A gas occupies $11.2 \mathrm{dm}^3$ at 105 kPa What is the volume if pressure is increased to 210 kPa ?
A gas occupies $11.2 \mathrm{dm}^3$ at 105 kPa What is the volume if pressure is increased to 210 kPa ?
- $5.6 \mathrm{dm}^3$
- $16.8 \mathrm{dm}^3$
- $22.4 \mathrm{dm}^3$
- $33.6 \mathrm{dm}^3$
Solution
According to Boyle's law,
$\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
\therefore \quad & \mathrm{~V}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3
\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 1)
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