xi_0)n$$

Obrigado por sua resposta!

 

[Philip Koeck]

TL;DR Summary: A monatomic and a diatomic gas, initially at the same temperature and pressure, are separated by a movable, heat-conducting wall in an adiabatic container. How do they reach equilibrium?

Consider an adiabatic container divided into two chambers by a movable, heat-conducting wall. One side contains a monatomic ideal gas, while the other contains a diatomic ideal gas. Initially, both gases are at the same temperature and pressure. Over time, the wall moves until a new thermodynamic equilibrium is reached.

• How can the final position of the wall be determined based on the thermodynamic properties of the gases?
• Will the final temperatures of the two gases be the same? If not, what physical principle justifies this difference?
• If the gases were real instead of ideal, how might the system’s behavior deviate from the predictions of the ideal gas model?

I'm also confused. Why are they not already in equilibrium?
Both gases have the same temperature so there should be no heat transfer through the wall.
Both gases have the same pressure so the wall shouldn't move.
What am I missing?

 

[Swamp Thing]

I'm also confused. Why are they not already in equilibrium?
Both gases have the same temperature so there should be no heat transfer through the wall.
Both gases have the same pressure so the wall shouldn't move.
What am I missing?

As Chestermiller said,

I'm assuming that the wall is forced to move slowly by a rod passing through the wall of the container.

So Question #1 from the op becomes,

Q #1: if you can't see the moving rod, and someone told you the final pressures and temperatures, then could you figure out how much the rod moved?

Q #2 : Would you expect them to report the same temperature for both sides of the container?

Q #3: What if they were real gases? How would things change?

 

[Philip Koeck]

As Chestermiller said,


So Question #1 from the op becomes,

Q #1: if you can't see the moving rod, and someone told you the final pressures and temperatures, then could you figure out how much the rod moved?

Q #2 : Would you expect them to report the same temperature?

Q #3: What if they were real gases? How would things change?

The problem statement confused me.
The wall doesn't move by itself "until they reach equilibrium".
The system is already in equilibrium and the wall is moved by an external force, so work is done on the system and it moves to a new equilibrium.