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

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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)

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