QJEEM24Ph7e13: JEE Main 2024 Physics - Class 11 JEEM24Ph7e13 Solution: If the root mean square velocity of hydrogen molecule at a…

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To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:

$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$

where:

• $v_{\text{rms}}$ is the root mean square velocity of the gas molecules,
• $k$ is the Boltzmann constant,
• $T$ is the absolute temperature in Kelvin, and
• $m$ is the mass of one molecule of the gas.

Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.

Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen. The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.

We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules. Using the rms velocity formula and considering the ratio of masses:

$\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{m_{\text{O₂}}}{m_{\text{H₂}}}}$

We know the mass m is proportional to the molecular weight for each gas, so we can substitute:

$\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{M_{\text{O₂}}}{M_{\text{H₂}}}}$

Now we plug in the values:

$\frac{2 \text{ km/s}}{v_{\text{rms, O₂}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

Solving for $v_{\text{rms, O₂}}$ :

$v_{\text{rms, O₂}} = \frac{2 \text{ km/s}}{4} = 0.5 \text{ km/s}$

Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.

The correct answer is Option D : 0.5.