Considering a P-V diagram of an ideal gas, the slope of isothermal and adiabatic curves are related as

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isothermal curve slope = adiabatic curve slope

isothermal cure slope = $γ$ x adiabatic curve slope

adiabatic curve slope = $γ$ x isothermal curve slope

adiabatic curve slope = ½ isothermal curve slope

answer is C.

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Detailed Solution

At a given point (P₀, V₀) on PV diagram
${(\frac{dP}{dV})}_{isothermal} = - \frac{P_{0}}{V_{0}} and (\frac{dP}{dV}) adiabatic = - \frac{{γP}_{0}}{V_{0}}$

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