1. Write the standard formulas

For an ideal gas at equilibrium:

• Root mean square speed: $U_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$ where $T$ is absolute temperature and $M$ is molar mass.

  For a single molecule form, this is also $U_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}}$ so clearly $U_{\mathrm{rms}} \propto \sqrt{\frac{T}{m}}$.

• Average translational kinetic energy per molecule: $E_{\mathrm{avg}} = \frac{3}{2}kT$

  Hence, it depends only on temperature, not on molecular mass.

2. Check option A

Option A says: $U_{\mathrm{rms}}$ is inversely proportional to the square root of its molecular mass.

From $U_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}}$ we get $U_{\mathrm{rms}} \propto \frac{1}{\sqrt{m}}$ (at constant temperature).

So, A is correct.

3. Check option B

Option B says: $U_{\mathrm{rms}}$ is doubled when its temperature is increased four times.

Since $U_{\mathrm{rms}} \propto \sqrt{T}$ if temperature becomes $4T$, then $U'_{\mathrm{rms}} = \sqrt{4}\,U_{\mathrm{rms}} = 2U_{\mathrm{rms}}$

So, B is correct.

4. Check option C

Option C says: $E_{\mathrm{avg}}$ is doubled when its temperature is increased four times.

But $E_{\mathrm{avg}} = \frac{3}{2}kT \propto T$ If temperature becomes $4T$, then $E'_{\mathrm{avg}} = 4E_{\mathrm{avg}}$ not double.

So, C is incorrect.

5. Check option D

Option D says: $E_{\mathrm{avg}}$ at a given temperature does not depend on its molecular mass.

From $E_{\mathrm{avg}} = \frac{3}{2}kT$ it depends only on $T$.

So, D is correct.

6. Final answer

The correct statements are: $\boxed{A,\ B,\ D}$

7. Comparison with stored answer

Stored correct answer: $A, B, D$

My derived answer matches the stored answer exactly.