How to Calculate the Correct Fraction of Monatomic Molecules in a Gas Mixture?

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Jahromi24
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Homework Statement



An experiment you're designing needs a gas with γ = 1.49. 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 γ = 1.49 by mixing together a monatomic gas and a diatomic gas.
What fraction of the molecules need to be monatomic?

Homework Equations



[itex]\gamma[/itex]=5/3 for Monatomic
[itex]\gamma[/itex]=7/5 for Diatomic

The Attempt at a Solution



[itex]\gamma[/itex]=5x/3+7/5(1-x)=1.49
I solve for x and get 0.3375, but this is wrong.
I can't figure out what I'm doing wrong because everywhere I look it seems correct to me.
 
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The 7/5 for the diatomic gas is a simplified version of reality where you're saying it has 5 degrees of freedom. x,y,z and then it can rotate around the x and z axis. In reality it also rotates around the y-axis but the number is very small. Just an idea.
 
Cp and Cv of a gas mixture are proportional to the number of moles of each gas, but the ratio of Cp to Cv for the mixture is not. You need to treat Cp and Cv separately, and also make use of the condition that Cp - Cv of each species is equal to R.

chet
 
Chestermiller said:
Cp and Cv of a gas mixture are proportional to the number of moles of each gas, but the ratio of Cp to Cv for the mixture is not. You need to treat Cp and Cv separately, and also make use of the condition that Cp - Cv of each species is equal to R.

chet
Could you please go through the steps? I still do not understand, and I have my final tomorrow.
 
##\gamma=\frac{C_P}{C_V}##

Can you use that relation to work out the value of ##\gamma## if the mixture is half monatomic and half diatomic?
 
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Mister T said:
##\gamma=\frac{C_P}{C_V}##

Can you use that relation to work out the value of ##\gamma## if the mixture is half monatomic and half diatomic?

I think you're supposed to use that, but I'm not sure how.
Also, I found this, and I think it helps, but I'm not sure what to do with this either.
xawc9l.png
 
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Mister T said:
Can you answer my question? Yes or no?

(Your images don't show up).
Okay, I have actually figured it out! Thank you for leading me to find the right path, Mister T!
My work is below! :) I was actually working with a gamma that was 1.52 for the problem, but the math should still be correct.
dgt9br.jpg