MHT CET · Chemistry · Thermodynamics (C)
100 ml of \(\mathrm{H}_{2(\mathrm{~g})}\) and 100 ml of \(\mathrm{Cl}_{2(\mathrm{~g})}\) were allowed to react at 1 bar pressure as \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\) What will be the PV type of work done diuring reaction?
- A Zero J
- B +10 J
- C -10 J
- D \(\quad-100 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) Zero J
Step-by-step Solution
Detailed explanation
1 mole of \(\mathrm{H}_2\) reacts with 1 mole of \(\mathrm{Cl}_2\) to produce 2 moles of HCl.
Hence, 100 mL of \(\mathrm{H}_2\) would react with 100 mL of \(\mathrm{Cl}_2\) to produce 200 mL of HCl.
\(\mathrm{V}_1 =100+100=200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \)
\( \mathrm{V}_2 =200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \)
\( \mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{~V}_2-\mathrm{V}_1\right)\)\(\left(\because 100 \mathrm{~J}=1 \mathrm{dm}^3 \text { bar }\right).\)
\( \mathrm{W} =-1(0.2-0.2) \)
\( =\text { zero } \mathrm{J}\)
Hence, 100 mL of \(\mathrm{H}_2\) would react with 100 mL of \(\mathrm{Cl}_2\) to produce 200 mL of HCl.
\(\mathrm{V}_1 =100+100=200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \)
\( \mathrm{V}_2 =200 \mathrm{~mL}=0.2 \mathrm{dm}^3 \)
\( \mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{~V}_2-\mathrm{V}_1\right)\)\(\left(\because 100 \mathrm{~J}=1 \mathrm{dm}^3 \text { bar }\right).\)
\( \mathrm{W} =-1(0.2-0.2) \)
\( =\text { zero } \mathrm{J}\)
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