Answer
$-1.04\times10^{3}\,J$
Work Step by Step
$n=\frac{\text{mass in grams}}{\text{Molar mass}}=\frac{50.0\,g}{118.71\,g/mol}=0.421\,mol$
$pV=nRT$
$\implies V=\frac{nRT}{p}=\frac{0.421\,mol\times0.0821\,L\,atm\,K^{-1}mol^{-1}\times(273+25)\,K}{1.00\,atm}$
$=10.3\,L$
This volume is the change in volume.
$w=-P\Delta V=-(1.00\,atm)(10.3\,L)=-10.3\,L\cdot atm$
$=-10.3\,L\cdot atm\times\frac{101.3\,J}{1\,L\cdot atm}=-1.04\times10^{3}\,J$
