Given: rms velocity of hydrogen at 300K is `1.9 xx 10^3` m/s. The rms velocity of oxygen at 1200K will be

Options

A. `7.6 xx 10^3 m//s`
B. `3.8 xx 10^3 m//s`
C. `0.95 xx 10^3 m//s`
D. `0.475 xx 10^3 m//s `

Answer

To find the RMS velocity of oxygen at 1200 K given the RMS velocity of hydrogen at 300 K, we can use the formula for RMS velocity:

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

Where:
- \( V_{rms} \) is the root mean square velocity,
- \( R \) is the gas constant,
- \( T \) is the absolute temperature (in Kelvin),
- \( M \) is the molar mass of the gas.

### Step-by-step Solution:

1. **Identify the given values:**
- RMS velocity of hydrogen (\( V_{rms, H_2} \)) at 300 K = \( 1.9 \times 10^3 \, \text{m/s} \)
- Temperature of hydrogen (\( T_H \)) = 300 K
- Temperature of oxygen (\( T_O \)) = 1200 K
- Molar mass of hydrogen (\( M_H \)) = 2 g/mol
- Molar mass of oxygen (\( M_O \)) = 32 g/mol

2. **Write the relationship between RMS velocities:**
Using the proportionality of RMS velocities:
\[ \frac{V_{rms, H_2}}{V_{rms, O_2}} = \sqrt{\frac{T_H \cdot M_O}{T_O \cdot M_H}} \]

3. **Substitute the known values into the equation:**
\[ \frac{1.9 \times 10^3}{V_{rms, O_2}} = \sqrt{\frac{300 \cdot 32}{1200 \cdot 2}} \]

4. **Simplify the right side:**
- Calculate the numerator:
\[ 300 \cdot 32 = 9600 \]
- Calculate the denominator:
\[ 1200 \cdot 2 = 2400 \]
- Thus, we have:
\[ \frac{9600}{2400} = 4 \]
- Therefore:
\[ \sqrt{4} = 2 \]

5. **Substitute back into the equation:**
\[ \frac{1.9 \times 10^3}{V_{rms, O_2}} = 2 \]

6. **Solve for \( V_{rms, O_2} \):**
\[ V_{rms, O_2} = \frac{1.9 \times 10^3}{2} = 0.95 \times 10^3 \, \text{m/s} \]

7. **Final result:**
The RMS velocity of oxygen at 1200 K is:
\[ V_{rms, O_2} = 0.95 \times 10^3 \, \text{m/s} \]

### Conclusion:
The correct answer is \( 0.95 \times 10^3 \, \text{m/s} \).

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