Physics Circle home Student Login

Home›Chapters›Thermodynamics PYQs›Thermodynamic Processes›Question 161

JEE Main 2022 (29 Jun, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (29 Jun, Shift 2)

Q161JEE Main 2022 · 29 JUN · SHIFT 2Medium

Thermodynamic Processes

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option

Show answer and solution

Answer: (d)

  1. Work by the gas = area under the PP–VV curve from V1V_1 to V2V_2.
  2. Isobaric: pressure stays at P1P_1 (largest area). Isothermal: PP falls as 1/V1/V. Adiabatic: PP falls fastest, as 1/Vγ1/V^{\gamma} (smallest area).
  3. W2<W1<W3W_2 < W_1 < W_3, so the answer is (d).
Report a mistake

More questions on Thermodynamic Processes