Isothermal Versus Adiabatic Curves

Question

On a single pressure-volume diagram an isothermal curve and an adiabatic curve for the same ideal gas pass through one common point. Which statement correctly compares their slopes at that intersection?

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Differentiating the isothermal condition $PV = 	ext{constant}$ gives $P\,dV + V\,dP = 0$, so the isothermal slope is $\left(\dfrac{dP}{dV}ight)_T = -\dfrac{P}{V}$. Differentiating the adiabatic condition $PV^{\gamma} = 	ext{constant}$ gives $\gamma P V^{\gamma-1}dV + V^{\gamma}dP = 0$, so the adiabatic slope is $\left(\dfrac{dP}{dV}ight)_{	ext{adia}} = -\gamma\dfrac{P}{V}$. At the shared point $P$ and $V$ are identical, so the adiabatic slope is exactly $\gamma$ times the isothermal slope in magnitude. Since $\gamma > 1$ for any real gas, the adiabatic curve falls more steeply. The option that the isothermal is steeper reverses the ratio. The option of identical slopes ignores the factor $\gamma$ entirely. The option that the adiabatic is flatter by $\gamma$ contradicts $\gamma > 1$. Physically, during adiabatic expansion the gas also cools, so its pressure drops faster than in the isothermal case where temperature is held fixed. This steeper slope has a practical consequence: in the sound-propagation context, Newton's isothermal assumption underestimated the speed of sound until Laplace corrected it using the adiabatic relation, restoring agreement with measurement. A check confirms that with $\gamma = 1.4$ the adiabatic line is 1.4 times as steep at the crossing point, consistent with experiment.

Isothermal Versus Adiabatic Curves

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