The root mean square speed of hydrogen molecule at 300 K is 1930m/s. Then the root mean square speed of oxygen molecules at 900K will be………..

Answer

To find the root mean square (RMS) speed of oxygen molecules at 900 K, given that the RMS speed of hydrogen molecules at 300 K is 1930 m/s, we can use the formula for RMS speed:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:
- \( v_{rms} \) is the root mean square speed,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.

### Step 1: Write down the known values for hydrogen (H₂)
- RMS speed of hydrogen (\( v_{rms, H_2} \)): 1930 m/s
- Molar mass of hydrogen (\( M_{H_2} \)): 2 g/mol = 0.002 kg/mol
- Temperature for hydrogen (\( T_{H_2} \)): 300 K

Using the formula for hydrogen:
\[ v_{rms, H_2} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} \]

### Step 2: Set up the equation for oxygen (O₂)
- Molar mass of oxygen (\( M_{O_2} \)): 32 g/mol = 0.032 kg/mol
- Temperature for oxygen (\( T_{O_2} \)): 900 K

Using the formula for oxygen:
\[ v_{rms, O_2} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} \]

### Step 3: Formulate the ratio of the two equations
To find the RMS speed of oxygen, we can divide the equation for oxygen by the equation for hydrogen:

\[ \frac{v_{rms, O_2}}{v_{rms, H_2}} = \frac{\sqrt{\frac{3RT_{O_2}}{M_{O_2}}}}{\sqrt{\frac{3RT_{H_2}}{M_{H_2}}}} \]

This simplifies to:
\[ \frac{v_{rms, O_2}}{1930} = \sqrt{\frac{T_{O_2} \cdot M_{H_2}}{T_{H_2} \cdot M_{O_2}}} \]

### Step 4: Substitute the known values
Substituting the values we have:
- \( T_{O_2} = 900 \, K \)
- \( T_{H_2} = 300 \, K \)
- \( M_{H_2} = 0.002 \, kg/mol \)
- \( M_{O_2} = 0.032 \, kg/mol \)

\[ \frac{v_{rms, O_2}}{1930} = \sqrt{\frac{900 \cdot 0.002}{300 \cdot 0.032}} \]

### Step 5: Simplify the equation
Calculating the right side:
\[ \frac{v_{rms, O_2}}{1930} = \sqrt{\frac{900 \cdot 0.002}{300 \cdot 0.032}} = \sqrt{\frac{1.8}{9.6}} = \sqrt{0.1875} \]

### Step 6: Calculate the square root
Calculating the square root:
\[ \sqrt{0.1875} \approx 0.433 \]

### Step 7: Solve for \( v_{rms, O_2} \)
Now, multiply both sides by 1930:
\[ v_{rms, O_2} = 1930 \cdot 0.433 \approx 835.49 \, m/s \]

### Final Answer
The root mean square speed of oxygen molecules at 900 K is approximately **835.49 m/s**.