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Question

Two moles of a monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at a constant volume is

A
1.6R
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B
2.1R
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C
6.25R
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D
3.6R
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Solution

The correct option is B 2.1R
Given, no. of moles of monoatomic gas n1=2
& For a monoatomic gas, CV1=32R
No. of moles of diatomic gas, n2=3
& For a diatomic gas, CV2=52R
We know that,
Molar specific heat of the mixture
CV=n1×CV1+n2×CV2n1+n2
From the data given in the question,
CV=2×3R2+3×5R22+3=21R10=2.1R
Thus, option (b) is the correct answer.

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