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. `DeltaS=nR "ln" ((P_(f))/(P_(i)))`
B. `DeltaS=nR "In" ((P_(i))/(P_(f)))`
C. `DeltaS=nRT"In" ((P_(f))/(P_(i)))`
D. `DeltaS=RT "In" ((P_(i))/(P_(f)))`

Explanation

To find the entropy change 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. **Understand the Concept of Entropy Change**:
The change in entropy (\( \Delta S \)) for a reversible process can be expressed as:
\[ \Delta S = \frac{Q_{rev}}{T} \]
where \( Q_{rev} \) is the heat exchanged reversibly and \( T \) is the absolute temperature.

2. **Isothermal Process**:
Since the process is isothermal, the temperature \( T \) remains constant throughout the process.

3. **Relate Heat and Work**:
For an ideal gas undergoing an isothermal process, the heat added to the system can be related to the work done on/by the system. According to the first law of thermodynamics:
\[ dQ = dW \]
For an isothermal process, the work done \( dW \) can be expressed as:
\[ dW = nRT \frac{dV}{V} \]
where \( n \) is the number of moles and \( R \) is the universal gas constant.

4. **Substituting into the Entropy Change Formula**:
Substitute \( dQ \) into the entropy change formula:
\[ \Delta S = \frac{1}{T} \int_{V_i}^{V_f} nRT \frac{dV}{V} \]
Here, \( V_i \) and \( V_f \) are the initial and final volumes, respectively.

5. **Integrate**:
Since \( nR \) and \( T \) are constants, they can be taken out of the integral:
\[ \Delta S = nR \int_{V_i}^{V_f} \frac{dV}{V} = nR \left[ \ln V \right]_{V_i}^{V_f} = nR (\ln V_f - \ln V_i) \]
This simplifies to:
\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \]

6. **Using the Ideal Gas Law**:
From the ideal gas law, we know that:
\[ p_i V_i = nRT \quad \text{and} \quad p_f V_f = nRT \]
Therefore, we can express the volumes in terms of pressures:
\[ \frac{V_f}{V_i} = \frac{p_i}{p_f} \]

7. **Final Expression for Entropy Change**:
Substitute \( \frac{V_f}{V_i} \) back into the entropy change equation:
\[ \Delta S = nR \ln \left( \frac{p_i}{p_f} \right) \]

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