The average energy of molecules in a sample of oxygen gas at 300 K are `6.21xx10^(-21)J`. The corresponding values at 600 K are

Options

A. `12.12xx10^(-21)J`
B. `8.78xx10^(-21)J`
C. `6.21xx10^(-21)J`
D. `12.42xx10^(-21)J`

Explanation

To find the average energy of molecules in a sample of oxygen gas at 600 K, we can use the relationship between average kinetic energy and temperature from the kinetic theory of gases.

### Step-by-Step Solution:

1. **Understanding the Relationship**: The average kinetic energy (KE) of gas molecules is given by the formula:
\[ KE = \frac{3}{2} k_B T \]
where \( k_B \) is Boltzmann's constant and \( T \) is the temperature in Kelvin.

2. **Given Values**: We know that at \( T_1 = 300 \, K \), the average kinetic energy \( KE_1 = 6.21 \times 10^{-21} \, J \).

3. **Finding the Ratio**: The average kinetic energy is directly proportional to the temperature. Therefore, we can write:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]
where \( KE_2 \) is the average kinetic energy at \( T_2 = 600 \, K \).

4. **Substituting Known Values**:
\[ \frac{6.21 \times 10^{-21}}{KE_2} = \frac{300}{600} \]
Simplifying the right side, we get:
\[ \frac{6.21 \times 10^{-21}}{KE_2} = \frac{1}{2} \]

5. **Cross-Multiplying**:
\[ 6.21 \times 10^{-21} = \frac{1}{2} KE_2 \]
Therefore, multiplying both sides by 2 gives:
\[ KE_2 = 2 \times 6.21 \times 10^{-21} \]

6. **Calculating \( KE_2 \)**:
\[ KE_2 = 12.42 \times 10^{-21} \, J \]

7. **Final Answer**: The average energy of molecules in a sample of oxygen gas at 600 K is:
\[ KE_2 = 12.42 \times 10^{-21} \, J \]