If $100\mathrm{mol}$ of ${\mathrm{H}}_2{\mathrm{O}}_2$ decomposes at 1 bar and 300 K, the work done (kJ) by one mol of  ${\mathrm{O}}_2\left(\mathrm{g}\right)$ as it expands against 1 bar pressure is: $\left(R=8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)$

$2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\rightleftharpoons 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

 

Volume occupied by 1 mol of gas at $1\mathrm{bar}\left(={10}^2\mathrm{kPa}\right)$ and $300\mathrm{K}$ is 

$V=\frac{nRT}{p}=\frac{\left(1\mathrm{mol}\right)\left(8.3\mathrm{kPa}{\mathrm{dm}}^3{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\right)\left(300\mathrm{K}\right)}{\left({10}^2\mathrm{kPa}\right)}=24.9{\mathrm{dm}}^3$

Work of expansion of 1 mol of gas against 1 bar pressure is

$w=-p\left(\mathrm{\Delta }V\right)=-\left({10}^2\mathrm{kPa}\right)\left(24.9{\mathrm{dm}}^3\right)=-2490{\mathrm{kPadm}}^3\equiv -2490\mathrm{J}=-2.49\mathrm{kJ}$

Alternatively, $w=-p\mathrm{\Delta }V=-\left(\mathrm{\Delta }n\right)RT=-\left(1\mathrm{mol}\right)\left(8.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)\left(300\mathrm{K}\right)=-2490\mathrm{J}\equiv -2.49\mathrm{kJ}$