MCQMediumJEE Main 2025 · 29 January, Shift 2Isothermal & Adiabatic Processes

Physics Question from JEE Main 2025 · 29 January, Shift 2

A poly-atomic molecule (CV=3RC_V = 3R, CP=4RC_P = 4R, where RR is the gas constant) goes from phase space point A (PA=105Pa, VA=4×106m3P_A = 10^5 \, Pa,\ V_A = 4 \times 10^{-6} \, m^3) to point B (PB=5×104Pa, VB=6×106m3P_B = 5 \times 10^4 \, Pa,\ V_B = 6 \times 10^{-6} \, m^3) to point C (PC=104Pa, VC=8×106m3P_C = 10^4 \, Pa,\ V_C = 8 \times 10^{-6} \, m^3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:

Pressure-volume graph showing points A, B, and C with dashed verticals at V_A, V_B, and V_C, an adiabatic path from A to B, and an isothermal path from B to C labeled 500 K and 450 K.
  • A

    500R(ln3+ln4)500R(\ln 3 + \ln 4)

  • B

    450R(ln4ln3)450R(\ln 4 - \ln 3)

  • C

    500Rln2500R \ln 2

  • D

    400Rln4400R \ln 4

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Isothermal & Adiabatic Processes questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions