The temperature at which the root mean square velocity of a molecule will be doubled than at $100^{\circ} C$

1219°C

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1492°C

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400°C

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400 K

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Solution

The correct option is A
1219°C

$v = \sqrt{\frac{3 R T}{M}} . H e n c e v \propto \sqrt{T} 2 = \sqrt{\frac{T}{273 + 100}} \Rightarrow T = 1492 K = 1219^{\circ} C$

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The temperature at which the root mean square velocity of a molecule will be doubled than at 100∘C

The correct option is A 1219°C v=√3RTM. Hence v∝√T2=√T273+100⇒T=1492K=1219∘C

The temperature at which the root mean square velocity of a molecule will be doubled than at $100^{\circ} C$

The correct option is A
1219°C

$v = \sqrt{\frac{3 R T}{M}} . H e n c e v \propto \sqrt{T} 2 = \sqrt{\frac{T}{273 + 100}} \Rightarrow T = 1492 K = 1219^{\circ} C$