Q18 / 150 · MHT CET 2021 (20 Sep Shift 1) · Chemistry

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?

  • A
    $1.4 \mathrm{dm}^3$
  • B
    $7.0 \mathrm{dm}^3$
  • C
    $5.6 \mathrm{dm}^3$
  • Correct answer
    D
    $2.8\mathrm{dm}^3$

Solution

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

Hence, the correct answer is option: 'D'.

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