At $$300$$ K, the rms speed of oxygen molecules is $$\sqrt{\frac{a+5}{\alpha}}$$ times to that of its average speed in the gas. Then, the value of $$\alpha$$ will be (use $$\pi = \frac{22}{7}$$)

We need to determine $$\alpha$$ given that the rms speed of O$$_2$$ at 300 K is $$\sqrt{\frac{\alpha + 5}{\alpha}}$$ times the average speed.

Recalling that the rms speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ and the average speed by $$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$, we can form their ratio.

Thus, $$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT/M}{8RT/(\pi M)}} = \sqrt{\frac{3\pi}{8}}$$.

Upon substituting $$\pi = \frac{22}{7}$$, the expression becomes $$\frac{3\pi}{8} = \frac{3 \times 22}{7 \times 8} = \frac{66}{56} = \frac{33}{28}$$.

Since the problem states that this ratio equals $$\sqrt{\frac{\alpha + 5}{\alpha}}$$, we set $$\frac{\alpha + 5}{\alpha} = \frac{33}{28}$$ and solve. Multiplying both sides by 28 yields $$28(\alpha + 5) = 33\alpha$$, which simplifies to $$28\alpha + 140 = 33\alpha$$ and hence $$5\alpha = 140$$, giving $$\alpha = 28$$.

Accordingly, the correct choice corresponds to option 2 with a value of $$28$$.