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Physics Thermodynamics - Finding Work

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1.20 mol sample of an ideal diatomic gas at a pressure of 1.20 atm (P1) and temperature of 380 K (T1) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 680 K (T2) and 1.83 atm (P2).

    (b) Determine the work done by the gas.


    2. Relevant equations

    (T2 - T1) / (P2 - P1) = constant

    V2 / V1 = P1T2 / P2T1

    PV = nRT

    W = integral of PV

    3. The attempt at a solution

    The answer is supposed to be [nR(P1T2 - P2T1) ln(P2/P1)] / (P2 - P1)
    but I can't figure out how to get there despite the hours I have poured into this problem. All I know is that volume is not constant in order for there to be work done. Also, I have the vague idea of finding the linear equation or relationship between T and P using that equation to plug in T in PV = nRT. Then, technically, there would be a graph for PV and I could solve for V1 and V2, which can be the limits of integration for the integral of PV (to get work). But isn't that way too complicated, especially as an integral...so I was wondering whether someone else has any idea.

    Please help explain, and thank you!
     
  2. jcsd
  3. Feb 12, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I certainly understand your frustration!
    If P varies linearly with T, you would expect P = kT. Put that in V = nRT/P and you get that V is a constant, which is wrong. Looks like the question is inconsistent!

    The other meaning of "linear" would be P = kT + b so V = nRT/(kT+b) which doesn't seem likely. It does give quite a complex answer.
    Since P and V both vary, dW = PdV + VdP
    I worked the integral quickly and got
    W = -nRb(P2 - P1) + nR/k*[P2-P1 -ln(P2/P1)]
    Could well be errors in that! It doesn't look much like the given answer, does it?
     
  4. Feb 13, 2009 #3
    Ah, thank you so much for the insight. (I'll have to try and work that out when I can find the time.)
     
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