MHT CET · Chemistry · States of Matter
A certain sample of gas has a volume of \(0.2 \mathrm{~L}\) at one atmosphere pressure and 273.15 K. What is the volume of gas at \(273.15^{\circ} \mathrm{C}\) at same pressure?
- A \(2.703 \mathrm{~L}\)
- B \(0.2 \mathrm{~L}\)
- C \(0.4 \mathrm{~L}\)
- D \(5.406 \mathrm{~L}\)
Answer & Solution
Correct Answer
(C) \(0.4 \mathrm{~L}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_{1}=0.2 \mathrm{~L},\mathrm{~T}_{1}=237.15 \mathrm{~K}\)
\(\mathrm{~V}_{2}=?, \mathrm{~T}_{2}=273.15+273.15=546.30 \mathrm{~K}\)
\(\therefore\) According to Charle's law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \quad\) (at constant \(\mathrm{P}\) and \(\mathrm{n}\) )
\(\therefore \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \times \mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.2 \times 546.30}{273.15}=0.4 \mathrm{~L}\)
\(\mathrm{~V}_{2}=?, \mathrm{~T}_{2}=273.15+273.15=546.30 \mathrm{~K}\)
\(\therefore\) According to Charle's law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \quad\) (at constant \(\mathrm{P}\) and \(\mathrm{n}\) )
\(\therefore \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \times \mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.2 \times 546.30}{273.15}=0.4 \mathrm{~L}\)
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