At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at `47^(@)C`?

Options

A. 20 K
B. 80 K
C. `-73 K`
D. 3 K

Explanation

To find the temperature at which the root mean square (RMS) velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at \(47^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RMS Velocity Formula**:
The RMS velocity (\(v_{rms}\)) 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.

2. **Convert the Given Temperature**:
The temperature of oxygen is given as \(47^\circ C\). We need to convert this to Kelvin:
\[ T_{O_2} = 47 + 273 = 320 \, K \]

3. **Calculate the RMS Velocity for Oxygen**:
For oxygen (\(O_2\)), the molar mass \(M_{O_2} = 32 \, g/mol\). Thus, the RMS velocity for oxygen can be expressed as:
\[ v_{rms, O_2} = \sqrt{\frac{3R \cdot 320}{32}} \]

4. **Calculate the RMS Velocity for Hydrogen**:
For hydrogen (\(H_2\)), the molar mass \(M_{H_2} = 2 \, g/mol\). The RMS velocity for hydrogen can be expressed as:
\[ v_{rms, H_2} = \sqrt{\frac{3R \cdot T_{H_2}}{2}} \]

5. **Set the RMS Velocities Equal**:
Since we want the RMS velocities of hydrogen and oxygen to be equal:
\[ \sqrt{\frac{3R \cdot 320}{32}} = \sqrt{\frac{3R \cdot T_{H_2}}{2}} \]

6. **Square Both Sides**:
Squaring both sides to eliminate the square roots gives:
\[ \frac{3R \cdot 320}{32} = \frac{3R \cdot T_{H_2}}{2} \]

7. **Cancel Common Terms**:
We can cancel \(3R\) from both sides:
\[ \frac{320}{32} = \frac{T_{H_2}}{2} \]

8. **Solve for \(T_{H_2}\)**:
Simplifying the left side:
\[ 10 = \frac{T_{H_2}}{2} \]
Multiplying both sides by 2:
\[ T_{H_2} = 20 \, K \]

### Conclusion:
The temperature at which the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at \(47^\circ C\) is \(20 \, K\).