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The average speed of an ideal gas molecule at $27^{\large\circ}C$ is 0.3 m/s. Calculate the average speed at $927^{\large\circ}C$

$(a)\;1.3m/s\qquad(b)\;0.6m/s\qquad(c)\;2.1m/s\qquad(d)\;1.8m/s$

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$U_{AV} = \sqrt{(\large\frac{8RT}{\pi m})}$
Given
$U_{AV} = 0.3m/s$ at 300K
$\therefore U_1 = 0.3 = \sqrt{(\large\frac{8R\times300}{\pi m})}$-------(1)
At T = 273+927 = 1200K
$\therefore U_2 = \sqrt{(\large\frac{8R\times1200}{\pi m})}$-------(2)
$\therefore \large\frac{U_2}{0.3} = \sqrt{(\large\frac{1200}{300})}$
$\Rightarrow U_2 = 0.6m/s$
Hence answer is (b)
answered Feb 21, 2014 by sharmaaparna1
 

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