Chemistry - Thermodynamics - I Question with Solution | TestHub

${\text{P}}_1=\frac{\text{nRT}}{{\text{V}}_1}=\frac{2\times \text{0.0821}\times \text{243.6}}{20}=\text{2 atm}$

$\Delta \text{S}=\text{n}\text{R} \text{ℓ}\text{n}\left(\frac{{\text{P}}_1}{{\text{P}}_2}\right)$

$=2\times 2\text{ℓ}\text{n }2$

$=2\times 2\times \text{0.693}$

$\left[\because \text{ℓ}\text{n }\mathrm{2 }\text{or }{\text{log}}_{\text{e}}\mathrm{ 2}=\text{0.693}\right]$

$=\text{2.77 cal/K}$