An insulated container of gas has two chambers separated by an insulating partition. One of the chmabers 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_1T_2(P_1V_1+P_2V_2)/(P_1V_1T_2+P_2V_2T_1)`
B. `(P_1V_1T_1+P_2V_2T_2)/(P_1V_1+P_2V_2)`
C. `(P_1V_1T_2+P_2V_2T_1)/(P_1V_1+P_2V_2)`
D. `(T_1T_2(P_1V_1+P_2V_2))/(P_1V_1T_1+P_2V_2T_2)`

Answer

To solve the problem of finding the final equilibrium temperature of the gas in the insulated container after removing the partition, we can follow these steps:

### Step 1: Understand the System
We have an insulated container with two chambers. Chamber 1 has volume \( V_1 \), pressure \( P_1 \), and temperature \( T_1 \). Chamber 2 has volume \( V_2 \), pressure \( P_2 \), and temperature \( T_2 \). When the partition is removed, the gases will mix, and we need to find the final equilibrium temperature \( T_f \).

### Step 2: Apply the First Law of Thermodynamics
Since the container is insulated, there is no heat exchange with the surroundings, and no work is done on or by the gas. Therefore, the change in internal energy \( \Delta U \) is zero:
\[ \Delta Q = \Delta U + \Delta W \]
Given that \( \Delta Q = 0 \) and \( \Delta W = 0 \), we have:
\[ \Delta U = 0 \]

### Step 3: Relate Internal Energy to Temperature
For an ideal gas, the internal energy \( U \) is a function of temperature. Thus, the total internal energy before the partition is removed can be expressed as:
\[ U_{initial} = N_1 C_v T_1 + N_2 C_v T_2 \]
where \( N_1 \) 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.

### Step 4: Determine the Number of Moles
Using the ideal gas law \( PV = nRT \), we can express the number of moles for each chamber:
\[ N_1 = \frac{P_1 V_1}{R T_1} \]
\[ N_2 = \frac{P_2 V_2}{R T_2} \]

### Step 5: Set Up the Equation for Final Temperature
Since the internal energy remains constant, we can write:
\[ N_1 C_v T_f + N_2 C_v T_f = N_1 C_v T_1 + N_2 C_v T_2 \]
Factoring out \( C_v T_f \):
\[ (N_1 + N_2) C_v T_f = N_1 C_v T_1 + N_2 C_v T_2 \]
Dividing through by \( C_v \) (which is constant and non-zero):
\[ (N_1 + N_2) T_f = N_1 T_1 + N_2 T_2 \]

### Step 6: Solve for Final Temperature
Now, we can solve for \( T_f \):
\[ T_f = \frac{N_1 T_1 + N_2 T_2}{N_1 + N_2} \]
Substituting the expressions for \( N_1 \) and \( N_2 \):
\[ T_f = \frac{\left(\frac{P_1 V_1}{R T_1}\right) T_1 + \left(\frac{P_2 V_2}{R T_2}\right) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} \]
This simplifies to:
\[ T_f = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}} \]

### Final Answer
Thus, the final equilibrium temperature \( T_f \) is given by:
\[ T_f = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{P_1 V_1 + P_2 V_2} \]