Verified Solution • 2025–26 Session moderate

QJEEM18Phfab7: JEE Main 2018 Physics - Class 11 JEEM18Phfab7 Solution: One mole of an ideal monoatomic gas is taken along the path…

One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

JEE Main 2018 (Online) 16th April Morning Slot Physics - Heat and Thermodynamics Question 356 English

Multiple Choice Options

  • A 2516 PoVoR{{25} \over {16}}\,{{{P_o}{V_o}} \over R}
  • B 258 PoVoR{{25} \over {8}}\,{{{P_o}{V_o}} \over R}
  • C 254 PoVoR{{25} \over {4}}\,{{{P_o}{V_o}} \over R}
  • D 58 PoVoR{{5} \over {8}}\,{{{P_o}{V_o}} \over R}

Step-by-Step Verified Solution

Equation of line BC,

P = P0 −- 2P0V0(V−2V0){{2{P_0}} \over {{V_0}}}\left( {V - 2{V_0}} \right)

As      \,\,\,\,\, PV = nRT

∴    \therefore\,\,\,\, T = PVnR{{PV} \over {nR}}

= P0V−2P0V2V0+4P0V1×R{{{P_0}V - {{2{P_0}{V_2}} \over {{V_0}}} + 4{P_0}V} \over {1 \times R}}

(As given n = 1 mole gas)

⇒\Rightarrow    \,\,\, T = P0R[5V−2V2V0]{{{P_0}} \over R}\left[ {5V - {{2{V^2}} \over {{V_0}}}} \right]

For maximum value of T

dTdV=0{{dT} \over {dV}} = 0

∴   \therefore\,\,\, 5 −- 4VV0=0{{4V} \over {{V_0}}} = 0

⇒\Rightarrow    \,\,\, V = 54V0{5 \over 4}{V_0}

∴   \therefore\,\,\, Tmax = P0R[5×5V04−2V0×2516V02]{{{P_0}} \over R}\left[ {5 \times {{5{V_0}} \over 4} - {2 \over {{V_0}}} \times {{25} \over {16}}{V_0}^2} \right]

= 258×P0 V0R{{25} \over 8} \times {{{P_0}\,{V_0}} \over R}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring