JEE Advance - Physics (2020 - Paper 1 Offline - No. 9)
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 $${{{W_{iso}}} \over {{W_{adia}}}}$$ = f ln 2, then f is ______.
답변
1.77
설명
$${{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}$$
$${p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}$$
$${W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}$$
$$ = {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}$$
$$ = {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}$$
$${W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2$$
$$ \therefore $$ $${{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2$$
$$ \Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78$$