A thermally isolated sample of an ideal gas at a fixed temperature is confined to one half of a container by an impermeable membrane. The other half of the container is evacuated. The membrane is then pierced and the gas is allowed to expand freely and to double its volume as shown. Which one of the following statements is true concerning this situation?

Options

A. The process is reversible.
B. This is an isothermal process.
C. The entropy of the gas decreases.
D. The internal energy of the gas must decrease.

Answer

To solve the problem step by step, let's analyze the situation described:

### Step 1: Understand the Initial Conditions
- We have a thermally isolated container divided by an impermeable membrane.
- One half of the container contains an ideal gas at a fixed temperature, while the other half is a vacuum (evacuated).
- The initial volume of the gas is \( V_0 \) (in one half of the container).

**Hint:** Identify the properties of the system, including the fact that it is thermally isolated and that one side is a vacuum.

### Step 2: Analyze the Process of Expansion
- When the membrane is pierced, the gas expands freely into the vacuum.
- This is a free expansion process, meaning there is no external pressure opposing the expansion.

**Hint:** Recall that free expansion occurs when a gas expands into a vacuum, resulting in no work done against external pressure.

### Step 3: Apply the First Law of Thermodynamics
- The First Law of Thermodynamics states:
\[ \Delta Q = \Delta U + W \]
where \( \Delta Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system.
- Since the system is thermally isolated, \( \Delta Q = 0 \).
- The work done by the gas during free expansion into a vacuum is also \( W = 0 \).

**Hint:** Remember that in an isolated system, no heat is exchanged with the surroundings, and in free expansion, no work is done.

### Step 4: Determine the Change in Internal Energy
- From the First Law, we have:
\[ 0 = \Delta U + 0 \implies \Delta U = 0 \]
- The change in internal energy \( \Delta U \) being zero indicates that the internal energy of the gas remains constant.

**Hint:** Consider that for an ideal gas, the internal energy is a function of temperature. If \( \Delta U = 0 \), then the temperature must also remain constant.

### Step 5: Identify the Type of Process
- Since the temperature remains constant during the expansion, this process is classified as an isothermal process.
- Additionally, because the gas expands into a vacuum, this process is irreversible.

**Hint:** Isothermal processes occur at constant temperature, and free expansion is inherently irreversible.

### Step 6: Evaluate the Statements
Now, we can evaluate the statements provided in the question:
1. The process is isothermal (True).
2. The process is reversible (False).
3. The entropy of the system decreases (False).
4. The internal energy of the gas decreases (False).

**Hint:** Analyze each statement based on the conclusions drawn from the analysis of the process.

### Conclusion
The correct statement concerning the situation is that the process is isothermal.