Question

What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?

• A. The velocity of atomic oxygen remains same
• B. The velocity of atomic oxygen doubles
• C. The velocity of atomic oxygen becomes half
• D. The velocity of atomic oxygen becomes four times

Answer 1

The Correct Option is B

To determine the effect on the root mean square (rms) velocity of oxygen molecules when the temperature is doubled and the oxygen molecule dissociates into atomic oxygen, we need to understand the relationship between rms velocity, temperature, and molecular mass.

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

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

where:

• \(k\) is the Boltzmann constant,
• \(T\) is the temperature (in Kelvin),
• \(m\) is the mass of one molecule.

Step 1: Effect of Doubling the Temperature

When the temperature is doubled, i.e., \(T \to 2T\), the new rms velocity for molecular oxygen is:

\(v_{rms, new} = \sqrt{\frac{3k(2T)}{m}} = \sqrt{2} \times v_{rms, old}\).

Step 2: Effect of Dissociation of Oxygen Molecules

When an oxygen molecule (\(O_2\)) dissociates into two oxygen atoms, the mass of one atom becomes half the mass of the original molecule since

each \(O_2\) molecule divides into two \(O\) atoms.

\(m_{\text{atom}} = \frac{m_{\text{molecule}}}{2}\).

In this case, the rms velocity of atomic oxygen becomes:

\(v_{rms, atom} = \sqrt{\frac{3k(2T)}{m/2}} = \sqrt{\frac{6kT}{m}} = \sqrt{2} \cdot \sqrt{\frac{3kT}{m/2}}\).

This means:

\(v_{rms, atom} = \sqrt{2} \times \sqrt{2} \times v_{rms, old} = 2 \times v_{rms, old}\).

Thus, the velocity of atomic oxygen doubles.

Hence, the correct answer is: The velocity of atomic oxygen doubles.