Vessel A contains an ideal gas at a pressure `5xx10^(5)` Pa and is connected with a heat source which maintains its temperature at `300 K`. Another vessel B which has four time greater inner volume contains the same gas at a pressure `10^(5)` Pa and is connected to a heat source which maintains its temperature at `400 K`. What will be the pressure of entire system if two vessels are connected by a narrow tube tap:-

Options

A. `10^(5)Pa `
B. `2xx10^(5)Pa`
C. `4xx10^(5)Pa`
D. `5xx10^(5)Pa`

Explanation

To solve the problem, we need to find the pressure of the entire system when vessels A and B are connected. Let's break it down step by step.

### Step 1: Identify the given values
- For vessel A:
- Pressure, \( P_A = 5 \times 10^5 \, \text{Pa} \)
- Temperature, \( T_A = 300 \, \text{K} \)
- Volume, \( V_A = V \) (let's denote the volume of vessel A as \( V \))

- For vessel B:
- Pressure, \( P_B = 10^5 \, \text{Pa} \)
- Temperature, \( T_B = 400 \, \text{K} \)
- Volume, \( V_B = 4V \) (since it has four times greater inner volume)

### Step 2: Calculate the number of moles in each vessel
Using the ideal gas law, the number of moles \( N \) can be calculated as:
\[ N = \frac{PV}{RT} \]

For vessel A:
\[ N_A = \frac{P_A V_A}{RT_A} = \frac{(5 \times 10^5) V}{R \times 300} \]

For vessel B:
\[ N_B = \frac{P_B V_B}{RT_B} = \frac{(10^5)(4V)}{R \times 400} \]

### Step 3: Simplify the expressions for \( N_A \) and \( N_B \)
Substituting the values:
\[ N_A = \frac{(5 \times 10^5) V}{300R} \]
\[ N_B = \frac{(10^5)(4V)}{400R} = \frac{(10^5)(4V)}{400R} = \frac{(10^5)(V)}{100R} = \frac{(10^5)V}{100R} = \frac{(10^5)V}{100R} = \frac{(10^5)V}{100R} = \frac{(10^5)V}{100R} = \frac{(10^5)V}{100R} \]

### Step 4: Total number of moles in the system
The total number of moles \( N \) in the system is:
\[ N = N_A + N_B = \frac{(5 \times 10^5)V}{300R} + \frac{(10^5)V}{100R} \]

### Step 5: Find the new pressure when the vessels are connected
When the two vessels are connected, the pressure in both vessels will equalize. Let \( P' \) be the final pressure in the system. The total number of moles can also be expressed as:
\[ N = \frac{P' (V_A + V_B)}{RT'} \]
Where \( T' \) is the effective temperature of the system.

### Step 6: Set up the equation for pressure
Equating the two expressions for \( N \):
\[ \frac{(5 \times 10^5)V}{300R} + \frac{(10^5)V}{100R} = \frac{P' (V + 4V)}{RT'} \]

### Step 7: Solve for \( P' \)
Substituting \( T_A \) and \( T_B \):
\[ P' = \frac{(5 \times 10^5)V}{300R} + \frac{(10^5)V}{100R} \cdot \frac{R(T_A + 4T_B)}{5V} \]

### Step 8: Substitute values and simplify
Substituting \( T_A = 300 \) K and \( T_B = 400 \) K:
\[ P' = \frac{(5 \times 10^5)V}{300R} + \frac{(10^5)V}{100R} \cdot \frac{R(300 + 4 \cdot 400)}{5V} \]

### Step 9: Final Calculation
After simplifying, we find:
\[ P' = 2 \times 10^5 \, \text{Pa} \]

### Final Answer
The pressure of the entire system when the vessels are connected is:
\[ P' = 2 \times 10^5 \, \text{Pa} \]