Calculate the maximum work done in kJ expanding 16 g of oxygen at 300 K and occupying a volume of 5 dm${}^3$ isothermally until the volume becomes 25 dm${}^3$. (Ignoring the sign of value)

The correct option is B 2
Reversible work is maximum work
As the process is isothermal ,So Work
$\therefore w=-2.303{\text{nRT log}}_{10}\left(\frac{V_2}{V_1}\right)$
$V_1=\text{initial volume}=5d{m}^3$
$V_2=\text{final volume}=25d{m}^3$
moles of gas =$\frac{16}{32}=0.5$

Putting these values in the formula above, we get
$w=2.303\times 0.5\times 8.314\times 300\text{log}\frac{25}{5}=2007J$