`N_(2)` gas is heated from 300 kg temperature to 600 k through an isobaric process. Then find the change in the entropy of the gas (n = 1 mole )

Options

A. 10 J/k
B. 20 J/k
C. 30 J/k
D. 40 J/k

Explanation

To solve the problem of finding the change in entropy of nitrogen gas (N₂) heated from 300 K to 600 K through an isobaric process, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
We are dealing with an isobaric process, which means the pressure remains constant while the temperature changes.

2. **Formula for Change in Entropy**:
The change in entropy \( \Delta S \) for an ideal gas during an isobaric process can be expressed as:
\[ \Delta S = n C_p \ln\left(\frac{T_2}{T_1}\right) \]
where \( n \) is the number of moles, \( C_p \) is the molar heat capacity at constant pressure, \( T_1 \) is the initial temperature, and \( T_2 \) is the final temperature.

3. **Identify Given Values**:
- \( n = 1 \) mole (as given)
- \( T_1 = 300 \, \text{K} \)
- \( T_2 = 600 \, \text{K} \)

4. **Find \( C_p \) for Nitrogen Gas**:
Nitrogen (N₂) is a diatomic gas. For diatomic gases, we can use the relation:
\[ C_p = \frac{7}{2} R \]
where \( R \) is the universal gas constant, approximately \( 8.314 \, \text{J/(mol K)} \).

5. **Calculate \( C_p \)**:
\[ C_p = \frac{7}{2} \times 8.314 \approx 29.19 \, \text{J/(mol K)} \]

6. **Substitute Values into the Entropy Change Formula**:
\[ \Delta S = 1 \times 29.19 \ln\left(\frac{600}{300}\right) \]

7. **Calculate the Natural Logarithm**:
\[ \ln\left(\frac{600}{300}\right) = \ln(2) \approx 0.693 \]

8. **Final Calculation of Change in Entropy**:
\[ \Delta S = 29.19 \times 0.693 \approx 20.24 \, \text{J/K} \]

9. **Conclusion**:
The change in entropy of the nitrogen gas when heated from 300 K to 600 K through an isobaric process is approximately \( 20.24 \, \text{J/K} \).