The average translational energy and the rms speed of molecules in a sample of oxygen gas at `300K` are `6.21xx10^(-21)J` and `484m//s`, respectively. The corresponding values at `600K` are nearly (assuming ideal gas behaviour)

Options

A. `12.42 xx10^(-21)` J,968 m/s
B. `8.78 xx10^(-21)` J,968 m/s
C. `6.21 xx10^(-21)` J,968 m/s
D. `12.42xx10^(-21)` J,968 m/s

Explanation

To find the average translational energy and the root mean square (rms) speed of molecules in a sample of oxygen gas at 600 K, we can use the relationships between these quantities and temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Average translational energy at 300 K, \( E_1 = 6.21 \times 10^{-21} \, \text{J} \)
- rms speed at 300 K, \( V_1 = 484 \, \text{m/s} \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)
- Final temperature, \( T_2 = 600 \, \text{K} \)

2. **Use the Relationship for rms Speed:**
The rms speed \( V_{\text{rms}} \) is related to temperature by the formula:
\[ V_{\text{rms}} \propto \sqrt{T} \]
Therefore, we can write:
\[ \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} \]

3. **Substitute the Known Values:**
\[ \frac{V_2}{484} = \sqrt{\frac{600}{300}} = \sqrt{2} \]
\[ V_2 = 484 \cdot \sqrt{2} \approx 484 \cdot 1.414 \approx 684.7 \, \text{m/s} \]

4. **Calculate the Average Translational Energy:**
The average translational energy is directly proportional to temperature:
\[ \frac{E_2}{E_1} = \frac{T_2}{T_1} \]
Therefore, we can write:
\[ E_2 = E_1 \cdot \frac{T_2}{T_1} \]
Substituting the known values:
\[ E_2 = 6.21 \times 10^{-21} \cdot \frac{600}{300} = 6.21 \times 10^{-21} \cdot 2 = 12.42 \times 10^{-21} \, \text{J} \]

5. **Final Results:**
- The rms speed at 600 K, \( V_2 \approx 684.7 \, \text{m/s} \)
- The average translational energy at 600 K, \( E_2 = 12.42 \times 10^{-21} \, \text{J} \)

### Summary:
At 600 K, the average translational energy of the oxygen gas is approximately \( 12.42 \times 10^{-21} \, \text{J} \) and the rms speed is approximately \( 684.7 \, \text{m/s} \).