The maximum work done in Joules expanding 16g of oxygen at 300 K and occupying a volume of 7dm³ isothermally until the volume becomes 35 dm³ is $a \times {10}^3 J$. ''a'' value is ------

Maximum work is always obtained in a reversible process

$\begin{array}{l}\therefore \mathrm{w}=-2.303\mathrm{nRTlog}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\\ \because {\mathrm{n}}_{{\mathrm{O}}_2}=(\mathrm{w}/\mathrm{m})=(16/32);\mathrm{R}=8.314\mathrm{J}/\mathrm{K}\text{ mole }\end{array}$

$\begin{array}{l}\mathrm{T}=300\mathrm{K},{\mathrm{V}}_1=7{\mathrm{dm}}^3,{\mathrm{V}}_2=35{\mathrm{dm}}^3\\ \therefore \mathrm{w}=-2.303\times (16/32)\times 8.314\times 300\times \log \left(\frac{35}{7}\right)\\ =-2.01\times {10}^3\mathrm{J}\end{array}$