PVy=constant
TV−1=constant
300(V1)1.4−1=TB(16V1)2/5
TB=300×28/5
Now for BC process
TBVB=TcVc
Tc=vBVcTB=2×300×28/5
Tc=1818.859
Tc=1819K
JEE Main 2020 — Physics Thermodynamics
Starting at temperature 300K, one mole of an ideal diatomic gas (γ=1.4) is first compressed adiabatically from volume V1 to V2=16V1. It is then allowed to expand isobarically to volume 2V2 . If all the processes are the quasi-static then the final temperature of the gas (in oK ) is (to the nearest integer) ___________.
Held on 9 Jan 2020 · Verified 6 Jul 2026.
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