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Solving Enthaply by other ways?

  1. Feb 16, 2012 #1
    2 moles of an ideal gas at 300K and 5 bar is expanded adiabatically at a constant pressure of 1 bar till the volume doubles. Cv = 3R/2. Calculate w, q, dE, dH and change in T.

    I found that w = dE = -997.78J

    In class we solved dH by:

    dH = CpdT = -1.66kJ

    (we found Cp by Cp=Cv+nR ; we found dT by Cv=dE/dT)

    I was wondering if we could solve dH another way:
    dH = dE + d(PV)

    However, I get a different answer..
    dH = dE + Pext dV (because pressure is constant)
    dH = -997.78J +-997.78J (because Pext dV is work)
    dH = -2.00 kJ

    I also tried another way:
    dH = dE + d(nRT)
    dH = -977J + (2mol)(8.3145J/K)(-80K)
    dH = -1.08kJ

    I am not getting the same value for each of the different ways I am using.... Am I doing something wrong or am I not allowed to solve dH using the other methods?

    Also, at constant pressure, isn't dH = qrev.. and q=0, so shouldn't dH = 0?

    Thank you
  2. jcsd
  3. Feb 16, 2012 #2
    It is adiabatic or at constant pressure? How can be both?
  4. Feb 16, 2012 #3
    The external pressure is constant, but the system undergoes an adiabatic process
  5. Feb 16, 2012 #4

    Philip Wood

    User Avatar
    Gold Member

    I'm with nasu here. If the gas expands adiabatically pushing out a piston, then it will do work pushing out the piston, and its internal energy will fall. So, therefore, will its temperature, and its pressure (since P = nRT/V and T goes down and V goes up).
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