Two thermally insulated vessel 1 and 2 are filled with air at temperature `(T_1T_2), volume (V_1V_2)` and pressure `(P_1P_2)` respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

Options

A. `T_1+T_2`
B. `((T_1+T_2))/2`
C. `(T_1T_2(P_1V_1+P_2V_2))/(P_1V_1T_2+P_2V_2T_1)`
D. `T_1T_2(P_1V_1+P_2V_2)`

Answer

To find the equilibrium temperature \( T \) when two thermally insulated vessels are connected, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two thermally insulated vessels (1 and 2) filled with air at temperatures \( T_1 \) and \( T_2 \), volumes \( V_1 \) and \( V_2 \), and pressures \( P_1 \) and \( P_2 \) respectively.

2. **Applying the First Law of Thermodynamics**:
- Since the vessels are thermally insulated, there is no heat exchange (\( \Delta Q = 0 \)) and no work done (\( \Delta W = 0 \)). Therefore, the change in internal energy (\( \Delta U \)) is also zero.
- This implies that the total internal energy before opening the valve is equal to the total internal energy after opening the valve:
\[ U_1 + U_2 = U \]

3. **Internal Energy of Ideal Gases**:
- For an ideal gas, the internal energy \( U \) can be expressed as:
\[ U = n C_v T \]
- Where \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( T \) is the temperature.

4. **Calculating Moles**:
- The number of moles in each vessel can be calculated using the ideal gas law:
\[ n_1 = \frac{P_1 V_1}{R T_1} \quad \text{and} \quad n_2 = \frac{P_2 V_2}{R T_2} \]

5. **Setting Up the Equation**:
- The total internal energy before the valve is opened can be written as:
\[ U_1 + U_2 = n_1 C_v T_1 + n_2 C_v T_2 \]
- After opening the valve, the internal energy can be expressed as:
\[ U = (n_1 + n_2) C_v T \]
- Setting these equal gives:
\[ n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T \]

6. **Cancelling \( C_v \)**:
- Since \( C_v \) is a common factor, we can cancel it from both sides:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

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

8. **Substituting for \( n_1 \) and \( n_2 \)**:
- Substitute \( n_1 \) and \( n_2 \):
\[ T = \frac{\left(\frac{P_1 V_1}{R T_1}\right) T_1 + \left(\frac{P_2 V_2}{R T_2}\right) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} \]
- Simplifying gives:
\[ T = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}} \]

9. **Final Expression**:
- Thus, the equilibrium temperature \( T \) is given by:
\[ T = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{P_1 V_1 + P_2 V_2} \]