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Keeping temperature constant the pressure of $11.2 \mathrm{dm}^3$ of a gas was increased from $105 \mathrm{kPa}$ to $420 \mathrm{kPa}$. What is the new volume of gas?
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$2.8\mathrm{dm}^3$
$\begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ & 105 \times 11.2=420 \times \mathrm{V}_2 \\ & \mathrm{~V}_2=\frac{105 \times 11.2}{420} \\ & 2.8 \mathrm{dm}^3\end{aligned}$
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