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. `(2P_1 T_2)/(T_1 + T_2)`
B. `(2P_1)/(T_1 + T_2)`
C. `(2P_1 T_2)/(T_1 + T_2)`
D. none of these

Answer

To solve the problem step by step, we will use the Ideal Gas Law and the concept of conservation of moles. Here’s how we can derive the final pressure in the two connected vessels:

### Step 1: Understand the Initial Conditions
Initially, both vessels have the same pressure \( P_1 \) and temperature \( T_1 \). The volume of each vessel is \( V \).

### Step 2: Write the Expression for Initial Moles
Using the Ideal Gas Law, the number of moles \( n \) in each vessel can be expressed as:
\[ n = \frac{PV}{RT} \]
For both vessels at the initial state:
- For vessel 1:
\[ n_1 = \frac{P_1 V}{RT_1} \]
- For vessel 2:
\[ n_2 = \frac{P_1 V}{RT_1} \]
Thus, the total initial moles \( n_{\text{initial}} \) is:
\[ n_{\text{initial}} = n_1 + n_2 = \frac{P_1 V}{RT_1} + \frac{P_1 V}{RT_1} = \frac{2P_1 V}{RT_1} \]

### Step 3: Write the Expression for Final Moles
After connecting the vessels, one vessel is maintained at temperature \( T_1 \) and the other at \( T_2 \). Let the final pressure in both vessels be \( P \). The number of moles in each vessel now is:
- For vessel 1 (at \( T_1 \)):
\[ n_1' = \frac{PV}{RT_1} \]
- For vessel 2 (at \( T_2 \)):
\[ n_2' = \frac{PV}{RT_2} \]
Thus, the total final moles \( n_{\text{final}} \) is:
\[ n_{\text{final}} = n_1' + n_2' = \frac{PV}{RT_1} + \frac{PV}{RT_2} = PV \left( \frac{1}{RT_1} + \frac{1}{RT_2} \right) \]

### Step 4: Equate Initial and Final Moles
Since the number of moles is conserved, we can set the initial moles equal to the final moles:
\[ \frac{2P_1 V}{RT_1} = PV \left( \frac{1}{RT_1} + \frac{1}{RT_2} \right) \]

### Step 5: Simplify the Equation
Cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{2P_1}{RT_1} = P \left( \frac{1}{RT_1} + \frac{1}{RT_2} \right) \]
Now, multiply both sides by \( RT_1 RT_2 \):
\[ 2P_1 RT_2 = P \left( RT_2 + RT_1 \right) \]

### Step 6: Solve for \( P \)
Rearranging gives:
\[ P = \frac{2P_1 RT_2}{RT_1 + RT_2} \]
Since \( R \) is a constant, it cancels out:
\[ P = \frac{2P_1 T_2}{T_1 + T_2} \]

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