Keeping temperature constant the pressure of $11.2 \mathrm{dm}^3$ of a gas was increased from $105…

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?
  1. $1.4 \mathrm{dm}^3$
  2. $7.0 \mathrm{dm}^3$
  3. $5.6 \mathrm{dm}^3$
  4. $2.8\mathrm{dm}^3$

Solution

$\begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ & 105 \times 11.2=420 \times \mathrm{V}_2 \\ & \mathrm{~V}_2=\frac{105 \times 11.2}{420} \\ & 2.8 \mathrm{dm}^3\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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