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. `P_(A) lt P_(B)` and `T_(A) lt T_(B)`
B. `P_(A) gt P_(B)` and `T_(A) gt T_(B)`
C. `P_(A) gt P_(B)` and `T_(A) lt t_(B)`
D. `P_(A) lt P_(B)` and `T_(A) gt T_(B)`

Explanation

To solve the problem, we will analyze the behavior of two different ideal diatomic gases, A and B, during their respective processes: an adiabatic expansion for gas A and an isothermal expansion for gas B. We will derive the relationships between their final pressures and temperatures.

### Step-by-Step Solution:

1. **Understanding the Processes**:
- Gas A undergoes an **adiabatic expansion**, meaning there is no heat exchange with the surroundings. The internal energy change results in a change in temperature and pressure.
- Gas B undergoes an **isothermal expansion**, where the temperature remains constant throughout the process. The gas absorbs heat from the surroundings to maintain constant temperature while expanding.

2. **Final State**:
- Both gases are expanded to the **same final volume**. Let’s denote this final volume as \( V_f \).

3. **Using the Ideal Gas Law**:
- For both gases, we can use the ideal gas law, which states:
\[ PV = nRT \]
- Rearranging gives us:
\[ P = \frac{nRT}{V} \]
- Since both gases have the same number of moles \( n \) and the same final volume \( V_f \), we can express their pressures in terms of their temperatures:
\[ P_A = \frac{nRT_A}{V_f} \quad \text{(for gas A)} \]
\[ P_B = \frac{nRT_B}{V_f} \quad \text{(for gas B)} \]

4. **Analyzing the Adiabatic Process (Gas A)**:
- In an adiabatic process for an ideal gas, the relationship between pressure and temperature can be expressed as:
\[ P_A V_f^{\gamma} = \text{constant} \]
- Here, \( \gamma \) (gamma) is the heat capacity ratio \( C_p/C_v \). As gas A expands adiabatically, its temperature decreases, leading to a lower final pressure \( P_A \).

5. **Analyzing the Isothermal Process (Gas B)**:
- In the isothermal process, the temperature remains constant. Thus, the pressure can be expressed as:
\[ P_B = \frac{nRT_B}{V_f} \]
- Since \( T_B \) remains constant and is generally higher than \( T_A \) (due to the heat absorbed), the final pressure \( P_B \) will be higher than \( P_A \).

6. **Comparing Pressures**:
- From the analysis, we can conclude:
\[ P_B > P_A \]

7. **Comparing Temperatures**:
- Since \( P \) is directly proportional to \( T \) when \( V \) is constant, we can also conclude:
\[ T_B > T_A \]

### Final Relationships:
- Therefore, we have:
\[ P_B > P_A \quad \text{and} \quad T_B > T_A \]

### Conclusion:
The final results indicate that the pressure of gas B (isothermal process) is greater than the pressure of gas A (adiabatic process), and the temperature of gas B is also greater than that of gas A.

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