A sealed container with negiligible coefficient of volumetric expansion contains helium (a monatomic gas). When it is heated from 300 K to 600 K, the average KE of helium atoms is

Options

A. halved
B. unchanged
C. doubled
D. increased by factor `sqrt2`

Explanation

To solve the problem, we need to determine how the average kinetic energy (KE) of helium atoms changes when the temperature is increased from 300 K to 600 K.

### Step-by-Step Solution:

1. **Understand the relationship between temperature and kinetic energy**:
The average kinetic energy (KE) of gas particles is directly proportional to the absolute temperature (T) of the gas. The formula for average kinetic energy of a monatomic gas is given by:
\[ KE = \frac{3}{2} k_B T \]
where \( k_B \) is the Boltzmann constant.

2. **Identify the initial and final temperatures**:
- Initial temperature, \( T_1 = 300 \, K \)
- Final temperature, \( T_2 = 600 \, K \)

3. **Set up the ratio of average kinetic energies**:
Since the average kinetic energy is proportional to temperature, we can express the ratio of the final kinetic energy \( KE_2 \) to the initial kinetic energy \( KE_1 \) as:
\[ \frac{KE_2}{KE_1} = \frac{T_2}{T_1} \]

4. **Substitute the values of temperatures**:
Plugging in the values for \( T_1 \) and \( T_2 \):
\[ \frac{KE_2}{KE_1} = \frac{600}{300} = 2 \]

5. **Conclusion about the average kinetic energy**:
This means that the average kinetic energy of helium atoms at 600 K is twice that at 300 K:
\[ KE_2 = 2 \times KE_1 \]

6. **Final answer**:
Therefore, the average kinetic energy of helium atoms is doubled when the temperature is increased from 300 K to 600 K. The correct option is option 3.