Assertion : The ratio of frac{C_p}{C_v} for a | vedclass

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(A) If $f$ is the degree of freedom,then the ratio $\gamma = \frac{C_p}{C_v}$ is given by $\gamma = 1 + \frac{2}{f}$.For a monoatomic gas,$f = 3$.Ther

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) If $f$ is the degree of freedom,then the ratio $\gamma = \frac{C_p}{C_v}$ is given by $\gamma = 1 + \frac{2}{f}$.For a monoatomic gas,$f = 3$.Therefore,$\gamma_{	ext{mono}} = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.67$.For a diatomic gas,$f = 5$.Therefore,$\gamma_{	ext{dia}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4$.Since $1.4 The reason is correct because the degree of freedom $f$ is higher for diatomic molecules $(f=5)$ than for monoatomic atoms $(f=3)$,which leads to a smaller value of $\gamma$.

$Assertion :$ The ratio of $\frac{C_p}{C_v}$ for an ideal diatomic gas is less than that for an ideal monoatomic gas (where $C_p$ and $C_v$ have usual meaning).
$Reason :$ The atoms of a monoatomic gas have less degrees of freedom as compared to molecules of the diatomic gas.

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$Assertion :$ The ratio of $\frac{C_p}{C_v}$ for an ideal diatomic gas is less than that for an ideal monoatomic gas (where $C_p$ and $C_v$ have usual meaning).$Reason :$ The atoms of a monoatomic gas have less degrees of freedom as compared to molecules of the diatomic gas.