Cylinder w/Piston: Final Temp & Vol Answered in Terms of Nk, T0, V0

[S_Flaherty]

Homework Statement

A cylinder is closed at both ends and has thermally insulating walls. It is divided into 2 parts (left and right) by a movable, frictionless, thermally insulating piston. There is a heating coil in the left compartment, and there are N molecules of an ideal gas, with heat capacity C_v = 7/2 Nk, in each compartment. Originally, the volume and temperature on each side are V₀ and T₀. The left side is slowly warmed with the coil until its pressure doubles. Give answers to the following in terms of Nk, T₀, and V₀, only.

What are the final temperature and volume on the right side?

Homework Equations

PV = NkT

The Attempt at a Solution

From the value of C_v I know that the degrees of freedom are 7, so U = (7/2)NkT for each side before the piston moves. Since the piston is thermally insulated does that mean the temperature on the right side won't change? So T = T₀? If that's the case then V = V₀/2 wouldn't it? Is it that simple or am I completely missing something?

 

[Chestermiller]

Since the piston is thermally insulated does that mean the temperature on the right side won't change? So T = T₀? If that's the case then V = V₀/2 wouldn't it? Is it that simple or am I completely missing something?

No, the temperature on the right side will change, since the gas gets compressed. From the ideal gas law, what is the pressure in each of the chambers to begin with? What can you say about the relationship between the pressures in the two chambers after the system re-equilibrates? Let V be the final volume of the right chamber. In terms of V and V₀, what is the final volume of the left chamber. The gas in the right chamber gets compressed adibatically and reversibly. How is the final pressure and the final temperature related to V/V₀, T₀, and P₀? What ratio of V/V₀ is required to double the pressure in the right chamber? What is the final volume of the left chamber? If you know the volume of the left chamber and its pressure, what is its final temperature?

 

[S_Flaherty]

No, the temperature on the right side will change, since the gas gets compressed. From the ideal gas law, what is the pressure in each of the chambers to begin with? What can you say about the relationship between the pressures in the two chambers after the system re-equilibrates? Let V be the final volume of the right chamber. In terms of V and V₀, what is the final volume of the left chamber. The gas in the right chamber gets compressed adibatically and reversibly. How is the final pressure and the final temperature related to V/V₀, T₀, and P₀? What ratio of V/V₀ is required to double the pressure in the right chamber? What is the final volume of the left chamber? If you know the volume of the left chamber and its pressure, what is its final temperature?

The pressure of each side at first is P = NkT₀/V₀. The pressure has to be equal on both sides for the piston to stop moving right? So after the piston moves each side has a pressure of 2P. I'm not sure how to get the volume of the left side in terms of V and V₀.

 

[Chestermiller]

The pressure of each side at first is P = NkT₀/V₀. The pressure has to be equal on both sides for the piston to stop moving right? So after the piston moves each side has a pressure of 2P. I'm not sure how to get the volume of the left side in terms of V and V₀.

Have you answered the other questions I asked? The volume of the left side is 2V₀-V.

 

[S_Flaherty]

Have you answered the other questions I asked? The volume of the left side is 2V₀-V.

The final pressure is just 2P so 2NkT₀/V₀ and for adiabatic compression isn't V_f(T_f)^7/2 = V₀(T₀)^7/2 ?

 

[Chestermiller]

The final pressure is just 2P so 2NkT₀/V₀ and for adiabatic compression isn't V_f(T_f)^7/2 = V₀(T₀)^7/2 ?

Use PV^γ = Const. to get the final volume on the right side.

 

[S_Flaherty]

Use PV^γ = Const. to get the final volume on the right side.

How do I get the constant to be in terms I can use?

 

[Chestermiller]

How do I get the constant to be in terms I can use?

$$PV^\gamma=P_0V_0^\gamma$$
P=2P₀. Solve for V in terms of V₀.

 

[S_Flaherty]

$$PV^\gamma=P_0V_0^\gamma$$
P=2P₀. Solve for V in terms of V₀.

Ok, so V = V₀/2^7/9 which means T = 2^2/9T₀ for the right side?

And from this V_left = V₀(2 - 2^7/9) and T_left = 2^11/9T₀ ?

 

[Chestermiller]

Ok, so V = V₀/2^7/9 which means T = 2^2/9T₀ for the right side?

And from this V_left = V₀(2 - 2^7/9) and T_left = 2^11/9T₀ ?

The right side looks OK, but you did the algebra wrong for the left side (both results). Also, please calculate out the actual numbers. You get Tleft by using the ideal gas law.

 

[S_Flaherty]

The right side looks OK, but you did the algebra wrong for the left side (both results). Also, please calculate out the actual numbers. You get Tleft by using the ideal gas law.

I see what I did wrong, so for the left V = 2V₀ - V_right = 1.417V₀ so T = 2.834T₀

 

[Chestermiller]

I see what I did wrong, so for the left V = 2V₀ - V_right = 1.417V₀ so T = 2.834T₀

yes!

 

[S_Flaherty]

yes!

Thanks so much for your help

 

[S_Flaherty]

yes!

I have another question, would the work done on the right side just be W = -2P(0.583V_o - V_o) or would I have to do something else with the pressure since it increases to 2P instead of being 2P the whole time?

 

[Chestermiller]

I have another question, would the work done on the right side just be W = -2P(0.583V_o - V_o) or would I have to do something else with the pressure since it increases to 2P instead of being 2P the whole time?

You could get the work done on the right side in two different ways, but this was not one of them. You could integrate PdV from the initial to the final volume; or you could use the final temperature to determine the change in internal energy CvΔT, since no heat enters or leaves the right chamber. Both these methods will give the same result.

 

[S_Flaherty]

You could get the work done on the right side in two different ways, but this was not one of them. You could integrate PdV from the initial to the final volume; or you could use the final temperature to determine the change in internal energy CvΔT, since no heat enters or leaves the right chamber. Both these methods will give the same result.

I used a graphing method and got that the work done is 0.21NkT_o, but I guess that is incorrect because I tried CvΔT and got 0.58NkT_o. I use the change in temp of the right side right?

 

[Chestermiller]

I used a graphing method and got that the work done is 0.21NkT_o, but I guess that is incorrect because I tried CvΔT and got 0.58NkT_o. I use the change in temp of the right side right?

I got the 0.58NkT_o result doing it both ways.