Answer
7.50 L
Work Step by Step
$V_{1}=5.00\,L$
$n_{1}= 8.00\,g\,O_{2}\times\frac{1\,mol\,O_{2}}{32.0\,g\,\,O_{2}}=0.250\,mol$
$n_{2}=(8.00+4.00)g\,\,O_{2}\times\frac{1\,mol\,O_{2}}{32.0\,g}$
$=0.375\,mol$
Now if the temperature and pressure do not change, according to Avogadro's law
$\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}$
$\implies V_{2}=\frac{V_{1}\times n_{2}}{n_{1}}=\frac{5.00\,L\times0.375\,mol}{0.250\,mol}$
$=7.50\,L$
