A gas can expand through two processes : (i) isobaric, (ii) `p/V` =constant. Assuming that the initial volume is same in both processes and the final volume which is two times the initial volume is also same in both processes, which of the following is true?

Options

A. (a) Work done by gas in process (i) is greater than the work done by the gas in process (ii)
B. (b) Work done by gas in process (i) is smaller than the work done by the gas in process (ii)
C. (c) Final pressure is greater in process (i)
D. (d) Final temperature is greater in process (i)

Explanation

To solve the problem, we need to analyze the two processes of gas expansion: isobaric (constant pressure) and a process where \( P/V \) is constant. We will compare the work done and the final temperatures in both processes.

### Step-by-Step Solution:

1. **Define Initial and Final Conditions**:
- Let the initial volume \( V_i = V \).
- The final volume \( V_f = 2V \) (twice the initial volume).
- The initial pressure in both processes is \( P_i \).

2. **Process 1: Isobaric Expansion**:
- In an isobaric process, the pressure remains constant.
- Work done \( W_1 \) in an isobaric process can be calculated using the formula:
\[ W_1 = P \Delta V = P (V_f - V_i) = P (2V - V) = P V \]

3. **Process 2: \( P/V \) is Constant**:
- In this process, \( P/V = k \) (a constant).
- Therefore, \( P = kV \).
- Since \( P \) is inversely proportional to \( V \), as \( V \) increases, \( P \) decreases.
- The work done \( W_2 \) can be calculated as the area under the curve in the \( PV \) diagram. The area can be determined using the relationship \( P = kV \):
\[ W_2 = \int_{V}^{2V} P \, dV = \int_{V}^{2V} \frac{kV}{V} \, dV = \int_{V}^{2V} k \, dV = k(V_f - V_i) = k(2V - V) = kV \]
- Since \( k = P_i/V_i \), we can express \( W_2 \) in terms of \( P_i \):
\[ W_2 = P_i V \]

4. **Comparison of Work Done**:
- From the calculations, we see that:
\[ W_1 = P V \quad \text{and} \quad W_2 = P V \]
- Therefore, \( W_1 = W_2 \).

5. **Final Temperatures**:
- For the isobaric process:
- Using the ideal gas law \( PV = nRT \):
- \( T_f = \frac{P V_f}{nR} = \frac{P (2V)}{nR} = 2T_i \).
- For the \( P/V \) constant process:
- The final pressure \( P_f \) can be derived from the relationship \( P = kV \):
- If \( V_f = 2V \), then \( P_f = 2P_i \).
- Using the ideal gas law again:
- \( T_f = \frac{P_f V_f}{nR} = \frac{(2P_i)(2V)}{nR} = 4T_i \).

6. **Conclusion**:
- The work done in both processes is equal.
- The final temperature in the isobaric process is \( 2T_i \) while in the \( P/V \) constant process it is \( 4T_i \).
- Thus, the correct statement is that the work done by the gas in process 1 is equal to the work done in process 2, but the final temperature is greater in process 2.