MHT CET · Physics · Kinetic Theory of Gases
' \(N\) ' molecules of gas \(A\), each having mass ' \(m\) ' and ' 2 N ' molecules of gas B , each of mass ' 2 m ' are contained in the same vessel which is at constant temperature ' \(T\) '. The mean square velocity of \(B\) is \(V^2\) and mean square of x -component of A is \(\omega^2\). The value of \(\frac{\omega^2}{V^2}\) is
- A \(3: 2\)
- B \(2: 3\)
- C \(1: 2\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(B) \(2: 3\)
Step-by-step Solution
Detailed explanation
Mean square velocity of molecule \(=\frac{3 \mathrm{kT}}{\mathrm{m}}\)
For gas A,
\(x\) component of mean square velocity of molecule \(=\omega^2\)
\(\therefore \quad\) Mean square velocity \(=3 \omega^2=\frac{3 \mathrm{kT}}{\mathrm{m}}\)...(i)
For gas B,
Mean square velocity \(=\mathrm{V}^2=\frac{3 \mathrm{kT}}{2 \mathrm{~m}}\)...(ii)
From (i) and (ii)
\(\frac{3 \omega^2}{\mathrm{~V}^2}=\frac{3 \mathrm{kT}}{\mathrm{m}} \times \frac{2 \mathrm{~m}}{3 \mathrm{kT}}\)
\(\therefore \quad \frac{\omega^2}{\mathrm{~V}^2}=\frac{2}{3}\)
For gas A,
\(x\) component of mean square velocity of molecule \(=\omega^2\)
\(\therefore \quad\) Mean square velocity \(=3 \omega^2=\frac{3 \mathrm{kT}}{\mathrm{m}}\)...(i)
For gas B,
Mean square velocity \(=\mathrm{V}^2=\frac{3 \mathrm{kT}}{2 \mathrm{~m}}\)...(ii)
From (i) and (ii)
\(\frac{3 \omega^2}{\mathrm{~V}^2}=\frac{3 \mathrm{kT}}{\mathrm{m}} \times \frac{2 \mathrm{~m}}{3 \mathrm{kT}}\)
\(\therefore \quad \frac{\omega^2}{\mathrm{~V}^2}=\frac{2}{3}\)
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