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))`

Explanation

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

### Step 1: Understand the System
We have an insulated container divided into two chambers:
- Chamber 1: Volume \( V_1 \), Pressure \( P_1 \), Temperature \( T_1 \)
- Chamber 2: Volume \( V_2 \), Pressure \( P_2 \), Temperature \( T_2 \)

Since the container is insulated, there is no heat exchange with the surroundings (\( \Delta Q = 0 \)).

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Since \( Q = 0 \) (no heat exchange) and \( W = 0 \) (no work done), we have:
\[ \Delta U = 0 \]
This means that the change in internal energy of the system is zero.

### Step 3: Write the Expression for Internal Energy
The internal energy change can be expressed as:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = 0 \]
For the two chambers, the internal energy can be written as:
\[ U_{\text{initial}} = n_1 C_V T_1 + n_2 C_V T_2 \]
Where \( n_1 \) and \( n_2 \) are the number of moles in each chamber, and \( C_V \) is the molar specific heat at constant volume.

### Step 4: Express Moles in Terms of Pressure, Volume, and Temperature
Using the ideal gas equation \( PV = nRT \), we can express the number of moles:
\[ n_1 = \frac{P_1 V_1}{R T_1}, \quad n_2 = \frac{P_2 V_2}{R T_2} \]

### Step 5: Substitute into the Internal Energy Equation
Substituting \( n_1 \) and \( n_2 \) into the internal energy equation gives:
\[ U_{\text{initial}} = \left(\frac{P_1 V_1}{R T_1}\right) C_V T_1 + \left(\frac{P_2 V_2}{R T_2}\right) C_V T_2 \]
This simplifies to:
\[ U_{\text{initial}} = \frac{P_1 V_1 C_V}{R} + \frac{P_2 V_2 C_V}{R} \]

### Step 6: Write the Final Internal Energy Expression
After the partition is removed, the total number of moles is \( n_1 + n_2 \) and the final temperature is \( T \):
\[ U_{\text{final}} = (n_1 + n_2) C_V T \]
Substituting \( n_1 + n_2 \):
\[ U_{\text{final}} = \left(\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\right) C_V T \]

### Step 7: Set Initial and Final Internal Energies Equal
Since \( U_{\text{initial}} = U_{\text{final}} \):
\[ \frac{P_1 V_1 C_V}{R} + \frac{P_2 V_2 C_V}{R} = \left(\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\right) C_V T \]

### Step 8: Cancel \( C_V \) and Rearrange
Cancel \( C_V \) from both sides:
\[ \frac{P_1 V_1}{R} + \frac{P_2 V_2}{R} = \left(\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\right) T \]
Multiply through by \( R \):
\[ P_1 V_1 + P_2 V_2 = \left(\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}\right) T \]

### Step 9: Solve for Final Temperature \( T \)
Rearranging gives:
\[ T = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{P_1 V_1 + P_2 V_2} \]

### Final Answer
The final equilibrium temperature \( T \) is:
\[ T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1} \]