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Calculate the PV type of work for the following reaction at 1 bar pressure.
$\underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_{6(\mathrm{~g})}}+\underset{(150 \mathrm{~mL})}{\mathrm{HCl}_{(\mathrm{g})}} \longrightarrow \underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_7 \mathrm{Cl}_{(\mathrm{g})}}$
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$\underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_{6(\mathrm{~g})}}+\underset{(150 \mathrm{~mL})}{\mathrm{HCl}_{(\mathrm{g})}} \longrightarrow \underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_7 \mathrm{Cl}_{(\mathrm{g})}}$
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Verified Answer
The correct answer is:
$15.00 J$
$\begin{aligned} \mathrm{W} & =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ \mathrm{V}_1 & =150+150=300 \mathrm{~mL}=0.3 \mathrm{dm}^3 \\ \mathrm{~V}_2 & =150 \mathrm{~mL}=0.15 \mathrm{dm}^3 \\ \mathrm{P}_{\text {ext }} & =1 \mathrm{bar} \\ \mathrm{W} & =-1(0.15-0.3) \\ & =0.15 \mathrm{dm}^3 \text { bar } \\ & =15.0 \mathrm{~J} \quad\left(\because 100 \mathrm{~J}=1 \mathrm{dm}^3 \text { bar }\right)\end{aligned}$
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