Search any question & find its solution
Question:
Answered & Verified by Expert
What will be the pressure of the gaseous mixture when \(0.5 \mathrm{~L}\) of \(\mathrm{H}_2\) at \(0.8\) bar and \(2.0 \mathrm{~L}\) of dioxygen at \(0.7\) bar are introduced in a \(1 \mathrm{~L}\) vessel at \(27^{\circ} \mathrm{C}\) ?
Solution:
1405 Upvotes
Verified Answer
Partial pressure of \(\mathrm{H}_2\) in \(1 \mathrm{~L}\) vessel
\(\begin{aligned}
&\mathrm{P}_1=0.8 \mathrm{bar}, \mathrm{V}_1=0.5 \mathrm{~L} \\
&\mathrm{P}_2=? \mathrm{~V}_2=1.0 \mathrm{~L}
\end{aligned}\)
Temperature remaining constant,
Partial pressure of \(\mathrm{O}_2\) in \(1 \mathrm{~L}\) vessel
\(\mathrm{P}_1^{\prime}=0.7 \mathrm{bar}, \mathrm{V}_1^{\prime}=2.0 \mathrm{~L}, \mathrm{P}_2^{\prime}=? \mathrm{~V}_2^{\prime}=1.0 \mathrm{~L}\)
Temperature remaining constant
or, \(\mathrm{P}_2^1=1.4\) bar or, \(\mathrm{P}_{\mathrm{O}_2}=1.4\) bar
Total pressure \(=\mathrm{P}_{\mathrm{H}_2}+\mathrm{P}_{\mathrm{O}_2}=0.4 \mathrm{bar}+1.4 \mathrm{bar}=1.8 \mathrm{bar}\)
\(\begin{aligned}
&\mathrm{P}_1=0.8 \mathrm{bar}, \mathrm{V}_1=0.5 \mathrm{~L} \\
&\mathrm{P}_2=? \mathrm{~V}_2=1.0 \mathrm{~L}
\end{aligned}\)
Temperature remaining constant,
Partial pressure of \(\mathrm{O}_2\) in \(1 \mathrm{~L}\) vessel
\(\mathrm{P}_1^{\prime}=0.7 \mathrm{bar}, \mathrm{V}_1^{\prime}=2.0 \mathrm{~L}, \mathrm{P}_2^{\prime}=? \mathrm{~V}_2^{\prime}=1.0 \mathrm{~L}\)
Temperature remaining constant
or, \(\mathrm{P}_2^1=1.4\) bar or, \(\mathrm{P}_{\mathrm{O}_2}=1.4\) bar
Total pressure \(=\mathrm{P}_{\mathrm{H}_2}+\mathrm{P}_{\mathrm{O}_2}=0.4 \mathrm{bar}+1.4 \mathrm{bar}=1.8 \mathrm{bar}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.