The change in entropy of an ideal gas during reversible isothermal expansion is

Options

A. Negative
B. Positive
C. Zero
D. Infinite

Explanation

To find the change in entropy (ΔS) of an ideal gas during a reversible isothermal expansion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- We are dealing with an ideal gas undergoing a reversible isothermal expansion. In this process, the temperature (T) remains constant.

2. **Identify Key Relationships**:
- For an isothermal process, the change in internal energy (ΔU) is zero because the temperature does not change. This can be expressed as:
\[ \Delta U = 0 \]

3. **Apply the First Law of Thermodynamics**:
- The first law of thermodynamics states:
\[ \Delta U = Q + W \]
- Since ΔU = 0, we can rearrange this to find the relationship between heat (Q) and work (W):
\[ 0 = Q + W \implies Q = -W \]

4. **Work Done During Expansion**:
- For a reversible isothermal expansion of an ideal gas, the work done (W) can be expressed as:
\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]
- Here, \(n\) is the number of moles, \(R\) is the ideal gas constant, \(T\) is the temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume.

5. **Calculate Heat (Q)**:
- Since we established that \(Q = -W\), we can substitute the expression for work:
\[ Q = -nRT \ln\left(\frac{V_f}{V_i}\right) \]

6. **Calculate Change in Entropy (ΔS)**:
- The change in entropy (ΔS) for a reversible process is given by:
\[ \Delta S = \frac{Q_{\text{rev}}}{T} \]
- Substituting for \(Q\):
\[ \Delta S = \frac{-nRT \ln\left(\frac{V_f}{V_i}\right)}{T} \]
- Simplifying this gives:
\[ \Delta S = -nR \ln\left(\frac{V_f}{V_i}\right) \]

7. **Final Expression**:
- Since the volume increases during expansion (\(V_f > V_i\)), the logarithm will be positive, and thus:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]

### Final Answer:
The change in entropy of an ideal gas during reversible isothermal expansion is:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]