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Question

If α moles of a monoatomic gas are mixed with β moles of a polyatomic gas and mixture behaves like diatomic gas, then [ neglect the vibrational mode of freedom ]

A
2α=β
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B
α=2β
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C
α=3β
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D
3α=β
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Solution

The correct option is A 2α=β
Given: Moles of monoatomic gas = α, moles of diatomic gas is β.
Solution:
The mixture is behaving as diatomic gas. If we neglect the vibrational mode of freedom then degree of freedom (fmix) of mixture is 3 (for translational)+2(for rotational)=5.
We know that,
fmix=f1n1+f2n2n1+n2
5=3α+6βα+β
5α+5β=3α+6β
2α=β
Hence the correct option is A.

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