RMS velocity of one mole of an ideal gas was | vedclass

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(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M}$.Comparing this with the

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$1 	imes 10^{-1}$

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(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M}$.Comparing this with the equation of a straight line $y = mx + C$,where $y = v_{rms}^2$ and $x = T$:The slope $m = \frac{3R}{M}$.Given slope $m = 249 \ m^2 \ s^{-2} \ K^{-1}$ and $R = 8.3 \ J \ mol^{-1} \ K^{-1}$.Substituting the values: $249 = \frac{3 	imes 8.3}{M}$.$M = \frac{3 	imes 8.3}{249} = \frac{24.9}{249} = 0.1 \ kg \ mol^{-1}$.Thus,$M = 1 	imes 10^{-1} \ kg \ mol^{-1}$.

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