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Homework Help: Work done in Adiabatic, Quasistatic-compression

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    i have to find the W done in an Adiabatic, Quasistatic compression and im not having much luck.

    Ive calculated it using (delta)U = Q + W, since Q = 0, hence W = change in internal energy

    However i want to calculate it using W = -(delta)P*V

    3. The attempt at a solution

    My working is attached, my text just says to use W = delta(U), but i know there is a way to do it via pressure and volume, im just not sure where to go from where im at.

    Attempt is attached


    Has to be in terms of TA, n (moles) and R

    Cheers Trent
  2. jcsd
  3. May 16, 2009 #2
    I assume we are talking about perfect gases so that indeed [tex]p V^\gamma = const [/tex] for a quasistatic, adiabatic process :p


    -\int_{V_C}^{V_A} p dV =

    \left\{ p V^\gamma = const \right\}=

    -\int_{V_C}^{V_A} const \cdot \frac{dV}{V^\gamma}


    Determine [tex] const [/tex] from the conditions at A.
  4. May 16, 2009 #3
    I got it finally thanks heeps but i got it using conditions at "c" our starting point, (PCVCgamma = PVgamma, rearranged for P = and solved from ther) not "A", im assuming u meant "c"???

    Thanks :)
  5. May 16, 2009 #4
    Oopsie, yes I meant "C" , it's the point where you know P and V to start with. Sorry, I always think that A comes before C when I'm not thinking :P:P

    Though now that you said, . . . it's also true that you know P and V of point A too, so you could use any:) of course, they've given you two points such that :

    p V^\gamma = const1


    T V^{\gamma-1}= const2



    because if this was not so it wouldn't be a reversible, adiabatic process.
  6. May 17, 2009 #5
    thanks and good point, just trierd it the other way and it works :)
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