Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph43dc: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph43dc Solution: Consider one mole of helium gas enclosed in a container at…

Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It expands isothermally to volume 4V1. After this, the gas expands adiabatically and its volume becomes 32V1. The work done by the gas during isothermal and adiabatic expansion processes are Wiso and Wadia, respectively. If the ratio WisoWadia{{{W_{iso}}} \over {{W_{adia}}}} = f ln 2, then f is ______.

Step-by-Step Verified Solution

JEE Advanced 2020 Paper 1 Offline Physics - Heat and Thermodynamics Question 44 English Explanation

p14(4V1)5/3=p2(32V1)5/3{{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}

p2=p14(18)5/3=p1128{p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}

Wadi=p1V1−p2V2γ−1{W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}

=p1V1−p1128(32V1)53−1= {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}

=p1V1(3/4)2/3=98p1V1= {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}

Wiso=p1V1ln⁡(4V1V1)=2p1V1ln⁡2{W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2

∴\therefore WisoWadi=169ln⁡2{{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2

⇒f=169=1.7778≈1.78\Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78

Related Learning Resources & Exam Preparation

JEE Advanced 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