What is value of PV type of work for following reaction at 1 bar?

\(
\begin{aligned}
& 1 \text { mole of } \mathrm{C}_2 \mathrm{H}_4 \text { reacts with } 1 \text { mole of } \mathrm{HCl} \text { to } \\
& \text { produce } 1 \mathrm{~mole} \text { of } \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} \text {. } \\
& \text { Hence, } 150 \mathrm{~mL} \text { of } \mathrm{HCl} \text { would react with only } \\
& 150 \mathrm{~mL} \text { of } \mathrm{C}_2 \mathrm{H}_4 \text { to produce } 150 \mathrm{~mL} \text { of } \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} . \\
& \mathrm{V}_1=150 \mathrm{~mL}+150 \mathrm{~mL}=300 \mathrm{~mL}=0.3 \mathrm{dm}^3 \\
& \mathrm{~V}_2=150 \mathrm{~mL}=0.15 \mathrm{dm}^3 \\
& \mathrm{~W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\
& \therefore \quad \mathrm{W}=-1 \text { bar }\left(0.15 \mathrm{dm}^3-0.3 \mathrm{dm}^3\right) \\
& \therefore \quad \mathrm{W}=0.15 \mathrm{dm}^3 \text { bar } \\
& \therefore \quad W=0.15 \mathrm{dm}^3 \text { bar } \times 100 \frac{\mathrm{J}}{\mathrm{dm}^3 \text { bar }} \\
& \therefore \quad W=15 \mathrm{~J}
\end{aligned}
\)
The PV work in joules is \(15 \mathrm{~J}\).