For a sample of perfect gas when its pressure is changed isothermally from `p_(i) to p_(f)` the entropy change is given by

Options

A. `Delta S = nRIn(p_(f)/p_(i))`
B. `Delta S = nRIn(p_(i)/p_(f))`
C. `Delta S = nRTIn(p_(f)/p_(i))`
D. `Delta S = nRTIn(p_(i)/p_(f))`

Answer

To find the entropy change (\( \Delta S \)) for a sample of a perfect gas when its pressure changes isothermally from \( p_i \) to \( p_f \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Isothermal Process**:
- In an isothermal process, the temperature (\( T \)) remains constant. For an ideal gas, we can use the ideal gas law, which states that \( PV = nRT \).

2. **Relating Pressure and Volume**:
- From the ideal gas law, we can express pressure (\( P \)) in terms of volume (\( V \)):
\[ P = \frac{nRT}{V} \]
- As the pressure changes from \( p_i \) to \( p_f \), the volume will also change accordingly.

3. **Entropy Change Formula**:
- The change in entropy (\( \Delta S \)) for a reversible process can be expressed as:
\[ \Delta S = \int \frac{dq_{\text{rev}}}{T} \]
- For an isothermal process, the heat transfer (\( dq \)) can be related to the work done on/by the gas.

4. **Work Done in Isothermal Process**:
- The work done (\( W \)) in an isothermal process can be expressed as:
\[ W = \int P \, dV = \int \frac{nRT}{V} \, dV \]
- This integral will give us the work done as the volume changes.

5. **Calculating the Integral**:
- The integral of \( \frac{1}{V} \) is:
\[ \int \frac{1}{V} \, dV = \ln V \]
- Therefore, when we evaluate the work done from \( V_i \) to \( V_f \):
\[ W = nRT \left( \ln V_f - \ln V_i \right) = nRT \ln \left( \frac{V_f}{V_i} \right) \]

6. **Substituting into the Entropy Formula**:
- Now, substituting this back into the entropy change formula:
\[ \Delta S = \frac{W}{T} = \frac{nRT \ln \left( \frac{V_f}{V_i} \right)}{T} \]
- The \( T \) cancels out:
\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \]

7. **Relating Volume to Pressure**:
- Since \( V \) is inversely related to \( P \) in an isothermal process, we can express the volumes in terms of pressures:
\[ V_i = \frac{nRT}{p_i}, \quad V_f = \frac{nRT}{p_f} \]
- Substituting these into the entropy change equation gives:
\[ \Delta S = nR \ln \left( \frac{\frac{nRT}{p_f}}{\frac{nRT}{p_i}} \right) = nR \ln \left( \frac{p_i}{p_f} \right) \]

### Final Result:
The entropy change for the perfect gas when its pressure changes isothermally from \( p_i \) to \( p_f \) is given by:
\[ \Delta S = nR \ln \left( \frac{p_i}{p_f} \right) \]