Equation of the line BC, P=P0−V02P0(V−2V0), the temperature =1×RP0V−V02P0V2+4P0V⇒T=RP0[5V−V02V2]⇒dVdT=0
⇒5−V04V=0⇒V=45V0
⇒T=RP0[5×45V0−V02×1625V02]⇒T=825RP0V0.
JEE Main 2018 — Physics Thermodynamics
One mole of an ideal monatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by:

Held on 16 Apr 2018 · Verified 6 Jul 2026.
425RP0V0
85RP0V0
825RP0V0
1625RP0V0
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