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For a gas $\large\frac{R}{C_v}$$ = 0.67$ the gas may be

$(a)\;H_2\qquad(b)\;Ar.\qquad(c)\;O_3\qquad(d)\;H_2O_2$

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Given
$\large\frac{R}{C_v} = 0.67$
$(or) \large\frac{C_p - C_v}{C_v} = 0.67$
$(or) \large\frac{C_p}{C_v} - 1 = 0.67$ (or)
$\gamma = 1.67$
Thus gas is monoatomic , i.e Ar
Hence answer is (b)
answered Mar 9, 2014 by sharmaaparna1
 

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