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Equation of states for a gas that forms dimers

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data

    Show that to a first approximation the equation of state of a gas that dimerizes to a small extent is given by,

    ##\dfrac{PV}{RT} = 1 - \dfrac{K_c}{V}##

    Where ##K_c## is equilibrium constant for ##A + A \iff A_2##

    2. Relevant equations


    3. The attempt at a solution

    Using virial expansion, I get

    ##\dfrac{PV}{RT} = 1 + \dfrac{B}{V} + \cdots##

    Neglecting higher order terms,

    I need to prove ##B = -\dfrac{[A]^2}{[A_2]} = - K_c##,

    I know how to compute ##B## from already known equation of states but I don't know any relation between ##B## and ##K_c##.

    Any hints please :).
     
  2. jcsd
  3. Jun 2, 2017 #2
    Treat the gas as ideal, but with the number of moles changing.
     
  4. Jun 4, 2017 #3
    Let the amount of gas in moles be ##x## and intial amount be ##n## moles.

    Then ##[A] = x/V## and ##[A_2] = (n -x)/V##

    So, ##K_c = \dfrac{x^2}{V(n -x)}## --- (1)

    also ##\dfrac{PV}{RT} = x## --- (2)

    Should I solve for ##x## in ##(1)## and susbtitute it in ##(2)## ?

    This does not feel correct.
     
  5. Jun 4, 2017 #4
    What is Kc supposed to be? Is it supposed to be based on concentrations or partial pressures? The Kc as you have written it is based on $$A_2---> 2A_1$$. Is that what it is supposed to be?
     
  6. Jun 5, 2017 #5
    Concentrations.
     
  7. Jun 5, 2017 #6
    OK. Here goes. At very large specific volumes, the contents of the tank will be all A, but, as the specific volume is decreased, some A2 will be formed at the expense of A. Let ##V_T## be the current volume of the tank, and let ##n_0## be the number of moles of A that would be present in the tank if ##V_T## became very large. At volume ##V_T##, suppose that there are now x moles of ##A_2##. Then the moles of A will be ##n_0-2x##, and the total number of moles will be ##n=n_0-x##. So, the concentrations of A and ##A_2## when the volume is ##V_T## will be:

    $$[A]=\frac{(n_0-2x)}{V_T}$$
    $$[A_2]=\frac{x}{V_T}$$

    These concentrations are related by the equilibrium constant for the reaction:
    $$\frac{[A_2]}{[A]^2}=\frac{xV_T}{(n_0-2x)^2}=K_C$$If we neglect 2x in comparison to ##n_0##, we can solve this equation for x, and obtain:
    $$x=n_0^2\frac{K_C}{V_T}$$So the total number of moles at volume ##V_T## is given by:
    $$n=n_0-n_0^2\frac{K_C}{V_T}=n_0\left(1-\frac{K_C}{(V_T/n_0)}\right)=n_0\left(1-\frac{K_C}{V}\right)$$where V is the apparent specific volume ##V_T/n_0##
     
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