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.
E_th = (3/2)nRT
C_v = (3/2)R
E_th = (5/2)nRT
C_v = (5/2)R
(Change)E_th = W + Q = (n)(C_v)(change in T)
The Attempt at a Solution
Honestly, I'm not sure where to begin.
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)
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...
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?