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 $(R = 8.3\, J\, mol^{-1}\, K^{-1}$ )

• A. $H_2$
• B. $F_2$
• C. $Cl_2$
• D. $O_2$

Answer 1

Answer: (A)

$H_2$

Answer 2

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 $H_{2}.$