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Homework Help: Chemical Thermodynamics Proof

  1. May 11, 2010 #1

    danago

    User Avatar
    Gold Member

    Hey. Im currently studying chemical thermodynamics and have reached the section on solution thermodynamics.

    For a pure liquid:

    [tex]
    \left(\frac{\partial g}{\partial P}\right)_T=RT \left(\frac{\partial ln(f)}{\partial P}\right)_T
    [/tex]

    Where g is the molar gibbs energy
    P is pressure
    T is temperature
    R is the ideal gas constant
    f is the fugacity

    My first thought was to make use of one of the fundamental thermodynamic relations:

    [tex]dg = v dP - s dT \Rightarrow \left(\frac{\partial g}{\partial P}\right)_T=v[/tex]

    Where v is the molar volume and s is the molar entropy.

    Anybody have any suggestions for a next step? :smile:

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. May 11, 2010 #2

    danago

    User Avatar
    Gold Member

    Ok, just as i posted this i had an idea.

    Fugacity is defined by the equation:

    [tex]
    g-g^{o}=RT ln (\frac{f}{f^o})
    [/tex]

    Since the reference state is fixed (i.e. [tex]dg^o=df^o=0[/tex]):

    [tex]
    dg = RT d(ln f)
    [/tex]

    Equating this with dg from the fundamental thermodynamic relation i mentioned in the first post (with dT=0 since temperature is being held constant):

    [tex]RT d(ln f) = v dP \Rightarrow \frac{d(ln f)}{dP}=\frac{v}{RT}[/tex]

    Substituting in (from the first post)

    [tex]
    \left(\frac{\partial g}{\partial P}\right)_T=v
    [/tex]

    will give the required relationship.

    Does that look right?
     
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