QJEEAdv15Ph8274: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph8274 Solution: An ideal monoatomic gas is confined in a horizontal cylinder by …

Initially both the compartments has same pressure as they are in equilibrium.

Suppose spring is compressed by x on heating the gas.

Let A be the area of cross-section of piston. As gas is ideal monoatomic, so

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ...... (i)

Force on spring by gas = kx

$\therefore$ ${P_2} = {P_1} + {{kx} \over A}$ ...... (ii)

Case I : When ${V_2} = 2{V_1}$ , ${T_2} = 3{T_1}$

From Eqn. (i)

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(2{V_1})} \over {3{T_1}}} \Rightarrow {P_2} = {3 \over 2}{P_1}$

Putting this value in eqn. (ii) we get

${3 \over 2}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 2}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1} - {V_1}} \over A} = {{{V_1}} \over A}$

Energy stored in the spring

$= {1 \over 2}k{x^2} = {1 \over 2}(kx)(x) = {{{P_1}{V_1}} \over 4}$

So, option (a) is correct.

Change in internal energy,

$\Delta U = {f \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2}\left( {{3 \over 2}{P_1} \times 2{V_1} - {P_1}{V_1}} \right) = 3{P_1}{V_1}$

So, option (b) is correct.

Case II : When ${V_2} = 3{V_1}$ and ${T_2} = 4{T_1}$

From Eqn. (i),

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(3{V_1})} \over {4{T_1}}} \Rightarrow {P_2} = {4 \over 3}{P_1}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1}} \over A}$

From eqn. (ii),

${4 \over 3}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 3}$

Gas is heated very slowly so pressure on the other compartment remains same.

Work done by gas = Work done by gas on atmosphere + Energy stored in spring.

${W_g} = {P_1}Ax + {1 \over 2}k{x^2} = {P_1}(2{V_1}) + {1 \over 2}\left( {{{{P_1}A} \over 3}} \right)\left( {{{2{V_1}} \over A}} \right)$

$= 2{P_1}{V_1} + {1 \over 3}{P_1}{V_1} = {7 \over 3}{P_1}{V_1}$

So, option (c) is correct.

Heat supplied to the gas,

$\Delta Q = {W_g} + \Delta U$

$= {7 \over 3}{P_1}{V_1} + {3 \over 2}({P_2}{V_2} - {P_1}{V_1})$

$= {7 \over 3}{P_1}{V_1} + {3 \over 2}\left( {{4 \over 3}{P_1} \times 3{V_1} - {P_1}{V_1}} \right)$

$= {7 \over 3}{P_1}{V_1} + {9 \over 2}{P_1}{V_1} = {{41} \over 6}{P_1}{V_1}$

So, option (d) is incorrect.