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One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is.


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$C_v = \large\frac{3}{2}R$( for monoatomic )
$C_v = \large\frac{5}{2}R$(for diatomic)
Thus for mixture
$C_v = \large\frac{[\large\frac{3}{2}R+\large\frac{5}{2}R]}{2}$
$\;\;\;\;\; = 4 cal$
Hence answer is (b)
answered Mar 5, 2014 by sharmaaparna1

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