Find the temperature at which oxygen molecules would have the same rms speed as of hydrogen molecules at `300 K`.

Explanation

To find the temperature at which oxygen molecules would have the same root mean square (RMS) speed as hydrogen molecules at 300 K, we can follow these steps:

### Step 1: Write the formula for RMS speed
The formula for the root mean square speed (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,
- \( M \) is the molar mass of the gas.

### Step 2: Set the RMS speeds equal
Since we want the RMS speed of oxygen (\( v_{rms, O_2} \)) to equal the RMS speed of hydrogen (\( v_{rms, H_2} \)), we can set the two equations equal to each other:
\[ v_{rms, O_2} = v_{rms, H_2} \]

### Step 3: Substitute the RMS speed formulas
Substituting the formula for each gas, we have:
\[ \sqrt{\frac{3R T_{O_2}}{M_{O_2}}} = \sqrt{\frac{3R T_{H_2}}{M_{H_2}}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{3R T_{O_2}}{M_{O_2}} = \frac{3R T_{H_2}}{M_{H_2}} \]

### Step 5: Cancel out common terms
Since \( 3R \) is common on both sides, we can cancel it out:
\[ \frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}} \]

### Step 6: Rearrange to find \( T_{O_2} \)
Rearranging this equation to solve for \( T_{O_2} \) gives:
\[ T_{O_2} = T_{H_2} \cdot \frac{M_{O_2}}{M_{H_2}} \]

### Step 7: Substitute known values
We know:
- \( T_{H_2} = 300 \, K \)
- Molar mass of oxygen (\( M_{O_2} \)) = 32 g/mol
- Molar mass of hydrogen (\( M_{H_2} \)) = 2 g/mol

Substituting these values into the equation:
\[ T_{O_2} = 300 \cdot \frac{32}{2} \]

### Step 8: Calculate \( T_{O_2} \)
Calculating the above expression:
\[ T_{O_2} = 300 \cdot 16 = 4800 \, K \]

### Final Answer
The temperature at which oxygen molecules would have the same RMS speed as hydrogen molecules at 300 K is:
\[ \boxed{4800 \, K} \]