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Adiabatic Expansion: Constant pressure and temperature

  1. Sep 23, 2013 #1
    I'm new to the forum, so please be kind.

    I was reading through my pchem textbook, and I noticed something. We're given the equation:

    ΔU = q + w

    For an adiabatic expansion, we're told that q = 0. Fair enough, no heat transfer. But when there is a constant T and change in V, my book says:

    ΔU = 0 (U is constant)

    I don't understand. W = -∫pdV, doesn't it? Why is work zero?
    The book also says that for a perfect gas, U isn't dependent on volume. When they use this fact, it makes sense why ΔU = 0, but it seems like a contradiction when using the definition of work.

    Also, the book continues and says:

    ΔU = CvΔT when V is constant.

    Again, W = -∫pdV, right? So dV = 0, shouldn't ΔU = 0?

    Thanks in advance!
     
  2. jcsd
  3. Sep 24, 2013 #2
    Hi Hpatps1,
    For the adiabatic expansion we have the relation between T and V as T1V11-[itex]\gamma[/itex]=T2V21-[itex]\gamma[/itex].
    Where T1,V1 and T2,V2 are data of the system at two different states. If T is constant then T1=T2. Then by the above equation V1=V2. Hence dV is zero, and the W. That's why ΔU is zero.
    Now for the next. By the above data it is clear that dV is zero iff T is constant (ΔT=0). So in the equation ΔU=CvΔT, ΔU certainly becomes zero when ΔT=0.
    Regards.
     
  4. Sep 29, 2013 #3
    Thanks! That makes sense.
     
  5. Oct 1, 2013 #4
    As a gas expands or contracts adiabatically, although q=0, the temperature changes as a result of the particles doing work on the surroundings, so there is a change in internal energy equal to the work done by or to the system. U is only dependent on the kinetic energies of the molecules of the gas in an ideal system.
     
  6. Oct 1, 2013 #5

    Andrew Mason

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    I am not sure why you say W = 0. Why do you think W = 0?

    Why would ΔU = 0?

    AM
     
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