At a pressure of `24 xx 10^(5) dynecm^(-2)`. The volume of `O_(2)` is `10 litre` and mass is `20 g`. The rms velocity will be

Options

A. `800 ms^(-1)`
B. `400 ms^(-1)`
C. `600 ms^(-1)`
D. Data is incomplete

Explanation

To find the root mean square (rms) velocity of oxygen gas (O₂) under the given conditions, we will use the formula:

\[ V_{rms} = \sqrt{\frac{3PV}{m}} \]

where:
- \( P \) = pressure,
- \( V \) = volume,
- \( m \) = mass of the gas.

### Step-by-Step Solution:

1. **Convert the Pressure**:
Given pressure \( P = 24 \times 10^5 \, \text{dyne/cm}^2 \).
To convert this to SI units (Pascals), we use the conversion:
\[ 1 \, \text{dyne/cm}^2 = 0.001 \, \text{Pa} \]
Therefore,
\[ P = 24 \times 10^5 \, \text{dyne/cm}^2 = 24 \times 10^5 \times 0.001 \, \text{Pa} = 24 \times 10^2 \, \text{Pa} = 2400 \, \text{kPa} = 2400000 \, \text{Pa} \]

2. **Convert the Volume**:
Given volume \( V = 10 \, \text{litres} \).
To convert this to cubic meters:
\[ 1 \, \text{litre} = 10^{-3} \, \text{m}^3 \]
Therefore,
\[ V = 10 \, \text{litres} = 10 \times 10^{-3} \, \text{m}^3 = 0.01 \, \text{m}^3 \]

3. **Convert the Mass**:
Given mass \( m = 20 \, \text{g} \).
To convert this to kilograms:
\[ 1 \, \text{g} = 10^{-3} \, \text{kg} \]
Therefore,
\[ m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \]

4. **Substitute Values into the RMS Velocity Formula**:
Now we can substitute the values into the formula:
\[ V_{rms} = \sqrt{\frac{3PV}{m}} = \sqrt{\frac{3 \times 2400000 \, \text{Pa} \times 0.01 \, \text{m}^3}{0.02 \, \text{kg}}} \]

5. **Calculate the RMS Velocity**:
First, calculate the numerator:
\[ 3 \times 2400000 \times 0.01 = 72000 \]
Now divide by the mass:
\[ \frac{72000}{0.02} = 3600000 \]
Finally, take the square root:
\[ V_{rms} = \sqrt{3600000} \approx 600 \, \text{m/s} \]

### Final Answer:
The rms velocity of \( O_2 \) is approximately \( 600 \, \text{m/s} \).