Two samples `1` and `2` are initially kept in the same state. The sample `1` is expanded through an isothermal process where as sample `2` through an adiabatic process upto the same final volume. The final temperature in process `1` and `2` are `T_(1)` and `T_(2)` respectively, then

Options

A. `T_(1)gtT_(2)`
B. `T_(1)=T_(2)`
C. `T_(1)ltT_(2)`
D. The relation between `T_(1)` and `T_(2)` cannot be deduced

Answer

To solve the problem, we need to analyze the two processes: isothermal and adiabatic, and how they affect the temperature of the gas samples.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Both samples (1 and 2) are initially in the same state, which means they have the same initial temperature \( T \), pressure \( P \), and volume \( V \).

2. **Isothermal Process for Sample 1**:
- Sample 1 undergoes an isothermal expansion. In an isothermal process, the temperature of the gas remains constant throughout the process. Therefore, the final temperature \( T_1 \) of sample 1 will be equal to the initial temperature \( T \):
\[ T_1 = T \]

3. **Adiabatic Process for Sample 2**:
- Sample 2 undergoes an adiabatic expansion. In an adiabatic process, there is no heat exchange with the surroundings. As the gas expands, it does work on the surroundings, which results in a decrease in temperature.
- The relationship for an adiabatic process can be expressed as:
\[ TV^{\gamma - 1} = \text{constant} \]
- Since the volume of sample 2 increases during expansion, the temperature \( T_2 \) must decrease. Therefore, we can conclude that:
\[ T_2 < T \]

4. **Comparing Final Temperatures**:
- From the above analysis, we have:
\[ T_1 = T \quad \text{and} \quad T_2 < T \]
- This implies:
\[ T_2 < T_1 \]

5. **Conclusion**:
- The final temperature of sample 1 (isothermal process) is greater than the final temperature of sample 2 (adiabatic process). Thus, we can conclude:
\[ T_2 < T_1 \]

### Final Answer:
The relationship between the final temperatures is:
\[ T_2 < T_1 \]