MHT CET · Physics · Kinetic Theory of Gases
We have a sample of gas characterised by \(\mathrm{P}, \mathrm{V}\) and \(\mathrm{T}\) and another sample of gas characterised by \(2 \mathrm{P}, \mathrm{V} / 4\), and \(2 \mathrm{~T}\). What is the ratio of the number of molecules in the first and second samples?
- A \(2: 1\)
- B \(4: 1\)
- C \(8: 1\)
- D \(16: 1\)
Answer & Solution
Correct Answer
(B) \(4: 1\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}\text { PVT } & 2 \mathrm{P}, \mathrm{V} / 4,2 \mathrm{~T}\end{array}\)
\(\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{n}_{1} \mathrm{RT}_{1} ; \quad \mathrm{P}_{2} \mathrm{V}_{2}=\mathrm{n}_{2} \mathrm{RT}_{2}\)
\(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1} / \mathrm{RT}_{1}}{\mathrm{P}_{2} \mathrm{V}_{2} / \mathrm{RT}_{2}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{P}_{2} \mathrm{V}_{2}}\)
\(=\frac{\mathrm{PV}}{\mathrm{T}} \times \frac{2 \mathrm{T} \times 2}{2 \mathrm{P} \times \frac{\mathrm{V}}{4}}=4: 1\)
\(\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{n}_{1} \mathrm{RT}_{1} ; \quad \mathrm{P}_{2} \mathrm{V}_{2}=\mathrm{n}_{2} \mathrm{RT}_{2}\)
\(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1} / \mathrm{RT}_{1}}{\mathrm{P}_{2} \mathrm{V}_{2} / \mathrm{RT}_{2}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{P}_{2} \mathrm{V}_{2}}\)
\(=\frac{\mathrm{PV}}{\mathrm{T}} \times \frac{2 \mathrm{T} \times 2}{2 \mathrm{P} \times \frac{\mathrm{V}}{4}}=4: 1\)
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