Expansion of a perfect gas into vaccum is related with

Options

A. `Delta H = 0`
B. `q = 0`
C. `W= 0`
D. All of these

Answer

To solve the question regarding the expansion of a perfect gas into a vacuum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Process**: When a perfect gas expands into a vacuum, it does so without any external pressure acting against it. This is a free expansion process.

2. **Calculating Work Done**: The work done (W) during an expansion is given by the formula:
\[ W = - P_{\text{external}} (V_2 - V_1) \]
In this case, since the gas is expanding into a vacuum, the external pressure \( P_{\text{external}} = 0 \). Therefore:
\[ W = - 0 \times (V_2 - V_1) = 0 \]

3. **Heat Exchange (Q)**: In a vacuum, there is no medium to transfer heat. Therefore, the heat exchange with the surroundings is also zero:
\[ Q = 0 \]

4. **Change in Internal Energy (ΔU)**: According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
Substituting the values we found:
\[ \Delta U = 0 + 0 = 0 \]

5. **Change in Enthalpy (ΔH)**: The change in enthalpy is given by:
\[ \Delta H = \Delta U + P_{\text{external}} \Delta V \]
Since \( \Delta U = 0 \) and \( P_{\text{external}} = 0 \), we have:
\[ \Delta H = 0 + 0 = 0 \]

6. **Conclusion**: From the calculations, we can conclude that:
- Work done (W) = 0
- Heat exchange (Q) = 0
- Change in internal energy (ΔU) = 0
- Change in enthalpy (ΔH) = 0

### Final Answer:
The expansion of a perfect gas into a vacuum is related to:
- \( \Delta H = 0 \)
- \( Q = 0 \)
- \( W = 0 \)
- Therefore, the correct option is "All of these".