The ratio of the slopes of isothermal and adi | vedclass

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(C) For an isothermal process,the equation is $pV = K$. Differentiating both sides with respect to $V$,we get $p + V \frac{dp}{dV} = 0$,which implies

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$\frac{1}{\gamma}$

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(C) For an isothermal process,the equation is $pV = K$. Differentiating both sides with respect to $V$,we get $p + V \frac{dp}{dV} = 0$,which implies $\left(\frac{dp}{dV}ight)_{	ext{iso}} = -\frac{p}{V}$. For an adiabatic process,the equation is $pV^{\gamma} = K'$. Differentiating both sides with respect to $V$,we get $\frac{dp}{dV} V^{\gamma} + p \gamma V^{\gamma-1} = 0$. Rearranging this gives $\left(\frac{dp}{dV}ight)_{	ext{adia}} = -\gamma \frac{p}{V}$. Taking the ratio of the slope of the isothermal curve to the slope of the adiabatic curve: $\frac{(\frac{dp}{dV})_{	ext{iso}}}{(\frac{dp}{dV})_{	ext{adia}}} = \frac{-p/V}{-\gamma p/V} = \frac{1}{\gamma}$.

The ratio of the slopes of isothermal and adiabatic curves is

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