Starting with the same initial conditions, an ideal gas expands from volume $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$ in three different ways. The work done by the gas is $\mathrm{W}_{1}$ if the process is purely isothermal, $\mathrm{W}_{2}$ if the process is purely adiabatic and $\mathrm{W}_{3}$ if the process is purely isobaric. Then, choose the correct option

(D) $\mathrm{W}_{2}<\mathrm{W}_{1}<\mathrm{W}_{3}$

(A) $\mathrm{W}_{1}<\mathrm{W}_{2}<\mathrm{W}_{3}$

(B) $\mathrm{W}_{2}<\mathrm{W}_{3}<\mathrm{W}_{1}$

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Heat & Thermodynamics

Starting with the same initial conditions, an ideal gas expands from volume

V_{1}

V_{2}

in three different ways. The work done by the gas is

W_{1}

if the process is purely isothermal,

W_{2}

if the process is purely adiabatic and

W_{3}

if the process is purely isobaric. Then, choose the correct option

Answer: (D)

Work done by a gas equals the area under the P-V curve.

For the same initial state and final volume:

Isobaric process (constant pressure) has the highest pressure throughout, giving maximum area.
Isothermal process (PV = constant) lies between isobaric and adiabatic curves.
Adiabatic process (PV^γ = constant, γ > 1) has the steepest descent, giving minimum area.

Therefore:

W_{2} < W_{1} < W_{3}

Answer: (D)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option