If the root mean square velocity of hydrogen molecule at a given temperature and pressure is $$2 \text{ km s}^{-1}$$, the root mean square velocity of oxygen at the same condition in $$\text{km s}^{-1}$$ is:

For gases at same temperature,

$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

$$v_{rms}\propto\frac{1}{\sqrt{M}}$$

Therefore

$$\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$$

Molar masses:

$$M_{H_2}=2$$

$$M_{O_2}=32$$

Given

$$v_{H_2}=2\ \text{km s}^{-1}$$

2vO2=322\frac{2}{v_{O_2}}

\sqrt{\frac{32}{2}}vO22=232

$$=\sqrt{16}=4$$

Thus

$$v_{O_2}=\frac{2}{4}$$

$$=0.5\ \text{km s}^{-1}$$

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources