An ideal gas is allowed to expand freely against vacuum in a rigid insulated container. The gas undergoes

Options

A. An increase in its internal energy
B. A decrease in its internal energy
C. Neither an increase nor decrease in temperator internal energy
D. An increase in temperature.

Explanation

To solve the problem, we need to analyze the situation where an ideal gas expands freely against a vacuum in a rigid insulated container. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding the scenario**: The gas is expanding against a vacuum, which means there is no external pressure opposing the expansion. Therefore, the pressure exerted by the gas is effectively zero.

2. **Work Done (W)**: The work done by the gas during expansion can be calculated using the formula:
\[ W = P \cdot \Delta V \]
Since the external pressure (P) is zero (because it is expanding against a vacuum), the work done (W) is:
\[ W = 0 \]

3. **Heat Transfer (Q)**: The container is insulated, meaning no heat can enter or leave the system. Therefore, the heat transfer (Q) is also:
\[ Q = 0 \]

4. **Change in Internal Energy (ΔU)**: According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Substituting the values we found:
\[ \Delta U = 0 - 0 = 0 \]
This indicates that the internal energy of the gas does not change.

5. **Relationship between Internal Energy and Temperature**: For an ideal gas, the internal energy (U) is related to temperature (T) by the equation:
\[ U = \frac{f}{2} RT \]
where \(f\) is the degrees of freedom of the gas. Since ΔU is zero, it implies that the temperature must also remain constant:
\[ \Delta T = 0 \]

6. **Conclusion**: Since both the internal energy and temperature of the gas do not change during the free expansion against a vacuum, we conclude that:
- The gas undergoes neither an increase nor a decrease in temperature and internal energy.

### Final Answer:
The correct option is: **Neither increase nor decrease in temperature and internal energy.**