An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Options

A. `(T_(1)T_(2)(P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))`
B. `(P_(1)V_(1)T_(1) + P_(2)V_(2)T_(2))/(P_(1)V_(1) + P_(2)V_(2))`
C. `(P_(1)V_(1) T_(2) + P_(2) V_(2)T_(1))/(P_(1)V_(1) + P_(2)V_(2))`
D. `(T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1) T_(1) + P_(2)V_(2)T_(2))`

Answer

To find the final equilibrium temperature \( T \) of the gases in the insulated container after the partition is removed, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two chambers in an insulated container.
- Chamber 1 has volume \( V_1 \), pressure \( P_1 \), and temperature \( T_1 \).
- Chamber 2 has volume \( V_2 \), pressure \( P_2 \), and temperature \( T_2 \).
- The partition is insulating, meaning no heat transfer occurs between the two chambers.

2. **Applying the First Law of Thermodynamics**:
- The first law states that the change in internal energy \( \Delta U \) is equal to the heat added to the system \( \Delta Q \) minus the work done by the system \( \Delta W \):
\[ \Delta U = \Delta Q - \Delta W \]
- Since the system is insulated, \( \Delta Q = 0 \), and if no work is done on the gases, \( \Delta W = 0 \). Therefore, we have:
\[ \Delta U = 0 \]

3. **Setting Up the Energy Balance**:
- The total internal energy before mixing must equal the total internal energy after mixing:
\[ n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T \]
- Here, \( n_1 \) and \( n_2 \) are the number of moles of gas in chambers 1 and 2, respectively, and \( C_v \) is the specific heat at constant volume.

4. **Simplifying the Equation**:
- Since \( C_v \) is common to both sides, it can be canceled out:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

5. **Solving for Final Temperature \( T \)**:
- Rearranging gives:
\[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]

6. **Finding the Number of Moles**:
- Using the ideal gas law \( PV = nRT \), we can express \( n_1 \) and \( n_2 \):
\[ n_1 = \frac{P_1 V_1}{RT_1}, \quad n_2 = \frac{P_2 V_2}{RT_2} \]

7. **Substituting into the Temperature Equation**:
- Substitute \( n_1 \) and \( n_2 \) into the equation for \( T \):
\[ T = \frac{\left(\frac{P_1 V_1}{RT_1}\right) T_1 + \left(\frac{P_2 V_2}{RT_2}\right) T_2}{\frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2}} \]

8. **Simplifying the Expression**:
- This simplifies to:
\[ T = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{P_1 V_1 + P_2 V_2} \]

### Final Result:
The final equilibrium temperature \( T \) of the gases in the container is given by:
\[ T = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{P_1 V_1 + P_2 V_2} \]