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A vessel contains $3.2 \mathrm{~g}$ of dioxygen gas at STP (273.15 $\mathrm{K}$ and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in $\mathrm{L}$ is :
(Given - molar volume at STP is $22.4 \mathrm{~L}$ )
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(Given - molar volume at STP is $22.4 \mathrm{~L}$ )
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Verified Answer
The correct answer is:
6.72
No. of moles of $\mathrm{O}_2=\frac{\text { Given weight of } \mathrm{O}_2}{\text { Mol. weight of } \mathrm{O}_2}$
$=\frac{3.2}{32}=0.1$
Molar vol. at $\mathrm{STP}=0.1 \times 22.4=2.24 \mathrm{~L}$
$\mathrm{P}_1=1 \mathrm{~atm}, \mathrm{~V}_1=2.24 \mathrm{~L}, \mathrm{P}_2=\frac{1}{3} \mathrm{~atm}, \mathrm{~V}_2=$ ?
According to Boyle's law : $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$\begin{aligned}
\therefore 1 \times 2.24 & =\frac{1}{3} \times \mathrm{V}_2 \\
\mathrm{~V}_2 & =2.24 \times 3 \\
& =6.72 \mathrm{~L}
\end{aligned}$
$=\frac{3.2}{32}=0.1$
Molar vol. at $\mathrm{STP}=0.1 \times 22.4=2.24 \mathrm{~L}$
$\mathrm{P}_1=1 \mathrm{~atm}, \mathrm{~V}_1=2.24 \mathrm{~L}, \mathrm{P}_2=\frac{1}{3} \mathrm{~atm}, \mathrm{~V}_2=$ ?
According to Boyle's law : $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$\begin{aligned}
\therefore 1 \times 2.24 & =\frac{1}{3} \times \mathrm{V}_2 \\
\mathrm{~V}_2 & =2.24 \times 3 \\
& =6.72 \mathrm{~L}
\end{aligned}$
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