Which of the following is correct for the case of isothermal expansion of an ideal gas ?

Options

A. Q = 0
B. W = 0
C. `Delta U = 0`
D. `Delta U != 0`

Explanation

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

### Step 1: Understand the Process
In an isothermal process, the temperature of the system remains constant. For an ideal gas undergoing isothermal expansion, this means that the internal energy of the gas does not change.

### Step 2: Apply the First Law of Thermodynamics
The First Law of Thermodynamics states:
\[ Q = \Delta U + W \]
Where:
- \( Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the system.

### Step 3: Determine the Change in Internal Energy
For an ideal gas, the change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = \frac{F}{2} N R \Delta T \]
Where:
- \( F \) is the degrees of freedom,
- \( N \) is the number of moles,
- \( R \) is the gas constant,
- \( \Delta T \) is the change in temperature.

Since this is an isothermal process, \( \Delta T = 0 \). Therefore:
\[ \Delta U = 0 \]

### Step 4: Substitute into the First Law Equation
Now substituting \( \Delta U = 0 \) into the First Law equation:
\[ Q = 0 + W \]
This simplifies to:
\[ Q = W \]

### Step 5: Analyze Work Done
In an isothermal expansion, the gas does work on the surroundings as it expands. The work done by the gas is positive because the volume is increasing. Thus, \( W > 0 \).

### Step 6: Conclusion
From the analysis, we conclude:
- \( \Delta U = 0 \)
- \( Q \) is equal to \( W \) (both are positive values).

Thus, the correct statement for the case of isothermal expansion of an ideal gas is that the change in internal energy (\( \Delta U \)) is zero.

### Final Answer
The correct option is that the change in internal energy \( \Delta U = 0 \).

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