Paragraph A thermally insulating cylinder has a thermally...

Writing ${C}_{V}$ in terms of degree of freedom of the gas $f$, we get,

${C}_{V} = \dfrac{f}{2}R = 2R ⇒ f = 4,$

Therefore, adiabatic index of gas will be given by,$\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{4} = \dfrac{3}{2}$

The gas of the right portion has undergone through an adiabatic process so,

${T}_{i}{V}_{i}^{\gamma - 1} = {T}_{f}{V}_{f}^{\gamma - 1} ⇒ {T}_{0}{V}_{0}^{\dfrac{3}{2} - 1} = {T}_{R}{\left(\dfrac{{V}_{0}}{2}\right)}^{\dfrac{3}{2} - 1}$

$⇒ {T}_{R} = \sqrt{2}{T}_{0} ⇒ \dfrac{{T}_{R}}{{T}_{0}} = \sqrt{2}$

Work done by the gas \left(2\right)$,$= - ∆ U = - n{C}_{V}\left({T}_{f} - {T}_{i}\right)$</p><p>$= - \left(1\right)\left(2R\right)\left(\sqrt{2}{T}_{0} - {T}_{0}\right)$</p><p>$= - \left(2\sqrt{2} - 2\right)R{T}_{0}$</p><p>So work done by the gas \left(1\right)$,$= + \left(2\sqrt{2} - 2\right)R{T}_{0}$

$PV = nRT ⇒ P ∝ \dfrac{T}{V}$

For gas \left(2\right), T \to \sqrt{2}$times of initial value, Volume$\to \dfrac{1}{2}$times of initial value</p><p>So,$P ∝ \dfrac{\sqrt{2}}{\dfrac{1}{2}} = 2\sqrt{2}$times of initial value$⇒ {P}_{f} = 2\sqrt{2}{P}_{0}$</p><p>So pressure of the gas$( 1 )$will also be$2\sqrt{2}$times and volume of gas$( 1 )$becomes$\dfrac{3}{2}$times</p><p>$T ∝ PV$</p><p>$⇒ T ∝ \left(2\sqrt{2}\left(\dfrac{3}{2}\right)\right) = 3\sqrt{2}$times of initial value</p><p>${T}_{L} = 3\sqrt{2}{T}_{0}$</p><p>$∆ U$for the left gas$= n{C}_{V}\left({T}_{f} - {T}_{i}\right)$</p><p>$= \left(1\right)\left(2R\right)\left(3\sqrt{2}{T}_{0} - {T}_{0}\right) = \left(6\sqrt{2} - 2\right)R{T}_{0}$</p><p>Heat given to gas$( 1 )$,</p><p>$Q = W + ∆ U$</p><p>$= \left(2\sqrt{2} - 2\right)R{T}_{0} + \left(6\sqrt{2} - 2\right)R{T}_{0}$</p><p>$= 4{RT}_{0}\left(2\sqrt{2} - 1\right)$</p><p>Therefore,</p><p>$\dfrac{Q}{R{T}_{0}} = 4\left(2\sqrt{2} - 1\right)