1. Analyze Statement I

For an ideal gas in random thermal motion, molecular velocities are distributed symmetrically in all directions.

• Momentum of one molecule: $\vec p = m\vec v$
• Since the motion is random, the vector average velocity over all molecules is zero: $\langle \vec v \rangle = 0$
• Therefore, the average momentum is also zero: $\langle \vec p \rangle = m\langle \vec v \rangle = 0$

This does not depend on temperature.

So, Statement I is false.

Note: Quantities like average magnitude of momentum or rms momentum depend on temperature, but the statement says average momentum, which is zero for a gas at rest.

2. Analyze Statement II

Given initially for oxygen molecules $\mathrm{O_2}$: $v_{\mathrm{rms}} = v$

We use $v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}}$

Initial state:

• Mass of one $\mathrm{O_2}$ molecule $= m$
• Temperature $= T$
• So, $v = \sqrt{\frac{3kT}{m}}$

Final state:

• Temperature is doubled: $T' = 2T$
• $\mathrm{O_2}$ dissociates into oxygen atoms, so mass becomes half: $m' = \frac{m}{2}$

Thus, $v'_{\mathrm{rms}} = \sqrt{\frac{3k(2T)}{m/2}}$

Simplify: $v'_{\mathrm{rms}} = \sqrt{\frac{6kT}{m/2}} = \sqrt{\frac{12kT}{m}}$

Now compare with initial $v$: $v = \sqrt{\frac{3kT}{m}}$

Hence, $v'_{\mathrm{rms}} = \sqrt{4\cdot \frac{3kT}{m}} = 2\sqrt{\frac{3kT}{m}} = 2v$

So, Statement II is true.

3. Choose the correct option

• Statement I: False
• Statement II: True

Therefore, the correct option is: $\boxed{\text{D}}$

4. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.