An experiment you\'re designing needs a gas with gamma = 1.66. You recall from y

Question

An experiment you're designing needs a gas with gamma = 1.66. You recall from your physics class that no individual gas has this value, but it occurs to you that you could produce a gas with gamma = 1.66 by mixing together a monatomic gas and a diatomic gas.

What fraction of the molecules need to be monatomic?

I'm mostly confused on where they ask for the "fraction of the molecules." This question being answered with percentages but I still can't quite grasp the concept.

I apreciate the help! Thank You!

Explanation / Answer

g is also Cp/Cv

Cp is specific heat at constant pressure
Cv is specific heat at constant volume

Cp1 for monoatomic gas = 5R/2 ( R is universal gas constant)
Cv1 for monoatomic gas = 3R/2

Cp2 for diaatomic gas = 7R/2
Cv2 for diaatomic gas = 5R/2

Let n1 be the number of monoatomic molecules and n2 be diatomic.
now, Cv for the mixture will be
Cv = n1 Cv1 + n2 Cv2 / n1 + n2
Cv= (R/2) *((n1 * 3) + (n2*5))/(n1+n2)

we have g = Cp/Cv = 1.66
we have Cp = Cv + R
so, Cv + R / Cv = 1.66
Cv + R = 1.66 Cv
Cv = 1.5151 R
so, (R/2) *((n1 * 3) + (n2*5))/(n1+n2) = 1.5151 R
((n1 * 3) + (n2*5))/(n1+n2) = 3.0303
(n1 * 3) + (n2*5) = (3.0303 * n1) + (3.0303*n2)
1.9697*n2 = 0.0303 n1
so, n1/n2 = 65
n1 = 65 *n2
so, n1/ ( n1 + n2) = 65 * n2/ ( 65 + 1) *n2 = 0.984948
fraction of monoatomic molecules = 0.984948 is the answer.

In % , monoatomic = 98.4948 %

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