One mole of an ideal monoatomic gas is mixed with I mole of an ideal diatomic. gas. The molar specific heat of the mixture at constant volume is:

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3 cal

4 cal

8 cal

9 cal

answer is B.

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Detailed Solution

$C_{V} = \frac{3}{2} R$ ( for, monoatomic) and $\frac{5}{2} R$ (for diatomic).
Thus, for mixture, $C_{V} = \frac{[\frac{3}{2} R + \frac{5}{2} R]}{2} = 2 R = 4 cal$

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