Chemistry - Thermodynamics - 1 Question with Solution | TestHub

To calculate final volume${\mathrm{V}}_2=\left(\frac{\mathrm{n}\times \mathrm{R}\times \mathrm{T}}{{\mathrm{P}}_2}\right)$
${\text{V}}_{\text{2}}=\frac{2\times 0.0821\times 300}{4.926}=10\text{L}$
Now,${\text{V}}_{\text{1}}=1\text{L},{\text{V}}_{\text{2}}=10\text{L,}\text{T}=300\text{K,}\text{n}=2\text{ mole}$
$\text{W}=-2.303\text{nRT}\text{log}\left(\frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\right)$
$=-2.303\times 2\times 8.314\times 300\times \log \left(\frac{10}{1}\right)$
$=-11488.28 \mathrm{J}$