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 of finding the average kinetic energy of helium atoms when the temperature is increased from 300 K to 600 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between temperature and average kinetic energy**:
The average kinetic energy (KE) of a monatomic ideal gas can be expressed using the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.

2. **Calculate the initial average kinetic energy at 300 K**:
Using the formula mentioned above, we can calculate the initial kinetic energy (KE1) at \( T_1 = 300 \, K \):
\[ KE_1 = \frac{3}{2} k T_1 = \frac{3}{2} k (300) \]

3. **Calculate the final average kinetic energy at 600 K**:
Similarly, we can calculate the final kinetic energy (KE2) at \( T_2 = 600 \, K \):
\[ KE_2 = \frac{3}{2} k T_2 = \frac{3}{2} k (600) \]

4. **Set up the ratio of the two kinetic energies**:
To find how the kinetic energies change with temperature, we can set up the ratio:
\[ \frac{KE_1}{KE_2} = \frac{\frac{3}{2} k (300)}{\frac{3}{2} k (600)} \]
Here, the \( \frac{3}{2} k \) cancels out:
\[ \frac{KE_1}{KE_2} = \frac{300}{600} = \frac{1}{2} \]

5. **Determine the relationship between KE1 and KE2**:
From the ratio, we find that:
\[ KE_2 = 2 \times KE_1 \]
This means that the average kinetic energy of helium atoms doubles when the temperature is increased from 300 K to 600 K.

6. **Conclusion**:
Therefore, the average kinetic energy of helium atoms at 600 K is double that at 300 K.

### Final Answer:
The average kinetic energy of helium atoms when heated from 300 K to 600 K is **double**.

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