Question

If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2 km/s, the root mean square velocity of oxygen at the same condition in km/s is :

• A. 2.0
• B. 0.5
• C. 1.5
• D. 1.0

Answer 1

The Correct Option is B

To find the root mean square velocity of oxygen at the same temperature and pressure, we use the formula for root mean square velocity, which is given by:

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

where \(k\) is the Boltzmann's constant, \(T\) is the absolute temperature, and \(m\) is the mass of the molecule.

We know that the root mean square velocity for hydrogen is given as 2 km/s. Let this be \(v_{\text{rms, H}_2}\) and for oxygen it will be \(v_{\text{rms, O}_2}\).

We use the relation between the root mean square velocities and the molar masses:

\(\frac{v_{\text{rms, H}_2}}{v_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}}\)

Where:

• \(M_{\text{H}_2} = 2 \text{ g/mol}\) (molar mass of hydrogen)
• \(M_{\text{O}_2} = 32 \text{ g/mol}\) (molar mass of oxygen)

Substituting the values, we get:

\(\frac{2}{v_{\text{rms, O}_2}} = \sqrt{\frac{32}{2}}\)

\(\frac{2}{v_{\text{rms, O}_2}} = \sqrt{16}\)

\(\frac{2}{v_{\text{rms, O}_2}} = 4\)

Solving for \(v_{\text{rms, O}_2}\):

\(v_{\text{rms, O}_2} = \frac{2}{4} = 0.5 \text{ km/s}\)

Hence, the root mean square velocity of oxygen at the same condition is 0.5 km/s.

Answer 2

Given: - Root mean square (rms) velocity of hydrogen (\(v_{H_2}\)) = 2 km/s - Molecular mass of hydrogen (\(M_{H_2}\)) = 2 g/mol - Molecular mass of oxygen (\(M_{O_2}\)) = 32 g/mol

Step 1: Relationship for Root Mean Square Velocity

The root mean square velocity of a gas is given by:

\[ v_{\text{rms}} \propto \frac{1}{\sqrt{M}} \]

where \(M\) is the molar mass of the gas.

Step 2: Calculating the Ratio of Velocities

Using the inverse square root relationship for hydrogen and oxygen:

\[ \frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

Substituting the given values:

\[ \frac{v_{O_2}}{2} = \sqrt{\frac{2}{32}} \] \[ \frac{v_{O_2}}{2} = \sqrt{\frac{1}{16}} \] \[ \frac{v_{O_2}}{2} = \frac{1}{4} \]

Multiplying both sides by 2:

\[ v_{O_2} = \frac{1}{4} \times 2 = 0.5 \, \text{km/s} \]

Conclusion:

The root mean square velocity of oxygen at the same conditions is 0.5 km/s.