Question

$PV^{\gamma} = K_{1}$

$\Rightarrow P = K_{1}.V^{- \gamma} $$$\Rightarrow \frac{dP}{dV} = K_{1}.( - \gamma).V^{- \gamma - 1} = - \gamma.\frac{P}{V}$$

The gas having higher $\gamma$ will have higher magnitude of slope of P vs. V curve.

Now, $n.\frac{R}{\gamma - 1}.dT = - P.dV = - \frac{nRT}{V}.dV$

Or, $\frac{dV}{dT} = - \frac{1}{\gamma - 1}.\frac{V}{T}$

Gas having higher $\gamma$ will have lower magnitude of slope of V vs. T curve.

Now, $n.C_{V,m}.dT = - P.dV = - \lbrack nRdT - V.dP\rbrack$

Or, $n.C_{P,m}.dT = - V.dP \Rightarrow \frac{n.\gamma R}{\gamma - 1}.dT = \frac{nRT}{P}.dP$

$\therefore\frac{dP}{dT} = \frac{\gamma}{\gamma - 1}.\frac{P}{T}$

Gas having higher $\gamma$will have lower magnitude of slope of P vs. T curve.