Question:
At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be $1933\, m \,s^{-1}$. The gas is $(R = 8.3\, J\, mol^{-1}\, K^{-1}$ )
At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be $1933\, m \,s^{-1}$. The gas is $(R = 8.3\, J\, mol^{-1}\, K^{-1}$ )
Updated On: Jul 6, 2022
- $H_2$
- $F_2$
- $Cl_2$
- $O_2$
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The Correct Option is A
Solution and Explanation
The velocity of gas at temperature T is given by
$\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$
where, R is gas constant and M the molecular weight.
Given, $R = 8.3J\, mol^{-1} \,K^{-1}$
$T = 27^{\circ}C = 27 + 273 = 300 \,K$
$\upsilon^{2}_{rms} = 1933\,m\,s^{-1}$
$\therefore\quad M = \frac{3RT}{\upsilon ^{2}_{rms}} = \frac{3\times8.3\times 300}{\left(1933\right)^{2}} \therefore M \approx 2\times 10^{-3}\,kg$
which is molecular weight of $H_{2}.$
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