The root-mean-square speed of molecules of a given mass of a gas at 27°C and 1 atmosphere pressure is 200 m s$$^{-1}$$. The root-mean-square speed of molecules of the gas at 127°C and 2 atmosphere pressure is $$\frac{x}{\sqrt{3}}$$ m s$$^{-1}$$. The value of $$x$$ will be ______.

The root-mean-square (rms) speed of gas molecules is given by $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$, where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass of the gas. Notably, the rms speed depends only on the temperature and the molar mass of the gas, and is independent of pressure.

At $$T_1 = 27°\text{C} = 300$$ K, the rms speed is $$v_1 = 200$$ m s$$^{-1}$$. At $$T_2 = 127°\text{C} = 400$$ K, the rms speed is $$v_2$$. Since the gas is the same (same molar mass $$M$$), we have $$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$.

Substituting the values: $$\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$.

Therefore, $$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}$$ m s$$^{-1}$$.

Comparing with the given expression $$v_2 = \frac{x}{\sqrt{3}}$$ m s$$^{-1}$$, we get $$x = 400$$.

Therefore, the value of $$x$$ is $$400$$.