Equilibrium Temperature of Gases in Partitioned Container

[Raghav Gupta]

Homework Statement

A container has movable(without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container.
The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat.

The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K
And the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K.
The heat capacities per mole of an ideal monatomic gas are C_v = 3/2 R , C_p= 5/2 R and those for an ideal diatomic gas are C_v= 5/2 R, C_p= 7/2 R.

Q. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be
A. 550 K
B. 525 K
C. 513 K
D. 490 K

Homework Equations

ΔQ = nCΔt

The Attempt at a Solution

The lower one chamber will lose heat as,
ΔQ = 2* C (700 - T)
Upper one will gain as
ΔQ = 2 * C ( T - 400)
What C to take Cv or Cp?
Confusion here.

 

[mooncrater]

Actually at equilibrium they will have the same internal energy. Therefore C_v should be used( for ΔU, not ΔQ)

 

[Raghav Gupta]

C_v...
Since its internal energy...
It always has the formula:
nC_VΔT

But the answer is coming wrong if we take cv in both cases.

 

[mooncrater]

Is it 525 K?

 

[Raghav Gupta]

Is it 525 K?

No. I also did that initially but it is wrong.
It is a tricky question I guess.

 

[mooncrater]

By C_v its 513 K
And by C_P its 525 k...
Are both wrong?

 

[mooncrater]

What I think... : since there is no work done therefore ΔQ=ΔU... (since the partition is rigidly fixed)
And since ΔU=nC_vΔT
Therefore here C_v should be used. What is your thinking?

 

[Raghav Gupta]

By C_v its 513 K
And by C_P its 525 k...
Are both wrong?

Yeah

What I think... : since there is no work done therefore ΔQ=ΔU... (since the partition is rigidly fixed)
And since ΔU=nC_vΔT
Therefore here C_v should be used. What is your thinking?

I was also thinking the same way earlier.
But the question is correct.
There might be something that we must carefully observe in the question?

 

[mooncrater]

A container has movable(without friction) piston on top.

Consider the partition to be rigidly fixed so that it does not move.

What does that mean? One says it will move, other says it won't.

 

[mooncrater]

I think there is a change in volume in the upper compartment.

 

[Raghav Gupta]

See this diagram

 

[Raghav Gupta]

I think there is a change in volume in the upper compartment.

Hmm.. That could be possible.
Then what C we should take for upper compartment?

 

[mooncrater]

I don't think any C would work now..
I was thinking that :
Energy from the bottom part= work done to change volume of upper part+ energy to increase temperature of upper part.
To calculate the work done, we may take the final pressure of the upper part to be equal to atmospheric pressure. The problem here is that we don't know the initial pressure or initial volume. What do you think?

 

[Raghav Gupta]

I don't think any C would work now..
I was thinking that :
Energy from the bottom part= work done to change volume of upper part+ energy to increase temperature of upper part.
To calculate the work done, we may take the final pressure of the upper part to be equal to atmospheric pressure. The problem here is that we don't know the initial pressure or initial volume. What do you think?

That's true.
I saw the solution somewhere and they have taken cp for upper compartment and cv for lower compartment.
Don't know why?
That way the answer D will come.
I know there is constant volume in lower compartment but how constant pressure in upper one?

 

[mooncrater]

A reason for constant pressure can be that since its given that the piston is free therefore initially the pressure inside the upper part would have been atmospheric pressure since it has to stay in equilibrium with the pressure outside. Moreover since the piston is always free therefore the upper part has to stay in equilibrium with the outer pressure always, thus always having a pressure equal to the atmospheric pressure. What do you think?

 

[Raghav Gupta]

A reason for constant pressure can be that since its given that the piston is free therefore initially the pressure inside the upper part would have been atmospheric pressure since it has to stay in equilibrium with the pressure outside. Moreover since the piston is always free therefore the upper part has to stay in equilibrium with the outer pressure always, thus always having a pressure equal to the atmospheric pressure. What do you think?

Thanks, got it.

 

[Chestermiller]

Nice job guys. The only thing I would add is that the outside pressure does not necessarily have to be 1 atm, and the piston does not have to be massless in order for your analysis to work.

Chet