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If one mole of a monoatomic gas ($\gamma = \large\frac{5}{3}$) is mixed with one mole of a diatomic gas ($\gamma = \large\frac{7}{5}$) . The value of V for the mixture is :

$(a)\;1.4\qquad(b)\;1.5\qquad(c)\;1.53\qquad(d)\;3.07$

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For monoatomic gas;
$C_v = \large\frac{3}{2}RT$
$C_p =\large\frac{5}{2}RT$
For diatomic gas
$C_v = \large\frac{5}{2}RT$
$C_p = \large\frac{7}{2}RT$
Thus for mixture of 1 mole each
$C_v = \large\frac{\large\frac{3}{2}RT+\large\frac{5}{2}RT}{2}$
$C_p = \large\frac{\large\frac{5}{2}RT+\large\frac{7}{2}RT}{2}$
$\large\frac{C_p}{C_v} = \large\frac{3RT}{2RT}$
= 1.5
Hence answer is (b)
answered Mar 5, 2014 by sharmaaparna1
 

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