The r.m.s. speed of oxygen molecules at $$47^\circ$$ is equal to that of the hydrogen molecules kept at ________ $$C^\circ$$. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)

We need to find the temperature at which the r.m.s. speed of hydrogen molecules equals the r.m.s. speed of oxygen molecules at $$47°C$$.

The root mean square (r.m.s.) speed of gas molecules is given by:

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

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature in Kelvin, and $$M$$ is the molar mass of the gas.

Converting the temperature of oxygen to Kelvin gives

$$ T_{O_2} = 47 + 273 = 320 \text{ K} $$.

Setting the r.m.s. speeds equal, we have

$$ \sqrt{\frac{3R \times 320}{M_{O_2}}} = \sqrt{\frac{3R \times T_{H_2}}{M_{H_2}}} $$. Squaring both sides (the $$3R$$ cancels):

$$ \frac{320}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}} $$.

Given that $$\frac{M_{O_2}}{M_{H_2}} = \frac{32}{2} = 16$$, so $$M_{O_2} = 32$$ and $$M_{H_2} = 2$$:

$$ \frac{320}{32} = \frac{T_{H_2}}{2} $$. Hence

$$ 10 = \frac{T_{H_2}}{2} \quad\Longrightarrow\quad T_{H_2} = 20 \text{ K} $$.

Converting back to Celsius,

$$ T_{H_2} = 20 - 273 = -253°C $$.

The r.m.s. speed of oxygen at $$47°C$$ equals the r.m.s. speed of hydrogen at $$-253°C$$.

The correct answer is Option (3): $$-253$$.