MHT CET · Chemistry · States of Matter
A mas occupies \(11.2 \mathrm{dm}^3\) at \(105 \mathrm{kPa}\). What is its volume if pressure is increased to \(210 \mathrm{kPa}\) ?
- A \(22.4 \mathrm{dm}^3\)
- B \(33.6 \mathrm{dm}^3\)
- C \(5.6 \mathrm{dm}^3\)
- D \(16.8 \mathrm{dm}^3\)
Answer & Solution
Correct Answer
(C) \(5.6 \mathrm{dm}^3\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{P}_1=105 \mathrm{kPa}, \mathrm{V}_1=11.2 \mathrm{dm}^3
\)
\(
\mathrm{P}_2=210 \mathrm{kPa}, \mathrm{V}_2=?
\)
According to Boyle's law, \(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{P}_2\)
\(
\therefore \mathrm{V}_2 \frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3
\)
\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, \(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{P}_2\)
\(
\therefore \mathrm{V}_2 \frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3
\)
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