Deriving Molar Specific Heat Equation

  • Thread starter Funktimus
  • Start date
  • #1
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Homework Statement


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.

Homework 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)

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
 

Answers and Replies

  • #2
529
1
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
 
  • #3
13
0
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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