Root mean square velocity of `O_2` at STP is (in `cm//s`)

Options

A. `4.61xx10^(4)`
B. `2.6xx10^(4)`
C. `46.1xx10^(4)`
D. `26.0xx10^(4)`

Answer

To find the root mean square velocity (RMS velocity) of \( O_2 \) at STP, we can use the formula:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas in kilograms per mole.

### Step-by-Step Solution:

1. **Identify the values:**
- The universal gas constant \( R = 8.314 \, \text{J/(mol K)} \).
- The temperature at STP is \( 0^\circ C \), which is equivalent to \( 273 \, \text{K} \).
- The molar mass of \( O_2 \) is \( 32 \, \text{g/mol} \). To convert this to kilograms per mole, we divide by 1000:
\[ M = \frac{32 \, \text{g/mol}}{1000} = 0.032 \, \text{kg/mol} \]

2. **Substitute the values into the formula:**
\[ V_{rms} = \sqrt{\frac{3 \times 8.314 \, \text{J/(mol K)} \times 273 \, \text{K}}{0.032 \, \text{kg/mol}}} \]

3. **Calculate the numerator:**
\[ 3 \times 8.314 \times 273 = 6814.862 \, \text{J/mol} \]

4. **Calculate the division:**
\[ \frac{6814.862 \, \text{J/mol}}{0.032 \, \text{kg/mol}} = 213,401.94 \, \text{m}^2/\text{s}^2 \]

5. **Take the square root:**
\[ V_{rms} = \sqrt{213401.94} \approx 461.06 \, \text{m/s} \]

6. **Convert to cm/s:**
\[ V_{rms} = 461.06 \, \text{m/s} \times 100 = 46106 \, \text{cm/s} \approx 4.61 \times 10^4 \, \text{cm/s} \]

### Final Answer:
The root mean square velocity of \( O_2 \) at STP is approximately \( 4.61 \times 10^4 \, \text{cm/s} \).