6 g of O₂ gas at STP is expanded so that volume is doubled. Thus, work done is

$\begin{array}{l}\text{ Expansion work, }\mathrm{W}=-{\mathrm{p}}_{\text{ext }}\mathrm{\Delta V}={-}_{\text{ext }}\left({\mathrm{V}}_{\text{final }}-{\mathrm{V}}_{\text{intial }}\right)\\ {\mathrm{p}}_{\text{ext }}=1\mathrm{atm}\\ {\mathrm{V}}_{\text{inital }}=\text{ volume of }6\mathrm{g}\left(=\frac{1}{3}\mathrm{mol}\right){\mathrm{O}}_2\\ \text{ at }\mathrm{STP}=5.6\mathrm{L}\\ {\mathrm{V}}_{\text{final }}=11.2\mathrm{L}\\ \text{ Thus, }\mathrm{\Delta V}=5.6\mathrm{L}\\ \mathrm{W}=-(1\mathrm{atm}\times 5.6\mathrm{L})=-5.6\mathrm{L}\text{ atm }\end{array}$