Given below are two statements:
Statement I: The average momentum of a molecule in a sample of an ideal gas depends on temperature.
Statement II: The rms speed of oxygen molecules in a gas is $$v$$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $$2v$$.
In the light of the above statements, choose the correct answer from the options given below:

We need to evaluate two statements about kinetic theory of gases.

Statement I: The average momentum of a molecule in a sample of an ideal gas depends on temperature.

The molecules in an ideal gas move randomly in all directions. While the average speed depends on temperature, the average velocity (and hence average momentum) is zero because the molecular motion is random and symmetric in all directions. Since the average momentum is always zero regardless of temperature, it does not depend on temperature.

Statement I is FALSE.

Statement II: If the temperature is doubled and O$$_2$$ molecules dissociate into O atoms, the rms speed becomes 2v.

The rms speed is given by:

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

Initially, for O$$_2$$ (molar mass $$M = 32$$ g/mol) at temperature $$T$$:

$$v = \sqrt{\frac{3RT}{32}}$$

After dissociation, we have O atoms (molar mass $$M' = 16$$ g/mol) at temperature $$2T$$:

$$v' = \sqrt{\frac{3R(2T)}{16}} = \sqrt{\frac{6RT}{16}} = \sqrt{\frac{3RT}{8}}$$

Taking the ratio:

$$\frac{v'}{v} = \sqrt{\frac{3RT/8}{3RT/32}} = \sqrt{\frac{32}{8}} = \sqrt{4} = 2$$

So $$v' = 2v$$.

Statement II is TRUE.

Hence, the correct answer is Option D: Statement I is false but Statement II is true.