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DG is for isothermal thus dH =0?

  1. Jul 4, 2011 #1
    If U, H are functions of T only, then ΔU and ΔH should be zero for isothermal processes
    ΔG and ΔA are only defined at constant T (thus define isothermal processes)
    ΔG = ΔH – TΔS thus equals -TΔS always (since ΔH=0)?
    this makes no sense since from previous calculations I have done in many problems ΔH has nonzero value
    similarly, ΔA = ΔU - TΔS= -TΔS?
    Please help clear my confusion, thanks.
  2. jcsd
  3. Jul 5, 2011 #2


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    U and H are functions of only temperature if the gas is ideal (in which case ΔH=ΔU=0 for an isothermal process). For a non-ideal system, temperature is not the only relevant variable.
  4. Jul 5, 2011 #3
    The fundamental thermodynamic relation is:
    dU = T \, dS - P \, dV
    which would imply that the internal energy U is a function of entropy S and volume V as natural variables. The partial derivatives can be read off from this differential form:
    T = \left(\frac{\partial U}{\partial S}\right)_{V}, \; P = -\left(\frac{\partial U}{\partial V}\right)_{S}
    Because the mixed second partial derivatives have to be equal, we have the following identity:
    \left(\frac{\partial T}{\partial V}\right)_{S} = -\left(\frac{\partial P}{\partial S}\right)_{V}
    which is one of the Maxwell relations.

    The other therodynamic potentials are Legendre transfroms:
    H = U + P \, V & dH = T \, dS + V \, dP & H = H(S, P) \\

    A = U - T \, S & dA = -S \, dT - P \, dV & A = A(T, V) \\

    G = U - T \, S + P \, V & dG = -S \, dT + V \, dP & G = G(T, P)
    You should be able to deduce the other three Maxwell relations from these expressions.
  5. Jul 5, 2011 #4
    Wow is this really it?
    perhaps changes in moles of gas is also a factor?
  6. Jul 5, 2011 #5


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    If we are talking about the total enthalpy (i.e. in units of energy), then yes it will change with the number of moles of gas; however the specific enthalpy (i.e. energy/mole) will remain unchanged.
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