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