During the isothermal expansion of an ideal gas, its
internal energy increases
enthalpy decreases
enthalpy remains unaffected
enthalpy reduces to zero.

Options

A. internal energy increases
B. enthalpy decreases
C. enthalpy remains unaffected
D. enthalpy reduces to zero.

Explanation

To solve the question regarding the behavior of internal energy and enthalpy during the isothermal expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Isothermal Process**:
- An isothermal process is one in which the temperature (T) of the system remains constant throughout the process. For an ideal gas, this means that any energy added to the system as heat (Q) is used to do work (W) by the gas.

2. **Internal Energy of an Ideal Gas**:
- The internal energy (U) of an ideal gas depends only on its temperature. Since the temperature is constant during an isothermal process, the change in internal energy (ΔU) is zero.
- Mathematically, ΔU = 0 for an ideal gas in an isothermal process.

3. **Enthalpy Definition**:
- Enthalpy (H) is defined as H = U + PV, where P is pressure and V is volume. The change in enthalpy (ΔH) can be expressed as:
\[ \Delta H = \Delta U + \Delta(PV) \]

4. **Applying the Ideal Gas Law**:
- According to the ideal gas law, PV = nRT, where n is the number of moles and R is the universal gas constant. Since T is constant during an isothermal process, the product PV also remains constant.
- Therefore, Δ(PV) = 0.

5. **Calculate Change in Enthalpy**:
- Since both ΔU = 0 and Δ(PV) = 0, we can conclude:
\[ \Delta H = 0 + 0 = 0 \]
- This indicates that the enthalpy does not change during the isothermal expansion.

6. **Conclusion**:
- Based on the analysis, we can conclude that during the isothermal expansion of an ideal gas, the internal energy remains constant, and the enthalpy also remains unaffected (ΔH = 0).

### Final Answer:
- **Enthalpy remains unaffected**.