One mole of an ideal monoatomic gas undergoes two reversible...

${q}_{A \to C} = {RT}_{2}\ln 10$

${q}_{A \to B} = 0$ $\text{(} ∵ \text{adiabatic)}$

${q}_{A \to C} = {q}_{A \to B} + {q}_{B \to C}$

${q}_{A \to C} = {q}_{B \to C}$

${q}_{A \to c} = {nRT}_{2}ln$\left(\dfrac{{V}_{3}}{{V}_{2}}\right)$...(i)</p><p>For$B \to C$<br />$\Delta E = q + w$</p><p>$\Delta E = 0$(since isothermic)</p><p>$q = - w$</p><p>$= - \left(- {nRT}_{2}ln\dfrac{{V}_{3}}{{V}_{2}}\right)$</p><p>$= {nRT}_{2}\ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right)$</p><p>${q}_{B \to C} = {nRT}_{2}\ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right)$</p><p>${q}_{B \to C} = {RT}_{2}\ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right) [ Sincen = 1 ]$</p><p>From$A \to B$<br />${T}_{1}{V}_{1}^{\gamma - 1} = {T}_{2}{V}_{2}^{\gamma - 1}$</p><p>$600{V}_{1}^{\gamma - 1} = 60{V}_{2}^{\gamma - 1}$</p><p>$10 \times {10}^{\dfrac{5}{3} - 1} = {V}_{2}^{\gamma - 1}$</p><p>${10}^{5 / 3} = {V}_{2}^{\dfrac{5}{3} - 1}$</p><p>${10}^{5 / 3} = {V}_{2}^{2 / 3}$</p><p>${V}_{2} = {10}^{\dfrac{5}{3} \times \dfrac{3}{2}} = 10\dfrac{5}{2}$</p><p>${V}_{2} = 10\dfrac{5}{2} . . . ( 2 )$<br />From equation$( 1 )$<br />${q}_{A \to C} = {nRT}_{2}\ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right)$<br />Given,${q}_{A \to C} = {RT}_{2}\ln 10$</p><p>${RT}_{2}\ln 10 = {RT}_{2}\ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right)$</p><p>$\ln 10 = \ln \left(\dfrac{{V}_{3}}{{V}_{2}}\right)$</p><p>$\ln 10 = \ln \left(\dfrac{{V}_{3}}{10\dfrac{5}{2}}\right)$</p><p>$10 = \dfrac{{V}_{3}}{10\dfrac{5}{2}}$</p><p>${V}_{3} = {10}^{1 + \dfrac{5}{2}} = 10\dfrac{7}{2}$</p><p>$2{\log V}_{3} = 2log{10}^{7 / 2} = 7