A mass occupies $11.2 \mathrm{dm}^3$ at $105 \mathrm{kPa}$. What is its volume if pressure is increased to…

A mass occupies $11.2 \mathrm{dm}^3$ at $105 \mathrm{kPa}$. What is its volume if pressure is increased to $210 \mathrm{kPa}$ ?
  1. $22.4 \mathrm{dm}^3$
  2. $33.6 \mathrm{dm}^3$
  3. $5.6 \mathrm{dm}^3$
  4. $16.8 \mathrm{dm}^3$

Solution

$\mathrm{P}_1=105 \mathrm{kPa}, \mathrm{V}_1=11.2 \mathrm{dm}^3$ $\mathrm{P}_2=210 \mathrm{kPa}, \mathrm{V}_2=?$ According to Boyle's law, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{V}_2$ $\therefore \mathrm{V}_2= \frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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