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Cylinder with piston

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    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 Cv = 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?

    2. Relevant equations
    PV = NkT

    3. The attempt at a solution
    From the value of Cv 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?
  2. jcsd
  3. Nov 18, 2013 #2

    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?
  4. Nov 18, 2013 #3
    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.
  5. Nov 18, 2013 #4
    Have you answered the other questions I asked? The volume of the left side is 2V0-V.
  6. Nov 18, 2013 #5
    The final pressure is just 2P so 2NkT0/V0 and for adiabatic compression isn't Vf(Tf)7/2 = V0(T0)7/2 ?
  7. Nov 18, 2013 #6
    Use PVγ = Const. to get the final volume on the right side.
  8. Nov 18, 2013 #7
    How do I get the constant to be in terms I can use?
  9. Nov 18, 2013 #8
    P=2P0. Solve for V in terms of V0.
  10. Nov 18, 2013 #9
    Ok, so V = V0/27/9 which means T = 22/9T0 for the right side?

    And from this Vleft = V0(2 - 27/9) and Tleft = 211/9T0 ?
  11. Nov 18, 2013 #10
    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.
  12. Nov 18, 2013 #11
    I see what I did wrong, so for the left V = 2V0 - Vright = 1.417V0 so T = 2.834T0
  13. Nov 18, 2013 #12
  14. Nov 18, 2013 #13
    Thanks so much for your help
  15. Nov 19, 2013 #14
    I have another question, would the work done on the right side just be W = -2P(0.583Vo - Vo) or would I have to do something else with the pressure since it increases to 2P instead of being 2P the whole time?
  16. Nov 19, 2013 #15
    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.
  17. Nov 19, 2013 #16
    I used a graphing method and got that the work done is 0.21NkTo, but I guess that is incorrect because I tried CvΔT and got 0.58NkTo. I use the change in temp of the right side right?
  18. Nov 19, 2013 #17
    I got the 0.58NkTo result doing it both ways.
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