In an isobaric process, when temperature changes from `T_(1)" to "T_(2), DeltaS` is equal to

Options

A. `2.303 C_(p)log(T_(2)//T_(1))`
B. `2.303C_(p)ln(T_(2)//T_(1))`
C. `C_(p)ln(T_(1)//T_(2))`
D. `C_(V)ln(T_(2)//T_(1))`

Explanation

To solve the problem of finding the change in entropy (ΔS) during an isobaric process when the temperature changes from T1 to T2, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Isobaric Process**:
- In an isobaric process, the pressure remains constant. Therefore, we denote the initial and final pressures as P1 and P2, respectively, where P1 = P2.

2. **Entropy Change Formula**:
- The change in entropy (ΔS) for a process can be expressed as:
\[ \Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1}\right) \]
- Here, \(C_p\) is the heat capacity at constant pressure, \(T_1\) and \(T_2\) are the initial and final temperatures, and \(R\) is the universal gas constant.

3. **Apply the Isobaric Condition**:
- Since the process is isobaric, we have \(P_1 = P_2\). Therefore, the term \(\ln \left(\frac{P_2}{P_1}\right)\) becomes \(\ln(1)\), which is equal to 0.

4. **Simplify the Entropy Change**:
- With the above simplification, the equation for ΔS reduces to:
\[ \Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) \]

5. **Final Expression**:
- Thus, the change in entropy during the isobaric process when the temperature changes from T1 to T2 is given by:
\[ \Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) \]