An insulated container is divided into two equal portions.One portion contains an ideal monoatomic gasat pressue P and temperature. T while the other protion is a perfect vacuum. If a hole is opened between the two portion, the change in internal energy of the gas is

Options

A. zero
B. equal to work done by the gas
C. equal to work done on the gas
D. 3RT/2

Answer

To solve the problem, we will analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the System**:
We have an insulated container divided into two equal parts. One part contains an ideal monoatomic gas at pressure \( P \) and temperature \( T \), while the other part is a perfect vacuum.

2. **Identifying the Process**:
When a hole is opened between the two portions, the gas will expand into the vacuum. This process is known as free expansion, as the gas expands without doing any work against an external pressure.

3. **Work Done During Expansion**:
In free expansion, the gas expands into a vacuum, where the external pressure is zero. Therefore, the work done \( W \) by the gas during this expansion can be calculated as:
\[ W = \int P \, dV \]
Since the external pressure \( P = 0 \), we have:
\[ W = 0 \]

4. **Heat Exchange**:
The container is insulated, which means there is no heat exchange with the surroundings. Thus, the heat transfer \( Q \) is also zero:
\[ Q = 0 \]

5. **Applying the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Substituting the values we found:
\[ \Delta U = 0 - 0 = 0 \]

6. **Conclusion**:
The change in internal energy \( \Delta U \) of the gas after the expansion into the vacuum is:
\[ \Delta U = 0 \]

### Final Answer:
The change in internal energy of the gas is **0**.

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