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Ideal gas-thermodynamics, thermal equilibrium

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Two ideal gases are separated by a partition which does not allow molecules to pass from one volume to the other. Gas 1 has: N1, V1, T1, Cv1 for the number of molecules, volume it occupies, temperature in kelvin, and specific heat per molecule at constant volume respectively. Gas 2 has: N2, V2, T2, Cv2. The two gases are in thermal contact and reach a final temperature

    a) find the final temperature and the total change in energy of the combined system. Check your answer for the final temperature when N1=N2, V1=V2. Cv1=Cv2

    b)Evaluate the total change ina quantity H whose differential change is dH=dU+Vdp for each component and for the entire system

    c)evaluate the total change in a quantity A whose differential change is dA=(dU+pdV)/T for each component and for the entire system

    2. Relevant equations


    3. The attempt at a solution

    I already solved for the final temperature for part a, and when evaluated at equal N and V i got Tf=(T2+T1)/2

    i am having trouble finding b and c
  2. jcsd
  3. Apr 2, 2013 #2
    Note that the volume of each component as well as the entire system is constant. That means ## \frac p T = k = const ##. Use this to express both differential quantities via ## dT ##, then integrate, using the results of a) for the limits.
  4. Apr 2, 2013 #3
    thanks voko, ur the man, i solved a and b since ive posted this however, i have yet to solve c, one of the TAs told me that the value "pdV" goes to 0 in the expression dA=(dU+pdV)/T however from here i am lost. i know that dQ=dU+dW and if pdV= 0 than dQ=dU, and i know that dQ=CvdT....but he said something about how this is path dependent and i need to prove that, any help?
  5. Apr 2, 2013 #4
    Since ##U = cT##, where ##c## is some constant, ##dU = c dT##. ## p dV ## is indeed zero because the volume does not change. So all you have is ## dA = \frac {dU} T = \frac {c dT} T ##. I am sure you can integrate that.
  6. Apr 2, 2013 #5
    thanks again voko for being the man, i did do that and i got, A=Cvln(Tf/T1) however my TA said that regardless since heat and work are path dependent i need to prove that dU=cdT since that is only applicable at certain times he said...
  7. Apr 2, 2013 #6
    In #1, you gave the full expression for U. Differentiate it, paying attention to the conditions of the problem.
  8. Apr 2, 2013 #7
    well in step (a) i got, Tf=(N1Cv1T1+T2Cv2N2)/(N1Cv1+N2Cv2), and for dU1=N1Cv1(Tf-T1) dU2=N2Cv2(Tf-T2)


    From here i can see there is no volume dependence, but im not quite sure
  9. Apr 2, 2013 #8
    Even if there were a volume dependence, it would not matter: all volumes are constant in this problem.
  10. Apr 2, 2013 #9
    gotcha, thanks

    If i wanted to solve dA=dU/T is there a way to do it without using CvdT=dQ

    could i perhaps do:

    dA1=dU1/T=N1Cv1dT/T then take the integral from T1 to Tf, and then do the same for dA2 and then add the two together?
  11. Apr 2, 2013 #10
    That's exactly how you should solve this. All that alphabetical soup before dT/T is the constant c I mentioned earlier.
  12. Apr 2, 2013 #11
    oh ok thanks alot,

    essentially i will get A=A1+A2

    = c1(ln(Tf/T1)+c2ln(Tf/T2) where Tf=(c1T1+c2T2)/(c1+c2)

    seems ugly, but theres gotta be some way in which i can clean it up
  13. Apr 2, 2013 #12
    i cant say c1=c2=c can i? because N1 doesnt equal N2
    Last edited: Apr 2, 2013
  14. Apr 2, 2013 #13
    That is right, they are different. All of the gas constants in this problem seem to be different.
  15. Apr 2, 2013 #14

    so final answer i should get


    i dont see a way to break this up further or clean it up other than writing it as ln(Tf/T)
  16. Apr 2, 2013 #15
    thanks again for being so helpful, you truly are the man
  17. Apr 2, 2013 #16
    You are welcome.
  18. Apr 2, 2013 #17
    Your TA couldn't have meant that. He must know that for an ideal gas, dU is always equal to CvdT.
  19. Apr 2, 2013 #18
    thats what i thought, I think he might have not realized that it said ideal gases in the problem
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