A sample of gas is at $0^{\circ} C$ . To what temperature it must be raised in order to double the r.m.s. speed of the molecule

270°C

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819°C

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1090°C

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100°C

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Solution

The correct option is B
819°C

$v = \sqrt{\frac{3 R T}{M}}$

$v \propto \sqrt{T}$

$2 = \sqrt{\frac{T}{273}} \Rightarrow T = 1092 K = 819^{\circ} C$

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