One mole of an ideal gas is allowed to expand freely and adiabatically into vacuum until its volume has doubled. A statement which is not true concerning this expression is

Options

A. `DeltaH = 0`
B. `DeltaS = 0`
C. `DeltaE = 0`
D. W = 0

Explanation

To solve the question regarding the expansion of one mole of an ideal gas into a vacuum, we need to analyze the thermodynamic principles involved in free expansion and identify the statement that is not true.

### Step-by-Step Solution:

1. **Understanding Free Expansion**:
- When an ideal gas expands freely into a vacuum, it does so without any external pressure acting against it. This process is known as free expansion.
- In free expansion, the gas does not perform work on the surroundings because there is no opposing pressure.

2. **Thermodynamic Properties**:
- For an ideal gas undergoing free expansion, the change in internal energy (ΔU) can be expressed as:
\[ \Delta U = Q - W \]
where Q is the heat added to the system and W is the work done by the system.
- Since the process is adiabatic (no heat exchange with the surroundings), Q = 0.

3. **Calculating Work Done**:
- In free expansion, the work done (W) is zero because there is no external pressure to do work against. Thus, we have:
\[ \Delta U = 0 - 0 = 0 \]
- Therefore, the change in internal energy (ΔU) is zero.

4. **Change in Enthalpy (ΔH)**:
- The change in enthalpy (ΔH) for an ideal gas during free expansion can also be shown to be zero:
\[ \Delta H = \Delta U + P\Delta V \]
- Since ΔU = 0 and P (external pressure) is zero during free expansion, we conclude that ΔH = 0.

5. **Entropy Change (ΔS)**:
- The entropy of the gas increases because the gas molecules have more volume to occupy. The change in entropy (ΔS) can be calculated using:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]
- Here, \(V_f\) is the final volume (doubled) and \(V_i\) is the initial volume.

6. **Identifying the Incorrect Statement**:
- The statements regarding the process will typically include:
- ΔU = 0
- ΔH = 0
- W = 0
- ΔS > 0
- The statement that "the work done is not zero" is incorrect, as we have established that the work done during free expansion is indeed zero.

### Conclusion:
The statement that is not true concerning the expansion of one mole of an ideal gas into a vacuum is that "the work done is not zero."