Isothermal Versus Adiabatic Curves

Differentiating the isothermal condition $PV=\text{constant}$ gives $PdV+VdP=0$, so the isothermal slope is ${\left(\frac{dP}{dV}\right)}_T=-\frac{P}{V}$. Differentiating the adiabatic condition $P{V}^{\gamma }=\text{constant}$ gives $\gamma P{V}^{\gamma -1}dV+V^{\gamma }dP=0$, so the adiabatic slope is ${\left(\frac{dP}{dV}\right)}_{\text{adia}}=-\gamma \frac{P}{V}$. 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.