Expansion of gas into cylinder

[CAF123]

Homework Statement

Consider two rigid cylinders, P and Q, with respective volumes V_P and V_Q with V_P << V_Q. They are connected by a capillary. Initially cylinder Q is empty and cylinder P contains an ideal monatomic gas at 300K and pressure 1 bar. A moveable piston in P is used to maintain the pressure in P at 1 bar completely expelling all the gas into Q through the capillary. The system is thermally isolated and the walls/capillary have negligible heat capacity. What is the final temperature of the gas once it has reached thermal equilibrium in Q?

Homework Equations

Thermodynamic definition of work: Work done by piston W = -PdV
Internal energy of monatomic gas: U = (3/2)nRT

The Attempt at a Solution

Since the pressure in the cylinder P is unchanged, there is no pressure difference across the piston and so the external pressure supplied by the piston is equal to 1 bar. The piston does work on the n moles of gas moving them into the cylinder Q. Since the system is thermally isolated and the walls have negligible heat capacity, this work goes into increasing the internal energy of the gas. So write -P_i(V_Q-V_P) = (3/2)nR(T_Q-T_P), where the system is the n moles of gas originally in P ending up in Q.

The term V_Q-V_P ≈ V_Q by the question so sub this in, where V_Q = nRT_Q/P_i. Rearranging, I get that T_Q = (3/5)T_P, but I don't think this makes sense. I argued that the final pressure inside Q is the same as in P since if it wasn't the gas would flow back into P if it was greater and it wouldn't reach thermal equilibrium. So Pf = Pi of the n moles, and from ideal gas law, if V is increased, then T must go up to keep P constant.

Many thanks.

 

[TSny]

Hello, CAF123.

Your expression for the work done by the piston in P doesn't seem correct to me. What is the volume swept out by the piston by the time all the gas has left P?

Once all the gas is in Q, you can assume that the gas remains in Q because the piston in P will block the gas from returning to P. But that doesn't mean that the pressure in Q will equal the initial pressure in P.

 

[CAF123]

Hi TSny,

Your expression for the work done by the piston in P doesn't seem correct to me. What is the volume swept out by the piston by the time all the gas has left P?

The piston acts over a finite volume V_P, so the work it does is P_iV_P = nRT_P and I end up with T_Q=(5/3)T_P.
Is that better and if so, is my argument about the piston applying the external pressure P_i correct?

 

[TSny]

That looks good to me. Yes, the piston in P maintains a constant pressure of 1 bar inside P as the gas moves from P to Q. The pressure in Q is undefined until the gas settles down to thermal equilibrium in Q.

 

[CAF123]

The pressure in Q is undefined until the gas settles down to thermal equilibrium in Q.

Is that because the expansion of a portion of n moles of gas into Q is an irreversible process and only until the expansion has stopped, when all n moles are in Q and spread out uniformly, can we again define thermodynamic variables.

 

[TSny]

Yes. I think that's right.