Variation of chemical potential with T and P

  1. So the expression for Gibb's free energy is:

    dG = -SdT + VdP + μdN,

    Here, we see that the Gibb's free energy changes with temperature (dT), change in pressure (dP) and change in chemical potential (as a result of change in particle number).

    My question is: we know chemical potential varies with both change in temperature and pressure. So if we don't add/remove particles from the system, the chemical potential does change with variation of P and T...so is that already included in the above equation?

    (That is, in the above equation, are we accounting for the change in Gibb's free energy as a result of change in chemical potential as a result of variation of T and P, in addition to the change in chemical potential due to change in particle number).

    Further, when the number of particles changes, there might be a number of chemical reactions that take place, so the temperature T might change because of that also, which would change the sdT term at the beginning, right?

    I guess I'm just having problems understanding chemical potential :-/
     
  2. jcsd
  3. Chestermiller

    Staff: Mentor

    The answer to all your questions is "yes", the equation for dG takes all these things into account. The Gibbs Free Energy G can be expressed as a function of T, P, and N1, ..., Nm, where m is the number of species in the solution:

    [tex]G=G(T,P,N_1,...,N_m)[/tex]

    An infinitecimal change in G can be represented using the chain rule for partial differentiation:

    [tex]dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial N_1}dN_1+...+\frac{\partial G}{\partial N_m}dN_m[/tex]
    Each of the partial derivatives in this equation is a function of T, P, and the N's, with
    [tex]\frac{\partial G}{\partial T}=-S[/tex]
    [tex]\frac{\partial G}{\partial P}=V[/tex]
    and
    [tex]\frac{\partial G}{\partial N_i}=μ_i[/tex]

    I hope this helps.
     
  4. DrDu

    DrDu 4,247
    Science Advisor

    Of course mu is a function of T and P, also.
    Given that ##\mu=\partial G/\partial N## we have ##(\partial\mu /\partial T)_P=\partial^2 G/\partial N \partial T=\partial^2 G /\partial T \partial N =-(\partial S/\partial N)_P = S_m## i.e. the partial molar entropy and analogously
    ##(\partial \mu/\partial P)_T=V_m ## the partial molar volume.
    So for fixed N, ##d\mu=-S_mdT+V_m dP##
     
  5. Can I just go on to ask what the difference between chemical potential and chemical affinity is? They seem to , intuitively, mean the same thing but chemical potential is +ve for a reaction that's progressing and affinity is negative!

    Also, is A (affinity) always the same sign as the rate of reaction?
     
  6. DrDu

    DrDu 4,247
    Science Advisor

    ##A=-\Delta G_r=-\sum \nu_i \mu_i##
    were ##\nu_i## are the stochiometric coefficients of the reaction taking place.
    So basically A is a weighed sum of chemical potentials.
     
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