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_1 T_2(V_1 + V_2))/(p_1 V_1 T_2 + p_2 V_2 T_1)`.
D. `(T_1 T_2(p_1V_1 + p_2 V_2))/(p_1 V_1 T_2 + p_2 V_2 T_1)`.

Explanation

To find the equilibrium temperature \( T \) when two thermally insulated vessels are connected, we can use the principles of thermodynamics. Here’s a step-by-step solution:

### Step 1: Understand the System
We have two thermally insulated vessels (1 and 2) filled with air at different temperatures \( T_1 \) and \( T_2 \), volumes \( V_1 \) and \( V_2 \), and pressures \( P_1 \) and \( P_2 \). Since they are thermally insulated, no heat is exchanged with the surroundings.

### Step 2: Apply the First Law of Thermodynamics
The First Law of Thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
Since the vessels are insulated, \( \Delta Q = 0 \). Also, since the volume is constant, \( \Delta W = 0 \). Therefore:
\[ 0 = \Delta U + 0 \implies \Delta U = 0 \]

### Step 3: Express Internal Energy Change
The change in internal energy for an ideal gas can be expressed as:
\[ \Delta U = n_1 C_V T_1 + n_2 C_V T_2 \]
At equilibrium, the total internal energy before and after the valve is opened must be equal:
\[ n_1 C_V T_1 + n_2 C_V T_2 = (n_1 + n_2) C_V T \]

### Step 4: Simplify the Equation
Since \( C_V \) is a constant and appears in all terms, we can cancel it out:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

### Step 5: Express Number of Moles
Using the ideal gas law, we can express the number of moles \( n_1 \) and \( n_2 \) as:
\[ n_1 = \frac{P_1 V_1}{R T_1}, \quad n_2 = \frac{P_2 V_2}{R T_2} \]

### Step 6: Substitute Values
Substituting these expressions into the energy balance equation:
\[ \frac{P_1 V_1}{R T_1} T_1 + \frac{P_2 V_2}{R T_2} T_2 = \left(\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\right) T \]

### Step 7: Simplify Further
This simplifies to:
\[ \frac{P_1 V_1}{R} + \frac{P_2 V_2}{R} = \left(\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}\right) T \]
Multiplying 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 8: Solve for Equilibrium Temperature
Rearranging gives:
\[ T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1} \]

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