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Relationship between pressure & chem potential in ideal gas

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data

    (Excerpted from a longer, multipart problem but essentially)

    Show that for an ideal gas,
    $$ \frac{\partial p}{\partial T}\bigg)_\mu = \frac{S}{V}. $$

    2. Relevant equations

    • The ideal gas law, of course
    $$ pV = Nk_{\rm B}T $$
    • Pressure, temperature, and chemical potential are related in an ideal gas via
    $$ \mu-\mu_0=k_{\rm B}T\ln\frac{p}{p_0} $$
    • The Gibbs free energy
    $$ G = E - TS + pV = \mu N $$
    • Hopefully not the Sackur–Tetrode equation.

    3. The attempt at a solution

    Thoughts regarding left-hand side:

    So I rearranged the ##\mu## equation for ##p##:
    $$ p(T,\mu) = p_0e^{(\mu-\mu_0)/k_{\rm B}T}. $$
    Computing the partial derivative,
    $$ \frac{\partial p}{\partial T}\bigg)_\mu = -\frac{\mu-\mu_0}{k_{\rm B}T^2}p. $$
    By the ideal gas law ##p/k_{\rm B}T = N/V##, so
    $$ \frac{\partial p}{\partial T}\bigg)_\mu = \frac{(\mu_0-\mu)N}{VT}. $$

    Thoughts regarding the right-hand side:

    From the definition of the Gibbs free energy,
    $$ \frac{S}{V} = \frac{E + pV - \mu N}{VT}. $$

    These two would be equal if
    $$ E + pV = \mu_0N $$
    in an ideal gas. But I don't really understand what the definition of the "reference" pressure and chem. potential ##p_0## and ##\mu_0## are. So maybe this is true, but if it is I don't know why!

    …Or perhaps there's an altogether different starting point or process that gives the result. I am by no means married to my approach here.
  2. jcsd
  3. Feb 2, 2015 #2
    Hi. Chemical potential is the partial molar Gibbs free energy, so for a pure substance (ideal gas) it is equal to the molar Gibbs free energy and you have:
    dμ = dG = –SdT + VdP

    Now examine this for fixed μ...
  4. Feb 2, 2015 #3
    I agree. I knew I was going to get myself in trouble for excerpting this out of a larger question.

    The greater context of the problem is:

    (a) Derive the formula ##N{\rm d}\mu=-S{\rm d}T+V{\rm d}p##.

    (b) Derive the general formulas ##\dfrac{\partial p}{\partial T}\bigg)_{\mu}=\dfrac{S}{V}## and ##\dfrac{\partial p}{\partial\mu}\bigg)_{T}=\dfrac{N}{V}##.

    (c) Find the pressure p of an ideal gas as a function of µ and T. Use this to verify the two equations from part (b) for the specific case of an ideal gas.​

    So your response is very relevant to parts (a) and (b), but I am now working on part (c).
  5. Feb 2, 2015 #4
    Ok. To get to (a) and (b), remember that: G = μN = U–TS + PV and also: dU = TdS–PdV + μdN
    Play with the differential dG...
    (Isn't that the relation you were trying to use in your first post? How did you obtain it then?)
    Hint: use the relation in (a) at fixed temperature and the equation of state for an ideal gas...
  6. Feb 3, 2015 #5
    Okay, for posterity's sake, here's what I ultimately came up with. I suspect I go around in circles a little and could streamline some parts, but it'll do.$$\require{cancel}$$(a) The Gibbs free energy is
    G = E - TS + pV = \mu N,
    Consider its total differential:
    {\rm d}G = {\rm d}E - T{\rm d} S - S{\rm d} T + p{\rm d} V + V{\rm d} p = \mu{\rm d} N + N{\rm d}\mu.
    Solve for ##N{\rm d}\mu##:
    N{\rm d}\mu = {\rm d} E - T{\rm d}S - S{\rm d}T + p{\rm d} V + V{\rm d}p - \mu{\rm d}N.
    Substituting for ##{\rm d}E##,
    {\rm d} E = T{\rm d}S - p{\rm d} V + \mu{\rm d} N,
    we find
    N{\rm d}\mu = (\cancel{T{\rm d} S} - \cancel{p{\rm d} V} + \cancel{\mu{\rm d} N}) - \cancel{T{\rm d} S} - S{\rm d} T + \cancel{p{\rm d} V} + V{\rm d} p - \cancel{\mu{\rm d}N}
    = -S{\rm d} T + V{\rm d} P.

    (b) Our result from part (a) above reads
    N{\rm d}\mu = -S{\rm d}T + V{\rm d}p.
    If we suppose ##\mu## to be held constant, this reads
    0 = -S{\rm d} T + V{\rm d} p \qquad\Rightarrow\qquad S = V\frac{\partial p}{\partial T}\bigg)_{\mu}.
    Or if we suppose instead that ##T## is held constant,
    N{\rm d}\mu = 0 + V{\rm d} p \qquad\Rightarrow\qquad N = V\frac{\partial p}{\partial\mu}\bigg)_{T}.

    (c) The energy of a monatomic ideal gas is
    E = \frac32Nk_{\rm B}T.
    The entropy of a monatomic ideal gas is
    S = Nk_{\rm B}\bigg(\ln\frac{V}{Nh^3}(2\pi mk_{\rm B}T)^{\frac32}+\frac52\bigg).
    The volume of an ideal gas is
    V = \frac{Nk_{\rm B}T}{p}.
    Combine all these expressions into the relation ##E - TS + pV = \mu N##:
    \frac32Nk_{\rm B}T - TNk_{\rm B}\bigg(\ln\frac{k_{\rm B}T}{h^3p}(2\pi mk_{\rm B}T)^{\frac32}+\frac52\bigg) + Nk_{\rm B} T = \mu N.
    Divide through by ##Nk_{\rm B}T##:
    \cancel{\frac32} - \ln\frac{k_{\rm B}T}{h^3p}(2\pi mk_{\rm B} T)^{\frac32} - \cancel{\frac52} + \cancel{1} = \frac{\mu}{k_{\rm B}T}.
    Solving for ##p##,
    p(\mu, T) = \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}e^{\mu/k_{\rm B}T}.

    The derivatives under consideration are
    \frac{\partial p}{\partial T}\bigg)_{\mu} = \frac{(2\pi m)^{\frac32}}{h^3}\bigg(\frac52k_{\rm B}^{\frac52}T^{\frac32}\bigg)e^{\mu/k_{\rm B}T} + \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}\bigg({-\frac{\mu}{k_{\rm B}T^2}}e^{\mu/k_{\rm B}T}\bigg)=\bigg(\frac52\frac1T - \frac{\mu}{k_{\rm B}T^2}\bigg)p
    \frac{\partial p}{\partial\mu}\bigg)_{T} = \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}\bigg(\frac{1}{k_{\rm B}T}e^{-\mu/k_{\rm B}T}\bigg) = \frac{1}{k_{\rm B}T}p.
    V \frac{\partial p}{\partial T}\bigg)_{\mu}= V\bigg(\frac52\frac1T - \frac{\mu}{k_{\rm B}T^2}\bigg)p = \frac52Nk_{\rm B}- \frac{\mu N}{T};
    but we want to get the entropy back in there somehow, so
    V \frac{\partial p}{\partial T}\bigg)_{\mu}= \frac52Nk_{\rm B}- \frac{E-TS+pV}{T} = \cancel{\frac52Nk_{\rm B}} - \cancel{\frac32Nk_{\rm B}} + S - \cancel{Nk_{\rm B}} = S.
    And, more straightforwardly,
    \frac{\partial p}{\partial\mu}\bigg)_{T} = V\frac{1}{k_{\rm B}T}p = N.
  7. Feb 3, 2015 #6
    That's one way to go... i'm guessing there's a quicker way directly from the partition function to the derivatives of U or F (Helmholtz free energy), but it really comes down to the same calculations minus the detour so your answer works regardless.
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