The root mean square (r.m.s.) velocity \( v_{\text{rms}} \) of gas molecules is given by the formula: \[v_{\text{rms}} = \sqrt{\frac{3RT}{M}},\] where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas. Step 1: Equalizing \( v_{\text{rms}} \) for Hydrogen and Oxygen For hydrogen (\( H_2 \)) and oxygen (\( O_2 \)), we set their r.m.s. velocities equal: \[v_{\text{rms}}(H_2) = v_{\text{rms}}(O_2).\] Substituting the formula for \( v_{\text{rms}} \): \[\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}.\] Squaring both sides yields: \[\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}.\] Rearranging to solve for \( T_{H_2} \): \[T_{H_2} = T_{O_2} \cdot \frac{M_{H_2}}{M_{O_2}}.\] Step 2: Substituting Known Values The temperature of oxygen gas is converted to Kelvin: \[T_{O_2} = 47^\circ \text{C} + 273 = 320 \, \text{K}.\] The molar masses are: \[M_{H_2} = 2, \quad M_{O_2} = 32.\] Substituting these values into the equation for \( T_{H_2} \): \[T_{H_2} = 320 \cdot \frac{2}{32}.\] Simplifying the expression: \[T_{H_2} = 320 \cdot \frac{1}{16} = 20 \, \text{K}.\] Final Answer: \[\boxed{20 \, \text{K}}\]