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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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At room temperature , the rms speed of the molecules of a diatomic gas is found to be 1933 m/s . The gas is

$(a)\;H_{2} \qquad(b)\;F_{2}\qquad(c)\;Cl_{2}\qquad(d)\;O_{2}$

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Answer : $\;H_{2}$
Explanation :
$V_{rms} = \sqrt{\large\frac{3RT}{M}}$
or $\;M=\large\frac{3RT}{V_{rms}^{2}}= \large\frac{3 \times 8.31 \times T}{1933 \times 1933}$
$= \large\frac{3 \times 8.31 \times 300 K}{ 1933 \times 1933}$
$= 0.002 Kg$
$M = 2 gms$
Hence $\;H_{2}$
answered Mar 31, 2014 by yamini.v

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