# Isothermal and Adiabatic piston problems

1. Feb 11, 2013

### Tito_Tileto

1. The problem statement, all variables and given/known data

This is Problem 12.5 in Blundell's Thermal Physics book.

Two Thermally insulated cylinders, A and B, of equal volume, both equipped with pistons, are connected by a valve. Initially A has its piston fully withdrawn and contains a perfect monatomic gas at temperature T, while B has its piston fully inserted, and the valve is closed. Calculate the final temperature of the gas after the following operations, which each start with the same initial arrangement. The thermal capacity of the cylinders is to be ignored.

a) The valve is fully opened and the gas slowly drawn into B by pulling out the piston B; piston A remains stationary

b) Piston B is fully withdrawn and the valve is opened slightly; the gas is then driven as far as it will go into B by pushing home piston A at such a rate that the pressure in A remains constant: the cylinders are in thermal contact.

2. Relevant equations

TV^γ-1= Constant , Cv=$\frac{3}{2}$ NKb

pV^γ= Constant , Cp=$\frac{5}{2}$ NKb

ΔW=-ΔQ , γ=$\frac{5}{3}$

3. The attempt at a solution

This class is kicking my butt. I have a difficult time processing examples that do not have physical numbers in them.

For a) I believe you'd use TV^γ-1 Since you could relate temperature to volume. I'm not sure how exactly to calculate the temperature.

b) This is a quasistatic process. I guess since you drive the piston A down to force gas through the valve opening once the gas is completely in piston B, there is no way to distinguish which piston is A or B, the situation is the same as in the beginning so this would be a reversible process. I believe there would be no change in T and it would be the same as it was at the start. I do not know how to show/ prove this though.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 15, 2013

### rude man

(a) This is an adiabatic expansion from V0 to 2V0. So first determine the new pressure p1 in terms of the starting pressure p0, knowing that the volume has doubled.

Then what is the equation of state for the doubled volume, and then can you find suitable substitutions with what you've determined to this point to compute the final temperature?

(b) Think of the closed system of the two cylinders, isolated from the outside world.

What is the change in internal energy ΔU of the system by the 1st law?
What is ΔU as a function of temperature change ΔT for an ideal gas?
Then make those two equal each other.
Solve for ΔT, then final temperature = T + ΔT.

(OK CM, have at it ...)

3. Feb 15, 2013

### Staff: Mentor

Hi Rude Man. Sounds good. I like your thinking on part b.

Chet

4. Feb 15, 2013

### ChaseRLewis

1. You only need one equation. The only way to determine the final temperature is as a relationship to T-initial (To) which I don't see as a given but you can solve it symbolically.

if you have TVy-1 = constant, this equation assumes constant energy within the reaction volume which since you have stated the system is adiabatic is true. So you can equate the initial and final states since they have the same energy. No technical need for pressure in this equation since you have all the variables already right here. Also remember y is considered a constant value for monoatomic gases.

This is an adiabatic expansion problem. rude man's way works just extra steps.
2.This is the most magical valve ever. Which makes this problem kinda hard to understand. Seems like to drive gas through it you have to apply work. Such that the gas loss results in constant pressure. Until all the gas is out of A.

so what would be the total work to drive out all of the gas from A into B. So it will have the same volume and moles but T will be greater because of the work done. dU = dW = PdV = nCvdT remember constant A pressure (lucky dog).

Last edited: Feb 15, 2013
5. Feb 15, 2013

### Staff: Mentor

For part b, your suggestion sounds exactly like what Rude Man was suggesting, although, IMHO, not as clear. For part a, you are correct to say that there is no need to perform the intermediate step of calculating the final pressure, although, you will not get the wrong answer by doing so. So Rude Man was correct on this part too.

6. Feb 15, 2013

### rude man

Actually, folks, I didn't solve for the final pressure explicitly. It's just a parameter that is used and drops out to get the final temperatures.

Thanks for the thumbs-up, CM, you'll get me thinking thermo for real yet ...

I also don't see anything 'magical' about the valve in part b. It's just a small aperture that, were no pressure applied by the A piston, the gas would leak out at a slower rate and non-constant pressure. Easy to implement and visualize. Think throttling process, except that of course the additional requirements of cylinder thermal isolation and constant B pressure are not realized.

CM, question for you on part b: I don't understand the need for thermal contact between the cylinders. Seems like that's not necessary? Even if heat were exchaged between them, the only energy added to the total insulated system is still p0*V0 and the change in internal energy would still be ncVΔT = p0*V0?

Also, would you send me your answers via personal note? Thanks in any case.