Question

If \(r\) denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of an adiabatic and isothermal $\mathrm{P} \rightarrow \mathrm{V}$ curves at their point of intersection ? (A) \((1 / \gamma)\) (B) \(\gamma-1\) (C) \(\gamma\) (D) \(\gamma+1\)

Short answer

The ratio of the slopes of adiabatic and isothermal P-V curves at their point of intersection is \(\gamma\).

Step-by-step solution

Write down the equations of adiabatic and isothermal P-V curves

For an adiabatic process, we have the equation: \(PV^{\gamma}=K_{1}\), where K1 is a constant For an isothermal process, we have the equation: \(PV = K_{2}\), where K2 is a constant

Differentiate the equations with respect to V

Now we need to find the slope of both curves at any point, so we will differentiate them with respect to V: For adiabatic: \(\frac{dP}{dV} = -\frac{\gamma K_{1}}{V^{\gamma+1}}\) For isothermal: \(\frac{dP}{dV}=-\frac{K_{2}}{V^{2}}\)

Calculate the ratio of slopes

Now, let's find the ratio of slopes of adiabatic and isothermal processes: \(\frac{dP_{ad}/dV}{dP_{iso}/dV} = \frac{-\frac{\gamma K_{1}}{V^{\gamma+1}}}{-\frac{K_{2}}{V^{2}}}\)

Simplify the ratio

Simplifying the ratio, we get: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}V^{2}}{K_{2}V^{\gamma+1}}\) At the point of intersection of both curves, \(K_{1} = K_{2}\) and \(PV^{\gamma}=PV\) which gives us: \(V^{\gamma-1}=1\) This implies that: \(V^{\gamma+1}=V^{2}\) Now substituting this value in the ratio: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}V^{\gamma+1}}{K_{2}V^{\gamma+1}}\) \(\Rightarrow \frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}}{K_{2}}\) Since, at the point of intersection, \(K_{1}=K_{2}\), our final answer becomes: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\gamma\) The correct option is (C) \(\gamma\).