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(D) The root mean square (r.m.s) speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas const

Vedclass · Accepted Answer

$4:1$

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(D) The root mean square (r.m.s) speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass.For helium $(He)$: $M_{He} = 4$,$T_{He} = 327^{\circ} C = 327 + 273 = 600 \ K$.For oxygen $(O_2)$: $M_{O} = 32$,$T_{O} = 27^{\circ} C = 27 + 273 = 300 \ K$.The ratio of the r.m.s speeds is:$\frac{v_{He}}{v_{O}} = \sqrt{\frac{T_{He}}{T_{O}} 	imes \frac{M_{O}}{M_{He}}} = \sqrt{\frac{600}{300} 	imes \frac{32}{4}} = \sqrt{2 	imes 8} = \sqrt{16} = 4$.Thus,the ratio is $4:1$.

The molecular masses of helium and oxygen are $4$ and $32$ respectively. The ratio of r.m.s speed of helium at $327^{\circ} C$ to r.m.s speed of oxygen at $27^{\circ} C$ will be

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