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At $27^{\circ} \mathrm{C}$, a closed vessel contains a mixture of equal weights of helium (mol. wt. =4), methane (mol, wt. =16) and sulphur dioxide (mol. wt. = 64). The pressure exerted by the mixture is $210 \mathrm{~mm}$. If the partial pressures of helium, methane and sulphur dioxide are $p_1, p_1$ and $p_3$ respectively, which one of the following is correct?
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Verified Answer
The correct answer is:
$p_1>p_2>p_3$
Let us suppose each gas has $64 \mathrm{~g}$ weight.
\begin{array}{|c|c|c|c|c|}
\hline Gases & Weight & \begin{array}{c}\text { wt./at. } \\
\text { wt. }\end{array} & \begin{array}{c}\text { No. of } \\
\text { moles }\end{array} & \begin{array}{c}\text { Partial } \\
\text { pressure }\end{array} \\
\hline \mathrm{He} & 64 & \frac{64}{4} & 16 & \frac{16}{21} \times 210=160 \\
\mathrm{~mm}\left(P_1\right) \\
\hline \mathrm{CH}_4 & 64 & \frac{64}{16} & 4 & \frac{4}{21} \times 210=40 \\
\mathrm{~mm}\left(P_2\right) \\
\hline \mathrm{SO}_2 & 64 & \frac{64}{64} & 1 & \frac{1}{21} \times 210=10 \\
\mathrm{~mm}\left(P_3\right) \\
\hline
\end{array}
$\therefore \quad p_1>p_2>p_3$
\begin{array}{|c|c|c|c|c|}
\hline Gases & Weight & \begin{array}{c}\text { wt./at. } \\
\text { wt. }\end{array} & \begin{array}{c}\text { No. of } \\
\text { moles }\end{array} & \begin{array}{c}\text { Partial } \\
\text { pressure }\end{array} \\
\hline \mathrm{He} & 64 & \frac{64}{4} & 16 & \frac{16}{21} \times 210=160 \\
\mathrm{~mm}\left(P_1\right) \\
\hline \mathrm{CH}_4 & 64 & \frac{64}{16} & 4 & \frac{4}{21} \times 210=40 \\
\mathrm{~mm}\left(P_2\right) \\
\hline \mathrm{SO}_2 & 64 & \frac{64}{64} & 1 & \frac{1}{21} \times 210=10 \\
\mathrm{~mm}\left(P_3\right) \\
\hline
\end{array}
$\therefore \quad p_1>p_2>p_3$
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