The temperature at which the root mean square | vedclass

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(A) The root mean square $(R.M.S.)$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $R$,$T$,and $M$ are constants for a

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$1219^{\circ}C$

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(A) The root mean square $(R.M.S.)$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $R$,$T$,and $M$ are constants for a given gas,$v_{rms} \propto \sqrt{T}$.Let $v_1$ be the velocity at $T_1 = 100^{\circ}C = 373\; K$ and $v_2$ be the velocity at $T_2$.Given $v_2 = 2v_1$.Using the proportionality,$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.Substituting the values: $\frac{v_1}{2v_1} = \sqrt{\frac{373}{T_2}}$.$\frac{1}{2} = \sqrt{\frac{373}{T_2}}$.Squaring both sides: $\frac{1}{4} = \frac{373}{T_2}$.$T_2 = 373 	imes 4 = 1492\; K$.To convert the temperature to Celsius: $T(^{\circ}C) = T(K) - 273$.$T_2 = 1492 - 273 = 1219^{\circ}C$.

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

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