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Isobaric process

  1. Jan 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider 1 gram of an ideal gas undergoing isobaric process. Suppose del(H) be the amount of heat given to it. Then,
    del(H) = dU + del(W)
    del(H) = 1 x C(v)dT + (P.dV)/J
    But del(H) = C(p)dT
    P.dV = r.dT
    C(p)dT = C(v)dT + (r.dT)/J
    C(p) - C(v) = r/J


    2. Relevant equations



    3. The attempt at a solution

    In the above derivation, when volume is changing, how can they take dU = mC(v)dT?
    Here m = mass of gas,r = gas constant,J = Mechanical equivalent of heat
     
  2. jcsd
  3. Jan 2, 2007 #2

    Andrew Mason

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    U is a function of temperature only. It does not depend on volume or pressure (although those will affect temperature, of course). dU is always = mC(v)dT

    AM
     
  4. Jan 3, 2007 #3
    That was a nice piece of information.Thanks.
     
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