If gamma is the ratio of specific heats and R | vedclass

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(C) According to Mayer's relation for an ideal gas:$C_{p} - C_{V} = R$ $(i)$where $C_{p}$ is the molar specific heat at constant pressure and $C_{V}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) According to Mayer's relation for an ideal gas:$C_{p} - C_{V} = R$ $(i)$where $C_{p}$ is the molar specific heat at constant pressure and $C_{V}$ is the molar specific heat at constant volume.By definition,the ratio of specific heats is:$\gamma = \frac{C_{p}}{C_{V}}$This implies $C_{p} = \gamma C_{V}$ (ii)Substituting equation (ii) into equation $(i)$:$\gamma C_{V} - C_{V} = R$$C_{V}(\gamma - 1) = R$Therefore,$C_{V} = \frac{R}{\gamma - 1}$

If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant,then the molar specific heat at constant volume $C_{V}$ is given by

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