Chemistry - Thermodynamics - I Question with Solution | TestHub

For isothermal reversible expansion

$\mathrm{-w}=\mathrm{q}=\mathrm{n}\mathrm{R}\mathrm{T}\times 2.303\log \frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}$

$=2\mathrm{R}\mathrm{T}\times 2.303\log \frac{20}{2}$

$=2\times 2\times \mathrm{T}\times 2.303\times 1=9.2\mathrm{T}$

Entropy change,$\text{S}=\frac{\text{q}}{\text{T}}=\frac{9.2\text{T}}{\text{T}}=9.2{\text{ Cal K}}^{-1}$