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Why is enthalpy a function of Temperature and Pressure?

  1. Apr 28, 2012 #1
    right now, I'm following the MIT thermodynamics video lecture.

    I've gone thru

    dU = [STRIKE]d[/STRIKE]w + [STRIKE]d[/STRIKE]q
    ([STRIKE]d[/STRIKE] for "is path dependent")

    and

    H = U + pV

    But why is enthalpy a function of temperature and pressure?
    is it because pV = nRT and thus, V = nRT/p, so we only need p and T to get V?

    dH = ( [itex]\delta[/itex] H/ [itex]\delta[/itex] )TdT + ( [itex]\delta[/itex] H/ [itex]\delta[/itex] p)dp

    but why not:
    dH = ([itex]\delta[/itex]H/[itex]\delta[/itex] T)dT + ([itex]\delta[/itex]H/[itex]\delta[/itex]V)dV ???

    Thanks! : D
     
    Last edited: Apr 28, 2012
  2. jcsd
  3. Apr 29, 2012 #2
    You have (I won't write the path-dependent parts explicitly; just assume they are there :D ):
    [tex]
    \mathrm{d}E = T\mathrm{d}S - P\mathrm{d}V
    [/tex]

    We know that [itex]\mathrm{d}Q = T\mathrm{d}T[/itex], so that for a process at constant pressure we can rewrite the quantity of heat as the differential:
    [tex]
    \mathrm{d}Q = \mathrm{d}E + P\mathrm{d}V = \mathrm{d}(E + PV) = \mathrm{d}W
    [/tex]

    of some quantity
    [tex]W = E+PV[/tex]

    If you now want to find the total differential of the heat function itself (now no longer assuming constant pressure), you can easily get:
    [tex]
    \mathrm{d}W = T\mathrm{d}S + V\mathrm{d}P
    [/tex]

    For your notation: look at the definition of your [itex]U[/itex]. When you make the total differential of the heat function, you have to write explicitly the terms for the energy, where [itex]-P\mathrm{d}V[/itex] cancels with one of the differential you get from [itex]\mathrm{d}(PV)[/itex] in the heat function.
     
    Last edited: Apr 29, 2012
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