MHT CET · Chemistry · Thermodynamics (C)
Calculate the PV type of work for the following reaction at 1 bar pressure. \(\underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_{6(\mathrm{~g})}}+\underset{(150 \mathrm{~mL})}{\mathrm{HCl}_{(\mathrm{g})}} \longrightarrow \underset{(150 \mathrm{~mL})}{\mathrm{C}_3 \mathrm{H}_7 \mathrm{Cl}_{(\mathrm{g})}}\)
- A \(5.2 J\)
- B \(10.21 J\)
- C \(15.00 J\)
- D \(18.2 J\)
Answer & Solution
Correct Answer
(C) \(15.00 J\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{W} & =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ \mathrm{V}_1 & =150+150=300 \mathrm{~mL}=0.3 \mathrm{dm}^3 \\ \mathrm{~V}_2 & =150 \mathrm{~mL}=0.15 \mathrm{dm}^3 \\ \mathrm{P}_{\text {ext }} & =1 \mathrm{bar} \\ \mathrm{W} & =-1(0.15-0.3) \\ & =0.15 \mathrm{dm}^3 \text { bar } \\ & =15.0 \mathrm{~J} \quad\left(\because 100 \mathrm{~J}=1 \mathrm{dm}^3 \text { bar }\right)\end{aligned}\)
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