MHT CET · Physics · Kinetic Theory of Gases
The molecular masses of helium and oxygen are 4 and 32 respectively. The ratio of r.m.s speed of helium at \(327^{\circ} \mathrm{C}\) to r.m.s speed of oxygen at \(27^{\circ} \mathrm{C}\) will be
- A 1:6
- B 8:1
- C 1:8
- D 4:1
Answer & Solution
Correct Answer
(D) 4:1
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { R.M.S speed is given by, } \mathrm{c}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ & \mathrm{T}_{\mathrm{He}}=327^{\circ} \mathrm{C}=600 \mathrm{~K}, \mathrm{~T}_{\mathrm{o}}=27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ & \therefore \frac{\mathrm{C}_{\mathrm{He}}}{\mathrm{C}_{\mathrm{O}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{He}}}{\mathrm{T}_{\mathrm{O}}} \cdot \frac{\mathrm{M}_{\mathrm{O}}}{\mathrm{M}_{\mathrm{He}}}}=\sqrt{\frac{600}{300} \cdot \frac{32}{4}}=4\end{aligned}\)
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