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Thermodynamics in partition

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A container has movable(without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container.
    The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat.

    The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K
    And the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K.
    The heat capacities per mole of an ideal monatomic gas are Cv = 3/2 R , Cp= 5/2 R and those for an ideal diatomic gas are Cv= 5/2 R, Cp= 7/2 R.

    Q. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be
    A. 550 K
    B. 525 K
    C. 513 K
    D. 490 K

    2. Relevant equations
    ΔQ = nCΔt

    3. The attempt at a solution
    The lower one chamber will lose heat as,
    ΔQ = 2* C (700 - T)
    Upper one will gain as
    ΔQ = 2 * C ( T - 400)
    What C to take Cv or Cp?
    Confusion here.
     
  2. jcsd
  3. Apr 21, 2015 #2
    Actually at equilibrium they will have the same internal energy. Therefore Cv should be used( for ΔU, not ΔQ)
     
  4. Apr 21, 2015 #3
    But the answer is coming wrong if we take cv in both cases.
     
  5. Apr 21, 2015 #4
    Is it 525 K?
     
  6. Apr 21, 2015 #5
    No. I also did that initially but it is wrong.
    It is a tricky question I guess.
     
  7. Apr 21, 2015 #6
    By Cv its 513 K
    And by CP its 525 k....
    Are both wrong?
     
  8. Apr 21, 2015 #7
    What I think.... : since there is no work done therefore ΔQ=ΔU...... (since the partition is rigidly fixed)
    And since ΔU=nCvΔT
    Therefore here Cv should be used. What is your thinking?
     
  9. Apr 21, 2015 #8
    Yeah
    I was also thinking the same way earlier.
    But the question is correct.
    There might be something that we must carefully observe in the question?
     
  10. Apr 21, 2015 #9
    What does that mean? One says it will move, other says it won't.
     
  11. Apr 21, 2015 #10
    I think there is a change in volume in the upper compartment.
     
  12. Apr 21, 2015 #11
    See this diagram

    image.jpg
     
  13. Apr 21, 2015 #12
    Hmm.. That could be possible.
    Then what C we should take for upper compartment?
     
  14. Apr 21, 2015 #13
    I dont think any C would work now..
    I was thinking that :
    Energy from the bottom part= work done to change volume of upper part+ energy to increase temperature of upper part.
    To calculate the work done, we may take the final pressure of the upper part to be equal to atmospheric pressure. The problem here is that we dont know the initial pressure or initial volume. What do you think?
     
  15. Apr 21, 2015 #14
    That's true.
    I saw the solution somewhere and they have taken cp for upper compartment and cv for lower compartment.
    Don't know why?
    That way the answer D will come.
    I know there is constant volume in lower compartment but how constant pressure in upper one?
     
  16. Apr 21, 2015 #15
    A reason for constant pressure can be that since its given that the piston is free therefore initially the pressure inside the upper part would have been atmospheric pressure since it has to stay in equilibrium with the pressure outside. Moreover since the piston is always free therefore the upper part has to stay in equilibrium with the outer pressure always, thus always having a pressure equal to the atmospheric pressure. What do you think?
     
  17. Apr 21, 2015 #16
    Thanks, got it.
     
  18. Apr 21, 2015 #17
    Nice job guys. The only thing I would add is that the outside pressure does not necessarily have to be 1 atm, and the piston does not have to be massless in order for your analysis to work.

    Chet
     
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