At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at `47^(@)C`? a) 10 K b) 20 K c) 30 K d) 40 K

Options

A. 10 K
B. 20 K
C. 30 K
D. 40 K

Explanation

To find the temperature at which the root mean square (RMS) velocity of a hydrogen molecule (H₂) is equal to that of an oxygen molecule (O₂) at 47°C, we can follow these steps:

### Step 1: Write the formula for RMS velocity
The formula for the RMS velocity (v_rms) is given by:
\[ v_{rms} = \sqrt{\frac{RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Set the RMS velocities equal
According to the problem, we need to set the RMS velocity of hydrogen equal to that of oxygen:
\[ v_{rms, O_2} = v_{rms, H_2} \]
This gives us:
\[ \sqrt{\frac{R T_{O_2}}{M_{O_2}}} = \sqrt{\frac{R T_{H_2}}{M_{H_2}}} \]

### Step 3: Cancel out the gas constant \( R \)
Since \( R \) is the same for both gases, we can cancel it out:
\[ \sqrt{\frac{T_{O_2}}{M_{O_2}}} = \sqrt{\frac{T_{H_2}}{M_{H_2}}} \]

### Step 4: Square both sides
Squaring both sides eliminates the square root:
\[ \frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}} \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ T_{H_2} = T_{O_2} \cdot \frac{M_{H_2}}{M_{O_2}} \]

### Step 6: Substitute known values
- The molar mass of \( O_2 \) (oxygen) is \( 32 \, \text{g/mol} \).
- The molar mass of \( H_2 \) (hydrogen) is \( 2 \, \text{g/mol} \).
- Convert \( 47^\circ C \) to Kelvin:
\[ T_{O_2} = 47 + 273 = 320 \, K \]

Now substituting these values into the equation:
\[ T_{H_2} = 320 \cdot \frac{2}{32} \]

### Step 7: Calculate \( T_{H_2} \)
Calculating this gives:
\[ T_{H_2} = 320 \cdot \frac{1}{16} = 20 \, K \]

### Final Answer
Thus, the temperature at which the RMS velocity of a hydrogen molecule is equal to that of an oxygen molecule at \( 47^\circ C \) is:
\[ \boxed{20 \, K} \]