If the average velocity of N_2 molecule is 0. | vedclass

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(A) The average velocity $(V_{avg})$ of a gas molecule is given by the formula $V_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$,$M$,and $\pi$ are consta

Vedclass · Accepted Answer

$1200$

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(A) The average velocity $(V_{avg})$ of a gas molecule is given by the formula $V_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$,$M$,and $\pi$ are constants,$V_{avg} \propto \sqrt{T}$.Given,$V_1 = 0.3 \, m/s$ and $T_1 = 27 + 273 = 300 \, K$.We need to find $T_2$ when $V_2 = 0.6 \, m/s$.Using the relation $\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}$:$\frac{0.3}{0.6} = \sqrt{\frac{300}{T_2}}$$\frac{1}{2} = \sqrt{\frac{300}{T_2}}$Squaring both sides: $\frac{1}{4} = \frac{300}{T_2}$$T_2 = 300 	imes 4 = 1200 \, K$.

If the average velocity of $N_2$ molecule is $0.3 \, m/s$ at $27 \, ^\circ C$,then the velocity will be $0.6 \, m/s$ at ........... $K$.

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