Two different ideal diatomic gases `A` and `B` are initially in the same state. `A` and `B` are then expanded to same final volume through adiabatic and isothermal process respectively. If `P_(A), P_(B)` and `T_(A), T_(B)` represents the final pressure and temperature of `A` and `B` respectively then.

Options

A. (a)`P_A lt P_B` and `T_A lt T_B`
B. (b)`P_A P_B` and `T_A gt T_B`
C. (c)`P_A gt P_B` and `T_A lt T_B`
D. (d)`P_A lt P_B` and ` T_A gt T_B`

Explanation

To solve the problem, we need to analyze the behavior of two different ideal diatomic gases, A and B, when they undergo different processes (adiabatic for gas A and isothermal for gas B) while reaching the same final volume.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Both gases A and B are initially in the same state, which means they have the same initial pressure \( P_i \) and temperature \( T_i \).

2. **Final Volume**:
- Both gases are expanded to the same final volume \( V_f \).

3. **Process for Gas A (Adiabatic Expansion)**:
- In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas undergoing adiabatic expansion, the relationship between pressure and volume can be described by the equation:
\[ P V^{\gamma} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio \( C_p/C_v \). For diatomic gases, \( \gamma \) is typically \( \frac{7}{5} \) or \( 1.4 \).

4. **Final Pressure and Temperature for Gas A**:
- Since gas A expands adiabatically to a larger volume, its final pressure \( P_A \) will be lower than its initial pressure \( P_i \).
- The final temperature \( T_A \) will also decrease due to the adiabatic expansion, as the internal energy of the gas decreases.

5. **Process for Gas B (Isothermal Expansion)**:
- In an isothermal process, the temperature remains constant. The relationship between pressure and volume for an ideal gas is given by:
\[ P V = nRT \]
where \( n \) is the number of moles and \( R \) is the ideal gas constant.
- Since gas B is undergoing isothermal expansion, its final pressure \( P_B \) will be lower than its initial pressure \( P_i \), but the temperature \( T_B \) will remain equal to the initial temperature \( T_i \).

6. **Comparing Final States**:
- From the adiabatic expansion of gas A, we have:
\[ P_A < P_i \quad \text{and} \quad T_A < T_i \]
- From the isothermal expansion of gas B, we have:
\[ P_B < P_i \quad \text{and} \quad T_B = T_i \]

7. **Conclusion**:
- Since both gases are at the same final volume and gas A has undergone a temperature drop, we can conclude:
\[ P_A < P_B \quad \text{and} \quad T_A < T_B \]
- Therefore, the correct option is **C**: \( P_A < P_B \) and \( T_A < T_B \).