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B Isobaric formula

  1. Oct 20, 2016 #1
    In a isobaric process W=PΔV
    I
    Given the change in temperature and pressure is constant to find heat added to an mono-atomic ideal gas we use
    Q=ΔU + W = 3/2 nRΔT + PΔV = 3/2 nRΔT + nRΔT
    so Q=5/2 nRΔT correct?
    II
    But given the heat added to an ideal gas, constant pressure that acts on the gas, and change in volume
    To find ΔU we have to use ΔU = Q-W = Q-PΔV
    and using ΔU = 3/2 PΔV (based on PΔV=nRΔT ) is incorrect I was told this is so because P in PΔV=nRΔT is the pressure of the ideal gas and not the pressure acting on the ideal gas, is this explanation true?


    So why does PΔV= nRΔT hold true in the I and not in II ?
     
  2. jcsd
  3. Oct 21, 2016 #2

    Andrew Mason

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    The first law: Q = ΔU + W always holds when comparing two thermodynamic equilibrium states. But W is determined by the process between those two equilibrium states. If the process occurs (e.g. expansion) with the minimal difference between internal gas pressure and external pressure, the process will be quasi-static and the work done, W = PextΔV = PintΔV = nRΔT.

    This would occur, for example, when slowly adding heat to a gas in a balloon. The addition of heat causes the gas pressure to increase slightly above external pressure but expansion quickly causes the internal pressure to decrease so it never gets significantly higher than external pressure.

    But if you have a rapid process of expansion or compression the internal gas pressure and external gas pressure will not be the same during the process. In the case of a rapid expansion, the internal pressure will initially be significantly greater than the external pressure. The work done on the surroundings is determined by the external pressure x change in volume regardless of what the internal gas pressure is during the process. So W = PextΔV < PintΔV = nRΔT.

    AM
     
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