During the isothermal of an ideal gas :

Options

A. The internal energy remains unaffected
B. The temperature remains constant
C. The enthalpy remains unaffected
D. The enthalpy increases

Answer

### Step-by-Step Solution:

1. **Understanding Isothermal Process**:
- An isothermal process is defined as a thermodynamic process in which the temperature of the system remains constant throughout the process. This means that any heat added to the system is used to do work, and there is no change in internal energy.

2. **Internal Energy Change (ΔU)**:
- The change in internal energy (ΔU) for an ideal gas is given by the equation:
\[ \Delta U = n C_v \Delta T \]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.
- Since the process is isothermal, \( \Delta T = 0 \), therefore:
\[ \Delta U = n C_v \cdot 0 = 0 \]
- This indicates that the internal energy remains constant during an isothermal process.

3. **Temperature Change**:
- By definition of the isothermal process, the temperature remains constant. Thus, this statement is true.

4. **Enthalpy Change (ΔH)**:
- The change in enthalpy (ΔH) for an ideal gas is given by:
\[ \Delta H = n C_p \Delta T \]
where \( C_p \) is the molar heat capacity at constant pressure.
- Again, since \( \Delta T = 0 \) during an isothermal process:
\[ \Delta H = n C_p \cdot 0 = 0 \]
- This means that the enthalpy also remains constant during an isothermal process.

5. **Conclusion**:
- The correct statements regarding an isothermal process of an ideal gas are:
- Internal energy remains unaffected (ΔU = 0).
- Temperature remains constant.
- Enthalpy remains unaffected (ΔH = 0).
- The incorrect statement is that enthalpy increases.

### Final Answer:
- The true statements during the isothermal process of an ideal gas are:
1. Internal energy remains unaffected.
2. Temperature remains constant.
3. Enthalpy remains unaffected.
- The incorrect statement is that enthalpy increases.