To solve the problem, we need to analyze the four processes of the ideal gas expansion and determine the changes in average kinetic energy (E1, E2, E3, and E4) for each process.
### Step-by-Step Solution:
1. **Isothermal Process (Process 1)**:
- In an isothermal process, the temperature (T) of the gas remains constant.
- The change in average kinetic energy (E1) is given by the formula:
\[
E_1 = \frac{3}{2} k_B \Delta T
\]
- Since there is no change in temperature (ΔT = 0), we have:
\[
E_1 = 0
\]
2. **Adiabatic Process (Process 2)**:
- In an adiabatic process, there is no heat exchange with the surroundings (Q = 0).
- The work done on/by the gas results in a change in internal energy, which is related to temperature change.
- Since the gas expands, it does work on the surroundings, which decreases its internal energy and thus decreases the temperature.
- Therefore, the change in average kinetic energy (E2) will be negative:
\[
E_2 < 0
\]
3. **Free Expansion (Process 3)**:
- In free expansion, the gas expands into a vacuum without doing work and without heat exchange.
- There is no change in temperature (ΔT = 0), so:
\[
E_3 = \frac{3}{2} k_B \Delta T = 0
\]
4. **Isobaric Process (Process 4)**:
- In an isobaric process, the pressure remains constant.
- As the volume increases, the temperature must also increase (from the ideal gas law: PV = nRT).
- Thus, the change in average kinetic energy (E4) will be positive:
\[
E_4 = \frac{3}{2} k_B \Delta T \quad (\Delta T > 0)
\]
### Summary of Results:
- \(E_1 = 0\) (Isothermal)
- \(E_2 < 0\) (Adiabatic)
- \(E_3 = 0\) (Free Expansion)
- \(E_4 > 0\) (Isobaric)
### Conclusion:
- From the analysis, we can conclude:
- \(E_1 = E_3 = 0\)
- \(E_2 < 0\)
- \(E_4 > 0\)
### Correct Options:
1. \(E_1 = E_3\) (True)
2. \(E_2 < E_3\) (True, since \(E_2 < 0\) and \(E_3 = 0\))
3. \(E_1 > E_4\) (False, since \(E_1 = 0\) and \(E_4 > 0\))
4. \(E_4 > E_3\) (True, since \(E_4 > 0\) and \(E_3 = 0\))
### Final Choices:
- The correct options are: \(E_1 = E_3\), \(E_2 < E_3\), and \(E_4 > E_3\).
