MHT CET · Chemistry · States of Matter
Volume of a balloon at \(25^{\circ} \mathrm{C}\) and 1 bar pressure is \(2 \cdot 27 \mathrm{~L}\). If the pressure of the gas in balloon is reduced to \(0.227\) bar, what is the rise in volume of a gas?
- A \(12.27 \mathrm{~L}\)
- B \(7.73 \mathrm{~L}\)
- C \(10 \mathrm{~L}\)
- D \(\times 4.10 .227 \mathrm{~L}\)
Answer & Solution
Correct Answer
(B) \(7.73 \mathrm{~L}\)
Step-by-step Solution
Detailed explanation
\(P_{1}=1 \mathrm{~bar}, \quad V_{1}=2.27 \mathrm{~L}\)
\(\mathrm{P}_{2}=0.227 \mathrm{~bar}, \quad \mathrm{V}_{2}=?\)
According to Boyle's Law, \(P_{1} V_{1}=P_{2} V_{2}\)
\(\therefore V_{2}=\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.227}=10 \mathrm{~L}\)
\(\therefore\) The rise in volume of gas \(=10-2.27=7.73 \mathrm{~L}\)
\(\mathrm{P}_{2}=0.227 \mathrm{~bar}, \quad \mathrm{V}_{2}=?\)
According to Boyle's Law, \(P_{1} V_{1}=P_{2} V_{2}\)
\(\therefore V_{2}=\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.227}=10 \mathrm{~L}\)
\(\therefore\) The rise in volume of gas \(=10-2.27=7.73 \mathrm{~L}\)
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