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Homework Help: Deriving Molar Specific Heat Equation

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    n_1 moles of a monatomic gas and n_2 moles of a diatomic gas are mixed together in a container.

    Derive an expression for the molar specific heat at constant volume of the mixture.
    My answer can only use the variables n_1 and n_2, and I'm assuming constants.

    2. Relevant equations
    Monoatomic gas
    E_th = (3/2)nRT
    C_v = (3/2)R

    Diatomic gas
    E_th = (5/2)nRT
    C_v = (5/2)R

    First Law:
    (Change)E_th = W + Q = (n)(C_v)(change in T)

    3. The attempt at a solution
    Honestly, I'm not sure where to begin.

    I'm assuming:
    Since volume is constant, that means this is an Isochoric process.
    In which case W = 0
    Which means (change in E_th) = 0 + Q = n(C_v)(change in T)

    So...
    for the monoatomic gas
    E_1f = (n_1)(C_v)(T) = (3/2)(n_1)(R)(T_1f)

    for the diatomic gas
    E_2f = (n_2)(C_v)(T) = (5/2)(n_2)(R)(T_2f)

    Now I figure this is a process going for thermal equilibrium...
    so
    T_1f = T_2f = T_f, so I can make the temperature variable T_f from now on.
    E_1f = (n_1)/(n_1 + n_2)
    E_2f = (n_2)/(n_1 + n_2)
    E_tot = E_1f + E_2f
    E_tot = (3/2)(n_1)(R)(T_f) + (5/2)(n_2)(R)(T_f)

    I don't know where I'm going with this. Can someone tell me if I'm going in the right direction?

    Thank you
     
  2. jcsd
  3. Apr 29, 2007 #2
    What's wrong with the following?
    U=n1 3/2 kT + n2 5/2 kT

    C=dU/dT=n1 3/2 k +n2 5/2 k
     
  4. Apr 29, 2007 #3
    I dunno, apparently it's wrong. Cause that's what I initially plugged in. It told me k is not a variable, and I should use R, which is lame that it didn't say that in the first place. I converted k to R using avogaddies number and its still says its wrong.
     
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