One mole of an ideal gas left( frac{C_p}{C_v} | vedclass

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(D) The molar heat capacity $C$ is given by $C = C_V + \frac{dW}{n dT}$.For an ideal gas,$PV = nRT$. Given $P = \alpha V$,we substitute $P$ to get $\a

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$C = \frac{R}{2} \frac{(\gamma + 1)}{(\gamma - 1)}$

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(D) The molar heat capacity $C$ is given by $C = C_V + \frac{dW}{n dT}$.For an ideal gas,$PV = nRT$. Given $P = \alpha V$,we substitute $P$ to get $\alpha V^2 = nRT$.Differentiating both sides,we get $2\alpha V dV = nR dT$,so $dV = \frac{nR dT}{2\alpha V}$.The work done is $dW = P dV = (\alpha V) \left( \frac{nR dT}{2\alpha V} \right) = \frac{nR dT}{2}$.Thus,$\frac{dW}{n dT} = \frac{R}{2}$.Substituting this into the heat capacity formula: $C = C_V + \frac{R}{2}$.Since $C_V = \frac{R}{\gamma - 1}$,we have $C = \frac{R}{\gamma - 1} + \frac{R}{2}$.$C = R \left( \frac{1}{\gamma - 1} + \frac{1}{2} \right) = R \left( \frac{2 + \gamma - 1}{2(\gamma - 1)} \right) = \frac{R}{2} \frac{(\gamma + 1)}{(\gamma - 1)}$.

Column $I$	Column $II$
$(A)$ An insulated container has two chambers separated by a valve. Chamber $I$ contains an ideal gas and Chamber $II$ has a vacuum. The valve is opened.	$(p)$ The temperature of the gas decreases
$(B)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-2}$	$(q)$ The temperature of the gas increases or remains constant
$(C)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-4/3}$	$(r)$ The gas loses heat
$(D)$ An ideal monoatomic gas expands such that its pressure $P$ and volume $V$ follow the behavior shown in the graph	$(s)$ The gas gains heat

One mole of an ideal gas $\left( \frac{C_p}{C_v} = \gamma \right)$ is heated according to the law $P = \alpha V$,where $P$ is the pressure of the gas,$V$ is the volume,and $\alpha$ is a constant. What is the molar heat capacity of the gas in this process?

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