For isothermal expansion in case of an ideal gas `:`

Options

A. `DeltaG = DeltaS`
B. `DeltaG = DeltaH`
C. `DeltaG = - T. DeltaS`
D. None of these

Explanation

To solve the problem regarding isothermal expansion of an ideal gas, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Isothermal Process:**
In an isothermal process, the temperature (T) remains constant. For an ideal gas, this means that the internal energy (U) does not change since it is a function of temperature.

2. **Use the Gibbs Free Energy Equation:**
The Gibbs free energy (G) is related to enthalpy (H) and entropy (S) by the equation:
\[ G = H - T \Delta S \]
where \( \Delta S \) is the change in entropy.

3. **Determine Changes in Enthalpy (H) and Internal Energy (U):**
For an isothermal process, the change in internal energy (\( \Delta U \)) is zero:
\[ \Delta U = n C_V \Delta T \]
Since \( \Delta T = 0 \) (isothermal), it follows that:
\[ \Delta U = 0 \]

4. **Relate Enthalpy Change to Internal Energy and Pressure-Volume Work:**
The change in enthalpy can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]
For an ideal gas, \( PV = nRT \), and since T is constant, the change in \( PV \) during isothermal expansion is also zero:
\[ \Delta (PV) = 0 \]
Thus, we have:
\[ \Delta H = 0 + 0 = 0 \]

5. **Substitute into the Gibbs Free Energy Equation:**
Now substituting \( \Delta H = 0 \) into the Gibbs free energy equation:
\[ G = 0 - T \Delta S \]
This simplifies to:
\[ \Delta G = -T \Delta S \]

6. **Conclusion:**
Therefore, for an isothermal expansion of an ideal gas, the change in Gibbs free energy is given by:
\[ \Delta G = -T \Delta S \]

### Final Answer:
The correct expression for the change in Gibbs free energy during isothermal expansion of an ideal gas is:
\[ \Delta G = -T \Delta S \]