Slope Of Adiabatic Versus Isothermal Curve

Question

On a pressure-volume diagram an isothermal curve and an adiabatic curve pass through the same point for an ideal gas, so how do their slopes compare at that common point?

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Building on the process equations in NCERT Class 11, Chapter 12 (Thermodynamics), the slope of each curve is found by differentiating its governing relation. For an isothermal process an ideal gas obeys $PV = 	ext{constant}$, so differentiating gives the slope $\left(\dfrac{dP}{dV}ight)_{	ext{iso}} = -\dfrac{P}{V}$. For an adiabatic process the gas obeys $PV^{\gamma} = 	ext{constant}$, which on differentiation gives $\left(\dfrac{dP}{dV}ight)_{	ext{adi}} = -\gamma\dfrac{P}{V}$. At the common point the values of $P$ and $V$ are identical for both curves, so the ratio of the two slopes is exactly $\gamma$, the ratio of specific heats, which is always greater than 1 for any gas. Therefore the adiabatic slope has the larger magnitude and the adiabatic curve is the steeper of the two. The option saying the isothermal is steeper is wrong because that would require $\gamma < 1$, which never holds. The option of equal slopes is wrong because it would need $\gamma = 1$, impossible for any real gas. The option claiming a positive adiabatic slope is wrong because both slopes are negative; pressure falls as volume rises in both processes. A consistency check seals it: since $\gamma > 1$ for all gases, $|{-\gamma P/V}| > |{-P/V}|$ always, so the adiabatic curve must drop more sharply at the shared point.

Slope Of Adiabatic Versus Isothermal Curve

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