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Change in pressure and Internal Energy

  1. Mar 5, 2012 #1
    I'm aware of the equation
    ΔU = Q + W
    ΔU is the change in internal energy which equals 3/2 nRΔT
    Q is heat transfer
    W is workdone

    so PV = nRT
    would it be correct to say that
    ΔP V = nRΔT?

    For isovolumetric process (i.e. no volume change, no work done),
    can i say that
    change in internal energy = 3/2 nRΔT = 3/2 ΔP V?

    Pressure changes from 200Pa --> 100Pa
    Volume remains at 6 m^3
    Would it be right to conclude that the change in internal energy = 3/2 x 100 x 6 = 900J?

    **New Question: Is it possible (and if possible, under what condition) that the initial PV and final PV are known BUT the change in internal energy cannot be found directly by ΔU = 3/2 ΔPV?
    coz i always come across questions where a P-V graph is given but the internal energy has to be calculated by more complicated methods,
    i.e. ΔU = Q + W and not ΔU = 3/2 ΔPV

    Last edited: Mar 6, 2012
  2. jcsd
  3. Mar 5, 2012 #2
    I believe that is correct for a monatomic ideal gas. 900 J of heat flows into the system, which raises the temperature, which increases the pressure, which accounts for the increase in internal energy by the relation ΔU=3/2 ΔP V as you say.
  4. Mar 6, 2012 #3
    does the equation
    ΔU = 3/2 nRΔT = 3/2 PΔV = 3/2 ΔPV
    have its limitations?
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