One mole of an ideal monoatomic gas undergoes two reversible processes $(\mathrm{A}\to \mathrm{B}$ and $\mathrm{B}\to \mathrm{C})$ as shown in the given figure:

$\mathrm{A}\to \mathrm{B}$ is an adiabatic process. If the total heat absorbed in the entire process $(\mathrm{A}\to \mathrm{B}$ and $\mathrm{B}\to \mathrm{C})$ is ${\mathrm{RT}}_2\ln 10$, the value of $2{\mathrm{logV}}_3$ is
[Use molar heat capacity of the gas at constant pressure, ${\mathrm{C}}_{\mathrm{p},\mathrm{m}}=\frac{5}{2}\mathrm{R}$]

${\mathrm{q}}_{\mathrm{A}\to \mathrm{C}}={\mathrm{RT}}_2\ln 10$

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

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

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

${\mathrm{q}}_{\mathrm{A}\to \mathrm{c}}={\mathrm{nRT}}_2\ln \left(\frac{{\mathrm{V}}_3}{ {\mathrm{V}}_2}\right)$     ...(i)

For $\mathrm{B}\to \mathrm{C}$
$\mathrm{\Delta E}=\mathrm{q}+\mathrm{w}$

$\mathrm{\Delta E}=0$                  (since isothermic)

$\mathrm{q}=-\mathrm{w}$

$=-\left(-{\mathrm{nRT}}_2\ln \frac{{\mathrm{V}}_3}{ {\mathrm{V}}_2}\right)$

$={\mathrm{nRT}}_2\ln \left(\frac{{\mathrm{V}}_3}{ {\mathrm{V}}_2}\right)$

${\mathrm{q}}_{\mathrm{B}\to \mathrm{C}}={\mathrm{nRT}}_2\ln \left(\frac{{\mathrm{V}}_3}{ {\mathrm{V}}_2}\right)$

${\mathrm{q}}_{\mathrm{B}\to \mathrm{C}}={\mathrm{RT}}_2\ln \left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right) [\mathrm{Since} \mathrm{n} =1]$

From $\mathrm{A}\to \mathrm{B}$
${\mathrm{T}}_1 {\mathrm{V}}_1^{\mathrm{\gamma }-1}={\mathrm{T}}_2 {\mathrm{V}}_2^{\mathrm{\gamma }-1}$

$600 {\mathrm{V}}_1^{\gamma -1}=60 {\mathrm{V}}_2^{\gamma -1}$

$10\times {10}^{\frac{5}{3}-1}={\mathrm{V}}_2^{\mathrm{\gamma }-1}$

${10}^{5/3}={\mathrm{V}}_2^{\frac{5}{3}-1}$

${10}^{5/3}={\mathrm{V}}_2^{2/3}$

${\mathrm{V}}_2={10}^{\frac{5}{3}\times \frac{3}{2}}={10}^{\frac{5}{2}}$

${\mathrm{V}}_2={10}^{\frac{5}{2}} ...\left(2\right)$
From equation $\left(1\right)$
${\mathrm{q}}_{\mathrm{A}\to \mathrm{C}}={\mathrm{nRT}}_2\ln \left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right)$
Given, ${\mathrm{q}}_{\mathrm{A}\to \mathrm{C}}={\mathrm{RT}}_2\ln 10$

${\mathrm{RT}}_2\ln 10={\mathrm{RT}}_2\ln \left(\frac{{\mathrm{V}}_3}{ {\mathrm{V}}_2}\right)$

$\ln 10=\ln \left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right)$

$\ln 10=\ln \left(\frac{{\mathrm{V}}_3}{{10}^{\frac{5}{2}}}\right)$

$10=\frac{{\mathrm{V}}_3}{{10}^{\frac{5}{2}}}$

${\mathrm{V}}_3={10}^{1+\frac{5}{2}}={10}^{\frac{7}{2}}$

$2{\mathrm{logV}}_3=2\log {10}^{7/2}=7$