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Thermodynamics Constant Pressure Process

  1. May 1, 2014 #1
    Hi, I found a derivation for heat transfer in a constant pressure process. It goes as follows:

    Q + W = u2 - u1
    Q = u2 - u1 + p(v2 - v1)

    Since h = u + pv, Then

    Q = h2 - h1

    The first equation states that the sum of heat and work done is equal to the change in internal energy, I can comprehend up to there. But in the next line the work done is taken to the other side of the equation; both mathematically and physically I cannot understand how the work done (p(v2 - v1)) is still positive.

    Mathematically I would say that taking work to the opposite side would make work negative.
    Physically I would think that the amount of heat supplied/given off would be equal to the difference of initial and final internal energy minus the work done.

    Hope someone could clear this out for me. Thanks
  2. jcsd
  3. May 1, 2014 #2
    The First Law of Thermodynamics states that for a system ΔE = Q-W,
    which states that the change in energy of a system equals the heat added to the system minus the work done by the system. If we neglect changes in kinetic energy and potential energy of the system , then this can be written as,
    ΔU = Q - W

    where W = P(V2 - V1 )

    in which case you should have no problem determining that Q = h2-h1 for a constant pressure process.

    IF instead you consider the work done on the system by the surroundings as positive and write,
    ΔU = Q + W
    then the work is done by the system is -P(V2 - V1 ). Notice the minus sign.
    Note that for an expansion of the gas, V1 < V2, and for a compression V1 > V2.

    Take note that we are looking at the change in energy of the system.
    You have to keep in mind of how energy is crossing the state boundary of the system and whether you will consider work W entering the system as positive ( in which case work W leaving the system is negative ), or negative.

    Your statement should actually be, bold added,
    Q-W = u2-u1
    The first equation states that the sum of heat ADDED TO THE SYSTEM and work done BY THE SYSTEM is equal to the change in internal energy ...

    Notice the difference.
  4. May 2, 2014 #3
    I see, thanks for the detailed explanation :)
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