ΔQAB=0 adiabatic ΔQBC=ΔWBC=nRTℓn(VBVC)=450Rℓn(6×10−68×10−6)=450Rℓn(34)=450R(ln4−ℓn3)∴ΔQ=ΔQAB+ΔQBCΔQ=450R(ℓn4−ℓn3)
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
A poly-atomic molecule ( CV=3R,CP=4R, where R is gas constant) goes from phase space point A(PA=105 Pa, VA=4×10−6 m3) to point B(PB=5×104 Pa, VB=6×10−6 m3) to point C(PC=104 Pa,VC=8×10−6 m3). 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 :
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
500R(ln3+ln4)
- B
450R(ln4−ln3)
- C
500Rln2
- D
400Rln4
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