An ideal gas undergoes isothermal process from some initial state `i` to final state `f`. Choose the correct alternatives.

Options

A. 1.dU=dQ
B. 2. dU=-dW
C. 3. dU=0
D. 4. dU=dW

Explanation

To solve the question regarding an ideal gas undergoing an isothermal process from an initial state \(i\) to a final state \(f\), we will analyze the situation using the first law of thermodynamics and the properties of isothermal processes.

### Step-by-Step Solution:

1. **Understanding Isothermal Process**:
- An isothermal process is defined as a thermodynamic process that occurs at a constant temperature. For an ideal gas, this means that the internal energy of the gas does not change during the process.

2. **Applying the First Law of Thermodynamics**:
- The first law of thermodynamics is given by the equation:
\[ dQ = dU + dW \]
where \(dQ\) is the heat added to the system, \(dU\) is the change in internal energy, and \(dW\) is the work done by the system.

3. **Change in Internal Energy**:
- The change in internal energy \(dU\) for an ideal gas can be expressed as:
\[ dU = m C_v \Delta T \]
where \(m\) is the mass of the gas, \(C_v\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature.

4. **Temperature Change in Isothermal Process**:
- Since the process is isothermal, the temperature remains constant. Therefore, the change in temperature \(\Delta T\) is zero:
\[ \Delta T = T_f - T_i = 0 \]
This implies:
\[ dU = m C_v \cdot 0 = 0 \]

5. **Conclusion on Internal Energy**:
- Since \(dU = 0\), we can substitute this back into the first law of thermodynamics:
\[ dQ = 0 + dW \quad \Rightarrow \quad dQ = dW \]
This means that the heat added to the system is equal to the work done by the system.

6. **Evaluating the Options**:
- Now, we can evaluate the given options:
- **Option 1**: \(dU = dQ\) → Incorrect, since \(dU = 0\).
- **Option 2**: \(dU = -dW\) → Incorrect, since \(dU = 0\).
- **Option 3**: \(dU = 0\) → Correct, as derived above.

### Final Answer:
The correct alternative is that \(dU = 0\).