At which temperature the r.m.s.velocity of a hydrogen molecule equal to that of an oxygen molecule at `47^@C`?

Options

A. 4 K
B. 20 K
C. 80 K
D. `-73K`

Answer

To solve the problem, we need to find the temperature at which the root mean square (r.m.s) velocity of a hydrogen molecule is equal to that of an oxygen molecule at 47°C.

### Step-by-Step Solution:

1. **Understanding the r.m.s. velocity formula**:
The formula for the r.m.s. velocity \( v_{rms} \) of a gas 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.

2. **Setting up the equation**:
We know that the r.m.s. velocity of hydrogen \( v_{H_2} \) is equal to the r.m.s. velocity of oxygen \( v_{O_2} \):
\[ v_{H_2} = v_{O_2} \]
Thus, we can write:
\[ \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} \]

3. **Squaring both sides**:
Squaring both sides to eliminate the square root gives:
\[ \frac{3RT_{H_2}}{M_{H_2}} = \frac{3RT_{O_2}}{M_{O_2}} \]

4. **Canceling common terms**:
The \( 3R \) terms cancel out:
\[ \frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}} \]

5. **Rearranging the equation**:
Rearranging gives:
\[ T_{H_2} = T_{O_2} \cdot \frac{M_{H_2}}{M_{O_2}} \]

6. **Substituting known values**:
The molar mass of hydrogen \( M_{H_2} \) is approximately 2 g/mol, and the molar mass of oxygen \( M_{O_2} \) is approximately 32 g/mol. The temperature of oxygen \( T_{O_2} \) at 47°C must be converted to Kelvin:
\[ T_{O_2} = 47 + 273 = 320 \, K \]

7. **Calculating \( T_{H_2} \)**:
Now substituting the values:
\[ T_{H_2} = 320 \cdot \frac{2}{32} \]
Simplifying:
\[ T_{H_2} = 320 \cdot \frac{1}{16} = 20 \, K \]

### Final Answer:
The temperature at which the r.m.s. velocity of a hydrogen molecule equals that of an oxygen molecule at 47°C is **20 K**.