For the isothermal reversible expansion of an ideal gas

Options

A. `DeltaH gt 0 and Delta U = 0`
B. `DeltaH gt 0 and DeltaU lt 0`
C. `DeltaH = 0 and DeltaU = 0`
D. `DeltaH = 0 and DeltaU gt 0`

Explanation

To solve the question regarding the isothermal reversible expansion of an ideal gas, we will analyze the changes in internal energy (ΔU) and enthalpy (ΔH) during this process.

### Step-by-Step Solution:

1. **Understanding Isothermal Process**:
- An isothermal process is one in which the temperature (T) remains constant throughout the process. Therefore, the change in temperature (ΔT) is zero.
- **Hint**: Remember that in an isothermal process, temperature does not change.

2. **Internal Energy (ΔU)**:
- For an ideal gas, the internal energy (U) is a function of temperature only. This means that if the temperature does not change, the internal energy also does not change.
- Since ΔT = 0, we can conclude that ΔU = 0.
- **Hint**: Recall that for an ideal gas, internal energy is solely dependent on temperature.

3. **Enthalpy (ΔH)**:
- Enthalpy (H) is defined as H = U + PV. For an ideal gas, enthalpy is also a function of temperature alone.
- Since the temperature remains constant during an isothermal process, the change in enthalpy (ΔH) is also zero.
- Therefore, ΔH = 0.
- **Hint**: Enthalpy change is also linked to temperature; if temperature is constant, enthalpy change is zero.

4. **Conclusion**:
- From the above analysis, we find that both the change in internal energy (ΔU) and the change in enthalpy (ΔH) during the isothermal reversible expansion of an ideal gas are zero.
- Thus, we can summarize: ΔU = 0 and ΔH = 0.
- **Hint**: Summarize the findings clearly to reinforce understanding.

### Final Answer:
- For the isothermal reversible expansion of an ideal gas, both ΔU and ΔH are equal to zero (ΔU = 0, ΔH = 0).