Step 1: Understanding the Concept:
The root mean square (r.m.s.) velocity of gas molecules depends on the absolute temperature of the gas and its molecular weight.
We need to set up a ratio based on the given condition to find the unknown temperature.
Step 2: Key Formula or Approach:
The r.m.s. velocity is given by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant, $T$ is absolute temperature, and $M$ is molar mass.
Given condition: $(v_{\text{rms}})_{\text{H}_2} = 4 \times (v_{\text{rms}})_{\text{O}_2}$.
Remember to convert temperatures to Kelvin for the formula calculation.
Step 3: Detailed Explanation:
Let's denote Hydrogen as $1$ and Oxygen as $2$.
Given:
$M_1 (\text{H}_2) = 2$
$M_2 (\text{O}_2) = 32$
$T_2 = 47^\circ\text{C} = 47 + 273 = 320 \text{ K}$
Velocity condition: $v_1 = 4v_2$.
Substitute the $v_{\text{rms}}$ formula into the condition:
\[ \sqrt{\frac{3 R T_1}{M_1}} = 4 \cdot \sqrt{\frac{3 R T_2}{M_2}} \]
Square both sides to remove the square root:
\[ \frac{3 R T_1}{M_1} = 16 \cdot \frac{3 R T_2}{M_2} \]
Cancel the common '$3R$' term from both sides:
\[ \frac{T_1}{M_1} = 16 \cdot \frac{T_2}{M_2} \]
Substitute the known values ($M_1, M_2, T_2$):
\[ \frac{T_1}{2} = 16 \cdot \frac{320}{32} \]
Simplify the right side:
\[ \frac{T_1}{2} = 16 \cdot 10 \]
\[ \frac{T_1}{2} = 160 \]
Solve for $T_1$:
\[ T_1 = 320 \text{ K} \]
Convert the temperature back to Celsius to match the options:
\[ T_1 = 320 - 273 = 47^\circ\text{C} \]
Step 4: Final Answer:
The temperature of hydrogen molecules is $47^\circ\text{C}$.