Verified Solution • 2025–26 Session moderate

QJEEM21Ph1167: JEE Main 2021 Physics - Class 11 JEEM21Ph1167 Solution: If one mole of an ideal gas at (P 1 , V 1 ) is allowed to…

If one mole of an ideal gas at (P1, V1) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B →\to C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :

JEE Main 2021 (Online) 24th February Evening Shift Physics - Heat and Thermodynamics Question 249 English

Multiple Choice Options

  • A −RT2(γ−1)- {{RT} \over {2(\gamma - 1)}}
  • B RT(ln⁡2−12(γ−1))RT\left( {\ln 2 - {1 \over {2(\gamma - 1)}}} \right)
  • C RTln⁡2RT\ln 2
  • D 00

Step-by-Step Verified Solution

Let pi, pf, Vi and Vf be the initial and final pressure and volume.

Given, AB is isothermal ( Δ\Delta T = 0),

BC is isochoric ( Δ\Delta V = 0) and CA is adiabatic ( Δ\Delta Q = 0)

Since, isothermal work (WAB) = p1V1ln⁡VfVi{p_1}{V_1}\ln {{{V_f}} \over {{V_i}}}

JEE Main 2021 (Online) 24th February Evening Shift Physics - Heat and Thermodynamics Question 249 English Explanation
where, Vi and Vf are volume at A and B, respectively.

∴\therefore WAB=p1V1ln⁡2V1V1=p1V1ln⁡2{W_{AB}} = {p_1}{V_1}\ln {{2{V_1}} \over {{V_1}}} = {p_1}{V_1}\ln 2

Since, at constant volume, work done is zero.

∴\therefore WBC=0{W_{BC}} = 0

Since, WCA is an adiabatic work done, i.e.

WCA=11−γ(pfVf−piVi){W_{CA}} = {1 \over {1 - \gamma }}({p_f}{V_f} - {p_i}{V_i})

⇒WCA=11−γ(p1V1−p14×2V1)\Rightarrow {W_{CA}} = {1 \over {1 - \gamma }}({p_1}{V_1} - {{{p_1}} \over 4} \times 2{V_1})

=11−γ(p1V1−p1V1/2)=11−γp1V12= {1 \over {1 - \gamma }}({p_1}{V_1} - {p_1}{V_1}/2) = {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}

∴\therefore Net work done, Wnet=WAB+WBC+WCA{W_{net}} = {W_{AB}} + {W_{BC}} + {W_{CA}}

=p1V1ln⁡2+0+11−γp1V12= {p_1}{V_1}\ln 2 + 0 + {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}

=p1V1[ln⁡2+1/2(1−γ)]= {p_1}{V_1}[\ln 2 + 1/2(1 - \gamma )]

From ideal gas law, pV=nRTpV = nRT

∴\therefore Wnet=RT[ln⁡2−1/2(γ−1)]{W_{net}} = RT[\ln 2 - 1/2(\gamma - 1)] ( ∵\because n = 1)

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