Two closed vessels of equal volume containing air at pressure `P_(1)` and temperature `T_(1)` are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at `T_(1)` and that in the other at `T_(2)`, what will be the pressure in the vessels?

Options

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

Explanation

To solve the problem, we need to determine the final pressure in the two closed vessels after one vessel is maintained at temperature \( T_1 \) and the other at temperature \( T_2 \). We will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

1. **Initial Conditions**:
- Both vessels have the same volume \( V \).
- The initial pressure in both vessels is \( P_1 \).
- The initial temperature in both vessels is \( T_1 \).

2. **Number of Moles Calculation**:
- Using the Ideal Gas Law, we can calculate the number of moles in each vessel initially:
\[ n_1 = \frac{P_1 V}{RT_1} \]
Since both vessels are identical, the total number of moles in both vessels is:
\[ n_{total} = 2n_1 = 2 \cdot \frac{P_1 V}{RT_1} \]

3. **Final Conditions**:
- One vessel is maintained at temperature \( T_1 \) and the other at temperature \( T_2 \).
- Let \( P \) be the final pressure in both vessels.

4. **Final Number of Moles Calculation**:
- For the first vessel at temperature \( T_1 \):
\[ n_1' = \frac{PV}{RT_1} \]
- For the second vessel at temperature \( T_2 \):
\[ n_2' = \frac{PV}{RT_2} \]
- The total number of moles in both vessels is:
\[ n_{total} = n_1' + n_2' = \frac{PV}{RT_1} + \frac{PV}{RT_2} \]

5. **Equating the Total Moles**:
- Since the total number of moles must remain constant, we can equate the initial and final total moles:
\[ 2 \cdot \frac{P_1 V}{RT_1} = \frac{PV}{RT_1} + \frac{PV}{RT_2} \]

6. **Simplifying the Equation**:
- Cancel \( V \) from both sides (since \( V \) is the same):
\[ 2 \cdot \frac{P_1}{R} = \frac{P}{RT_1} + \frac{P}{RT_2} \]
- Multiply through by \( R \):
\[ 2P_1 = \frac{P}{T_1} + \frac{P}{T_2} \]

7. **Finding a Common Denominator**:
- Combine the fractions on the right-hand side:
\[ 2P_1 = P \left( \frac{T_2 + T_1}{T_1 T_2} \right) \]

8. **Solving for \( P \)**:
- Rearranging gives:
\[ P = \frac{2P_1 T_1 T_2}{T_1 + T_2} \]

### Final Result:
The final pressure \( P \) in the vessels is given by:
\[ P = \frac{2P_1 T_2}{T_1 + T_2} \]