MHT CET · Physics · Kinetic Theory of Gases
The molecular mass of a gas having r.m.s. speed four times as that of another gas having molecular mass 32 is
- A 2
- B 4
- C 26
- D 32
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}} \\ \therefore \quad & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{M}_1}{\mathrm{M}_2}} \\ & \text { Given that }\left(\mathrm{v}_{\mathrm{rms}}\right)_2=4\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ \therefore \quad & \sqrt{\frac{32}{\mathrm{M}_2}}=4 \\ \therefore \quad & \frac{32}{\mathrm{M}_2}=16 \\ \therefore \quad & \mathrm{M}_2=2\end{array}\)
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