JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Heat & Thermodynamics

On a P-V diagram, the steeper of the two curves through a common point is the adiabatic one, because an adiabatic process ($P{V}^{\gamma }=\text{const}$) falls faster than an isothermal one ($PV=\text{const}$) for the same increase in volume, since $\gamma >1$.

Curve B is the steeper curve, so it is adiabatic:


Using $PV=nRT$ to eliminate $V$ from this relation gives the equivalent form in terms of $P$ and $T$, but that alternate form is not what is being matched to B here — B is captured directly by $P{V}^{\gamma }=k$.

Curve A is the less steep curve, so it is isothermal:


Both statements — $P{V}^{\gamma }=k$ for B and $PV=k$ for A, and separately $\frac{T^{\gamma }}{P^{\gamma -1}}=k$ for B and $T=k$ for A — describe the same physical processes correctly.