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The mixture of two gases in gravitational field

  1. Sep 7, 2009 #1
    Two ideal gases are mixed in the box of infinite height placed in constant gravitational field.
    There are [tex]n_1[/tex] moles of the first gase and [tex]n_2[/tex] moles of the second. Their
    molar mases are [tex]M_1[/tex] and [tex]M_2[/tex] respectively. Let's assume that the
    temperature is constant. What's the molar concentration as a function of height of each gas.

    If in the box was only one gas the answer would be simple but when there are two different
    types of molecules we have to take into consideration the entropy of mixing. That's the
    problem because I don't know any formula that might be easily incorporated into differential
    equations.
     
  2. jcsd
  3. Sep 7, 2009 #2
    Try the following.
    First write the total density as a function of height. You get this solving
    [tex]
    \partial p_{total} / \partial y = \rho_{total}(y) g
    [/itex]
    (using the equation of state [itex]p_{total}=\rho_{total} R T / M_{total}[/itex])
    where [itex]M_{total}[/itex] is the total molar mass.
    Then write the total pressure as a function of height. Each of these expressions will have an unknown pre-factor.

    Next write
    [tex]
    \rho_{total} = N_1 M_1 + N_2 M_2
    [/tex]
    [tex]
    p_{total} = RT (N_1 + N_2)
    [/tex]
    where [itex]N_1[/itex] and [itex]N_2[/itex] are molar concentrations.

    You have two equations and two unknowns [itex]N_1[/itex] and [tex]N_2[/itex] which you can solve.

    Finally you can obtain the pre-factors by using the information given about the total amout of stuff you have. You will have to integrate your concentrations with respect to height and set them equal to [itex]n_1[/itex] and [itex]n_2[/itex].
     
    Last edited: Sep 7, 2009
  4. Sep 7, 2009 #3
    Correction.

    Instead of [itex]N_{total}[/itex] for the total molar mass I meant to write [itex]M_{total}[/itex] and I was unable to edit it for some reason.
     
  5. Sep 7, 2009 #4
    What exactly do you mean by [itex]M_{total}[/itex]. It should depend on [itex]N_{1}(y)[/itex] and [itex]N_{2}(y)[/itex] ([tex]M_{total} = (N_1(y) M_1+N_2(y)M_2)/(N_1(y)+N_2(y))[/tex]).

    So in fact there is only one equation:

    [tex]
    \frac{\partial \left( R T (N_1(y)+N_2(y)) \right)}{\partial y} = g (N_1(y) M_1 + N_2(y) M_2 )
    [/tex]
     
  6. Sep 7, 2009 #5
    Let me think about it. In the meanwhile you might want to re-post it to see if others can be of more immediate assistance.
     
  7. Sep 7, 2009 #6
    I think you have to write for each [itex]i[/itex]th constituent of the gas
    [tex]
    \frac{\partial p_i}{\partial y} = -\rho_i(y)g
    [/tex]
    and use the equation of state [itex]p_i = \rho_i R T/m_i[/itex] where [itex]m_i[/itex] is the molecular mass. Note that the molar density [itex]N_i = \rho_i/m_i[/itex].

    I assume you can do the rest. If not let me know.

    I apologize for the initial wrong answer.
     
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