If C_p and C_v are molar specific heats of an | vedclass

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(A) We know that for an ideal gas,the molar specific heat at constant pressure is $C_p = \left(1 + \frac{f}{2}\right)R$ and at constant volume is $C_v

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$\frac{R \gamma}{\gamma-1}$

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(A) We know that for an ideal gas,the molar specific heat at constant pressure is $C_p = \left(1 + \frac{f}{2}\right)R$ and at constant volume is $C_v = \frac{f}{2}R$.The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$,which implies $\frac{2}{f} = \gamma - 1$,or $\frac{f}{2} = \frac{1}{\gamma - 1}$.Substituting this into the expression for $C_p$:$C_p = \left(1 + \frac{1}{\gamma - 1}\right)R$$C_p = \left(\frac{\gamma - 1 + 1}{\gamma - 1}\right)R$$C_p = \frac{\gamma R}{\gamma - 1}$.

If $C_p$ and $C_v$ are molar specific heats of an ideal gas at constant pressure and volume respectively,if $\gamma$ is the ratio of the two specific heats and $R$ is the universal gas constant,then $C_p$ is equal to

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