If the average velocity of N_2 gas at 27 , ^o | vedclass

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(D) The average velocity $(v_{avg})$ of a gas is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$,$\pi$,and $M$ (molar mass of $N_

Vedclass · Accepted Answer

$1200$

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(D) The average velocity $(v_{avg})$ of a gas is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$,$\pi$,and $M$ (molar mass of $N_2$) are constant,we have $v_{avg} \propto \sqrt{T}$.Therefore,$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.Given: $v_1 = 0.3 \, m/sec$,$T_1 = 27 + 273 = 300 \, K$,$v_2 = 0.6 \, m/sec$.Substituting the values: $\frac{0.3}{0.6} = \sqrt{\frac{300}{T_2}}$.$\frac{1}{2} = \sqrt{\frac{300}{T_2}}$.Squaring both sides: $\frac{1}{4} = \frac{300}{T_2}$.$T_2 = 300 	imes 4 = 1200 \, K$.

If the average velocity of $N_2$ gas at $27 \, ^oC$ is $0.3 \, m/sec$,then at what temperature in $K$ will the average velocity be $0.6 \, m/sec$?

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