Answer
315 J
Work Step by Step
$\Delta V= (1.22-5.55)\,L=-4.33\,L=-4.33\times10^{-3}\,m^{3}$
$P=1.00\,atm=101325\,Pa=101325\,N/m^{2}$
$w= -P\Delta V=-101325\frac{N}{m^{2}}\times-4.33\times10^{-3}m^{3}=439\,J$
$q= -124\,J$ (q is negative as heat is released)
$\Delta E=q+w=-124\,J+439\,J=315\,J$
