The temperature at which the root mean square | vedclass

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(B) The root mean square speed $(v_{rms})$ of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have

Vedclass · Accepted Answer

$-\,204.75$

Vedclass · Answer

(B) The root mean square speed $(v_{rms})$ of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Given that $v_2 = \frac{1}{2}v_1$ and $T_1 = 0^{\circ}C = 273 \, K$.Substituting these values: $\frac{1}{2} = \sqrt{\frac{T_2}{273}}$.Squaring both sides: $\frac{1}{4} = \frac{T_2}{273}$.$T_2 = \frac{273}{4} = 68.25 \, K$.To convert this temperature to Celsius: $T(^{\circ}C) = T(K) - 273 = 68.25 - 273 = -204.75^{\circ}C$.

The temperature at which the root mean square speed of a gas will be half of its value at $0\,^{\circ}C$ is ....... $^{\circ}C$.

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