Question

Two boxes are at the same temperature. The first box contains gas with molecular mass \(m_1\) and rms speed \(v_1\). The second box contains gas with molecular mass \(m_2\) and average speed \(v_2\). If \(v_1 = 1.5 v_2\), find \(\frac{m_1}{m_2}\).

• A. 1.25
• B. 0.74
• C. 0.52
• D. 0.26

Hint:

For gases at same temperature, \(v_{\text{rms}} \propto 1/\sqrt{m}\). Use this to relate molecular masses when rms speeds are known.

Answer 1

The Correct Option is C

Step 1: Write both speeds using the kinetic theory formulas at the SAME temperature $T$.
\[ v_1 = v_{\text{rms},1} = \sqrt{\frac{3RT}{M_1}}, \qquad v_2 = v_{\text{avg},2} = \sqrt{\frac{8RT}{\pi M_2}} \]

Step 2: Square the given ratio to clear both square roots in one move.
\[ \left(\frac{v_1}{v_2}\right)^2 = (1.5)^2 = 2.25 = \frac{3RT/M_1}{8RT/(\pi M_2)} = \frac{3\pi M_2}{8M_1} \]

Step 3: Solve for the mass ratio.
\[ \frac{M_1}{M_2} = \frac{3\pi}{8\times2.25} = \frac{3\pi}{18} = \frac{\pi}{6} \]

Step 4: Evaluate numerically.
\[ \frac{M_1}{M_2} = \frac{3.1416}{6} \approx 0.52 \]

Step 5: Conclusion.
\[ \boxed{0.52} \]