Calculate the magnitude of maximum work done (in $kJ$) in expanding 6 g of hydrogen at 300 K and occupying a volume of $10d{m}^3$ isothermally untill the volume becomes $30d{m}^3$.
(Given $log5=0.6989$)

The correct option is C 8221 J
Reversible work is maximum work.
$\therefore W=-2.303nRT{\log }_{10}\left(\frac{V_2}{V_1}\right)$
where $V_2$ and $V_1$ are the final and initial volume
n = moles of the gas = $\frac{6}{2}$
$W=-2.303\times \frac{6}{2}\times 8.314\times 300\log \frac{30}{10}$
$W=-8221.95J$
so, magnitude of work done will be $8221J$