Calculate the change of temperature in terms of T

[Sam J]

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 Initially A has its piston fully withdrawn and contains a perfect monatomic gas at temperature T, while B has its piston fully inserted, and the valve is closed.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 operation. The thermal capacity of the cylinders is to be ignored.

Piston B is fully withdrawn and the valve is opened slightly; the gas is then driven as far as it will go into B by pushing home piston A at such a rate that the pressure in A remains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

My attempt:

From intuition, were the piston in cylinder A to remain stationary, then we would have a Joule expansion in which no work is done on/by the system. Therefore, the depression of the piston in A means that work must be being done on the system.

Given that the system is thermally isolated from any surroundings, there can be no gain/loss of heat energy. Therefore the first law of thermo reduces to:

dU=dW

We know that U is a function of T, therefore any work done on the system will increase its temperature.

Let T be the initial temperature of the system. Let T′ be the final temperature. Let p be the pressure. Let V be the volume of each piston.

Given T increases and p remains constant, there must be a corresponding increase in the volume of the gas.

Let V' be the final volume of the gas in cylinder B.

I write down expressions for the ideal gas equation for the initial and final thermodynamic equilibria:

Initial:

pV=nRT

Final:

p(V+V′)=nRT′

Combining and rearranging I derive:

T′=T(1+V′/V)

I become stuck when I am required to quantify the ratio of V' and V. Clearly I need to be able to express V' in terms of V, yet I cannot think of an equation with which this can be done.

 

[Chestermiller]

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 Initially A has its piston fully withdrawn and contains a perfect monatomic gas at temperature T, while B has its piston fully inserted, and the valve is closed.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 operation. The thermal capacity of the cylinders is to be ignored.

Piston B is fully withdrawn and the valve is opened slightly; the gas is then driven as far as it will go into B by pushing home piston A at such a rate that the pressure in A remains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

My attempt:

From intuition, were the piston in cylinder A to remain stationary, then we would have a Joule expansion in which no work is done on/by the system. Therefore, the depression of the piston in A means that work must be being done on the system.

Given that the system is thermally isolated from any surroundings, there can be no gain/loss of heat energy. Therefore the first law of thermo reduces to:

dU=dW

We know that U is a function of T, therefore any work done on the system will increase its temperature.

Let T be the initial temperature of the system. Let T′ be the final temperature. Let p be the pressure. Let V be the volume of each piston.

Given T increases and p remains constant, there must be a corresponding increase in the volume of the gas.

Let V' be the final volume of the gas in cylinder B.

I write down expressions for the ideal gas equation for the initial and final thermodynamic equilibria:

Initial:

pV=nRT

Final:

p(V+V′)=nRT′

Combining and rearranging I derive:

T′=T(1+V′/V)

I become stuck when I am required to quantify the ratio of V' and V. Clearly I need to be able to express V' in terms of V, yet I cannot think of an equation with which this can be done.

Very nicely done so far! You almost have it. You already said that $\Delta U=-W$, where W is the work done by the gas on the piston in cylinder A. This work is done at constant pressure P. I think you meant to say that V' should be the final volume of the gas in cylinder A (since the final volume of the gas in cylinder B is V). So, W = -P(V-V').

 

[Sam J]

Very nicely done so far! You almost have it. You already said that $\Delta U=-W$, where W is the work done by the gas on the piston in cylinder A. This work is done at constant pressure P. I think you meant to say that V' should be the final volume of the gas in cylinder A (since the final volume of the gas in cylinder B is V). So, W = -P(V-V').

Not clear as to how your expression for W helps me here. Without knowing the values of W and p.

I am also wondering why I cannot use the adiabatic constants to help me here. When I combine my expression for T' with

TV^γ-1 = constant

Therefore

T'(V+V')^γ-1 = TV^γ-1

I derive T=T' which I know to be false.

 

[Chestermiller]

Not clear as to how your expression for W helps me here. Without knowing the values of W and p.

P is an item of input data; it is the initial pressure in cylinder A. That's what the problem statement says. So you have:
$$\Delta U=nC_v(T'-T)=P(V-V')$$
where n is the number of moles, and, from the initial conditions, $$n=\frac{PV}{RT}$$
So,
$$\frac{PVC_v}{RT}(T'-T)=P(V-V')$$
Notice that P cancels out of this equation. So, you don't need to know P after all! Now you have two equations in the two unknowns T' and V'

I am also wondering why I cannot use the adiabatic constants to help me here. When I combine my expression for T' with

TV^γ-1 = constant

Therefore

T'(V+V')^γ-1 = TV^γ-1

I derive T=T' which I know to be false.

You can't use these equations because they don't apply to your system. Your system is not experiencing a reversible change, which is required by these equations. The change in your system is highly irreversible. Also, for those equations, all the gas is in one cylinder throughout and the pressure at the piston face is changing during the deformation. For your process, the pressure at the piston face (piston A) is constant throughout the deformation. This is the only place where work is being done on the gas.

 

[Chestermiller]

Is anyone interested in continuing the solution of this interesting problem? (Apparently, the OP has lost resolve.) If no one else responds within the next couple of days, I'm going to close this thread.

Chet