Related Formula
Isothermal: PV = constant$$\text{Isothermal: } PV = \text{constant}$$
Adiabatic: PVγ = constant$$\text{Adiabatic: } PV^{\gamma} = \text{constant}$$
Adiabatic T-P relation: Tγ P1-γ = constant TγPγ - 1 = constant$$\text{Adiabatic T-P relation: } T^{\gamma} P^{1-\gamma} = \text{constant} \implies \frac{T^{\gamma}}{P^{\gamma - 1}} = \text{constant}$$
Core Logic
In a P-V$P-V$ diagram starting from the same initial state and undergoing expansion, the adiabatic curve is steeper than the isothermal curve.
Looking at the graph, Curve (B) is steeper, so it represents an adiabatic process. Curve (A) is less steep, so it represents an isothermal process.
Step 1: Check Options
Process A (Isothermal): PV = k$PV = k$ or T = k$T = k$.
Process B (Adiabatic): PVγ = k$PV^{\gamma} = k$.
Option (1) states: PVγ = k$PV^{\gamma} = k$ for B and PV = k$PV = k$ for A. This is entirely correct.
Note: For an adiabatic process, Tγ P1-γ = k$T^{\gamma} P^{1-\gamma} = k$, which can be written as TγPγ-1 = k$\frac{T^{\gamma}}{P^{\gamma-1}} = k$. Option (3) proposes Pγ-1Tγ = k$\frac{P^{\gamma-1}}{T^{\gamma}} = k$ for B, which is mathematically equivalent, but Option (1) is the standard and most direct NTA choice.
Pattern Recognition
Slope of isothermal = -P/V$= -P/V$. Slope of adiabatic = -γ P/V$= -\gamma P/V$. Since γ > 1$\gamma > 1$, adiabatic is always steeper. In expansions, adiabatic lies below isothermal.
Chapter Mix
Class 11 Physics: Thermodynamics