# Two gases in a cylinder separated by an insulating piston.

• Tiberius47
In summary: If you consider the combination of the two chambers and the piston as your system, then the system does no work on the surroundings. Therefore, since no work is done on the surroundings, the p-v relationship is not constant.
Tiberius47

## Homework Statement

A cylinder of cross-sectional area A is divided into two chambers 1 and 2, by means of a frictionless piston. The piston as well as the walls of the chambers are heat-insulating, and the chambers initially have equal length L. Both chambers are filled with 1 mol of helium gas, with initial pressures 2Po and Po, respectively. The pisotn is then allowed to slide freely, whereupon the gas in chamber 1 pushes the piston a distance a to equalize the pressures to P.
(a) Find the distance a traveled by the piston.
(b) If W is the work done by the gas chamber 1, what are the final temperatures T1 and T2 in the two chambers? What is the final pressure P?
(c) Find W, the work done.

PV=NRT

## The Attempt at a Solution

I know that the pressures will be equal at equilibrium but I'm not sure how to set up the problem.

To tackle this problem, I believe you first need to identify what kind of process is taking place (isobaric, isothermal, isochoric, adiabatic, etc). Which process do you think this is?

Mute said:
To tackle this problem, I believe you first need to identify what kind of process is taking place (isobaric, isothermal, isochoric, adiabatic, etc). Which process do you think this is?

Well given that the two chambers are insulated I would say adiabatic. So the change in heat would be zero so:

ΔU+W=0

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Tiberius47 said:
Well given that the two chambers are insulated I would say adiabatic. So the change in heat would be zero so:

ΔU+W=0

Since mute is not on the air -
yes, adiabatic. So what p-v relationship can you ascribe to each half?

This is a very interesting and "meaty" problem. The first thing to do is to calculate the initial temperatures in each of the two chambers in terms of p0, V0=AL, and R. What are your results for this? The next step is to skip part (a) and most of part (b), and determine the final pressure p. To do this, you should first answer the question "How much work is done by the combined system in the two chambers?" Once you answer this question, you will be able to ascertain the answer to the question "What is the change in internal energy for the combined system?" Once you know the answer to this question, you should be able to answer the question "How is the sum of temperatures in the two chambers in the final equilibrium state related to the sum of the temperatures in the two chambers prior to release of the piston?" Next, express the volume of each chamber algebraically in terms of its temperature and pressure in the final state. Next, what is the sum of the two chamber volumes in the final state? Express this algebraically in terms of the temperatures and the pressure in the final state, and recognize that the algebraic expression involves the sum of the temperatures. Solve for the pressure.
Of course, this all is just the first step in the solution to the problem as posed. But, it gives you a good start.

Not necessary to solve for initial temperatures. 3 equations yield final pressure and both final volumes. Thus arriving at final temperatures. Then the 1st law gets us W.

I am assuming ideal gases.

EDIT: oops, we do too need the initial temperatures, but just to find W.

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rude man said:
Not necessary to solve for initial temperatures. 3 equations yield final pressure and both final volumes. Thus arriving at final temperatures. Then the 1st law gets us W.

I am assuming ideal gases.

Are you assuming that pVγ= constant? That is correct if the piston is forced to move slowly, but if the piston is free to slide (as specified in the problem statement), that is not the correct relationship to use. Better be very careful.

Chet

Chestermiller said:
Are you assuming that pVγ= constant? That is correct if the piston is forced to move slowly, but if the piston is free to slide (as specified in the problem statement), that is not the correct relationship to use. Better be very careful.

Chet

I understand that that relationship does not hold for the nonequilibrium states in transitioning from the initial to the final state, but I did think it applied to the initial and final equilibrium states.

I can't wait to see this thread through ...

rude man said:
I understand that that relationship does not hold for the nonequilibrium states in transitioning from the initial to the final state, but I did think it applied to the initial and final equilibrium states.

Actually, no. If you consider the combination of the two chambers and the piston as your system, then the system does no work on the surroundings. Therefore, since no heat enters or leaves the system, the change in internal energy of the system between the initial and final equilibrium states is zero. This means that, for an ideal gas, the sum of the temperatures in the two chambers is the same in the final equilibrium state as in the initial equilibrium state. This gives you enough information to calculate the final equilibrium pressure of the system. Of course, because of the fact that the piston is insulated, the final temperatures in the two chambers will not be equal.

Thanks for the revelation, Chester. But what if the piston were sufficiently massive to effect a quasi-static transformation - there would admittedly be kinetic energy of the piston to consider, but the beginning and end k.e.'s would both be zero?

EDIT: I guess thre would be no change in the system. The only way to effect a quasi-static process here is to allow the higher-pressure gas to apply work externally, right?

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rude man said:
Thanks for the revelation, Chester. But what if the piston were sufficiently massive to effect a quasi-static transformation - there would admittedly be kinetic energy of the piston to consider, but the beginning and end k.e.'s would both be zero?

EDIT: I guess thre would be no change in the system. The only way to effect a quasi-static process here is to allow the higher-pressure gas to apply work externally, right?

Yes. You're right. The piston ke would be zero at the beginning and end, so it would have no effect on the final equilibrium state.

Actually, if you did it quasi-staticaly, by,say, attaching a rod to the piston, you would have to take into account both the high pressure and the low pressure chamber pressures, and the work on the surroundings would be the difference between the two pressures integrated over the change in volume of one of the chambers.

## 1. What is the purpose of the insulating piston in a cylinder containing two gases?

The insulating piston acts as a barrier between the two gases, preventing them from mixing and allowing for the separate study of each gas's properties.

## 2. How does the pressure of the gases change when the insulating piston is pushed down?

When the insulating piston is pushed down, the volume of the gas decreases, causing an increase in pressure due to Boyle's Law (P1V1 = P2V2).

## 3. Can the gases in a cylinder separated by an insulating piston reach thermal equilibrium?

No, the insulating piston prevents heat transfer between the two gases, so they cannot reach thermal equilibrium.

## 4. What factors determine the behavior of the gases in a cylinder separated by an insulating piston?

The behavior of the gases is determined by the initial temperature, volume, and pressure of each gas, as well as the properties of the insulating piston itself.

## 5. How can the insulating piston be used to study the properties of gases?

The insulating piston allows for the separate manipulation and study of each gas, allowing scientists to observe how changes in volume, pressure, and temperature affect the behavior of each gas individually.

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