In the isothermal expansion of an ideal gas

Options

A. `Delta U= 0`
B. `Delta T=0`
C. q=0
D. `W= -q`

Answer

### Step-by-Step Solution:

1. **Understanding the Process**:
- The question refers to the isothermal expansion of an ideal gas. In an isothermal process, the temperature (T) of the system remains constant.

2. **Applying the First Law of Thermodynamics**:
- The first law of thermodynamics is given by the equation:
\[ \Delta U = Q + W \]
where:
- \(\Delta U\) = Change in internal energy
- \(Q\) = Heat exchanged
- \(W\) = Work done by the system

3. **Identifying Changes in Internal Energy**:
- For an ideal gas, the internal energy (U) is a function of temperature. Since the process is isothermal (constant temperature), the change in temperature (\(\Delta T\)) is zero:
\[ \Delta T = 0 \]
- Therefore, the change in internal energy (\(\Delta U\)) is also zero:
\[ \Delta U = 0 \]

4. **Substituting into the First Law**:
- Substituting \(\Delta U = 0\) into the first law equation:
\[ 0 = Q + W \]
- Rearranging gives:
\[ W = -Q \]
- This indicates that the work done by the gas is equal in magnitude but opposite in sign to the heat absorbed by the gas.

5. **Conclusion**:
- In an isothermal expansion of an ideal gas:
- The change in internal energy (\(\Delta U\)) is zero.
- The heat exchanged (\(Q\)) is not zero.
- The work done (\(W\)) is equal to \(-Q\).

### Summary of Results:
- \(\Delta U = 0\) (Change in internal energy is zero)
- \(Q \neq 0\) (Heat exchanged is not zero)
- \(W = -Q\) (Work done is equal to negative heat exchanged)