At what temperature is the "effective" speed of gaseous hydrogen molecules (molecular weight = 2) equal to that of oxygen molecules (molecular weight = 32) at `47^@ C`?

Explanation

To solve the problem, we need to find the temperature at which the root mean square speed (VRMS) of hydrogen molecules is equal to that of oxygen molecules at 47°C.

### Step-by-Step Solution:

1. **Understand the formula for VRMS**:
The root mean square speed (VRMS) of a gas is given by the formula:
\[ V_{RMS} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas in kg/mol.

2. **Set up the equation**:
According to the problem, we need to equate the VRMS of hydrogen (H₂) to the VRMS of oxygen (O₂) at 47°C.
- For hydrogen (molecular weight = 2 g/mol = 0.002 kg/mol):
\[ V_{RMS, H_2} = \sqrt{\frac{3RT_H}{0.002}} \]
- For oxygen (molecular weight = 32 g/mol = 0.032 kg/mol) at 47°C (which is 320 K):
\[ V_{RMS, O_2} = \sqrt{\frac{3R \cdot 320}{0.032}} \]

3. **Set the two equations equal**:
\[ \sqrt{\frac{3RT_H}{0.002}} = \sqrt{\frac{3R \cdot 320}{0.032}} \]

4. **Simplify the equation**:
Since \( \sqrt{3R} \) is common in both sides, we can cancel it out:
\[ \sqrt{\frac{T_H}{0.002}} = \sqrt{\frac{320}{0.032}} \]

5. **Square both sides**:
\[ \frac{T_H}{0.002} = \frac{320}{0.032} \]

6. **Cross-multiply to solve for \( T_H \)**:
\[ T_H = 0.002 \cdot \frac{320}{0.032} \]

7. **Calculate the right-hand side**:
\[ T_H = 0.002 \cdot 10 = 0.02 \text{ K} \]

8. **Convert to Celsius**:
Since the temperature in Kelvin is 0.02 K, we convert it to Celsius:
\[ T_H = 0.02 - 273.15 = -273.13 °C \]

### Final Answer:
The temperature at which the effective speed of gaseous hydrogen molecules is equal to that of oxygen molecules at 47°C is approximately -273.13 °C.