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Heat and Thermodynamics question

2025 · 29 Jan · Shift 2 · Q65
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Heat and Thermodynamics question

2025 · 29 Jan · Shift 2 · Q65

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
JEE Main 2025 (Online) 29th January Evening Shift Physics - Heat and Thermodynamics Question 47 EnglishA poly-atomic molecule (CV=3R,CP=4R\left(C_V=3 R, C_P=4 R\right.(CV​=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=104Pa,VC=8×10−6 m3\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.\mathrm{Pa}, \mathrm{V}_C=8 \times 10^{-6} \mathrm{~m}^3C(PC​=104Pa,VC​=8×10−6 m3). A to BBB is an adiabatic path and BBB to CCC is an isothermal path. The net heat absorbed per unit mole by the system is :
  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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