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JEE Main 2025 Physics, Thermodynamics — question figure A poly-atomic molecule ( C_V=3 R, C_P=4 R, where R is gas constant) goes from phase space…

JEE Main 2025 — Physics Thermodynamics

2025mcqhard

JEE Main 2025 Physics, Thermodynamics — question figure

A poly-atomic molecule ( CV=3R,CP=4RC_V=3 R, C_P=4 RCV​=3R,CP​=4R, where RRR is gas constant) goes from phase space point A(PA=105 Pa, VA=4×10−6 m3)\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)A(PA​=105 Pa, VA​=4×10−6 m3) to point B(PB=5×104 Pa, VB=6×10−6 m3)\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)B(PB​=5×104 Pa, VB​=6×10−6 m3) to point C(PC=104\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.C(PC​=104 Pa,VC=8×10−6 m3)\left.\mathrm{Pa}, \mathrm{V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^3\right)Pa,VC​=8×10−6 m3). A to B is an adiabatic path and BBB to CCC is an isothermal path. The net heat absorbed per unit mole by the system is :

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    500R(ln⁡3+ln⁡4)500 \mathrm{R}(\ln 3+\ln 4)500R(ln3+ln4)

  2. B

    450R(ln⁡4−ln⁡3)450 \mathrm{R}(\ln 4-\ln 3)450R(ln4−ln3)

  3. C

    500Rln⁡2500 \mathrm{R} \ln 2500Rln2

  4. D

    400Rln⁡4400 \mathrm{R} \ln 4400Rln4

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