The volumes of two vessels at same temperature are in the ratio of `2 : 3`. One vessel contains `H_(2)` and other `N_(2)` at 600 mm and 900 mm respectively. The final pressure when they are connected together is : (Assume that `N_(2) and H_(2)` react to form `NH_(3)`)

Options

A. 620 mm
B. 760 mm
C. 780 mm
D. 800 mm

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given data
We have two vessels:
- Vessel 1 contains \( H_2 \) at a pressure of 600 mm and volume \( V_1 = 2V \).
- Vessel 2 contains \( N_2 \) at a pressure of 900 mm and volume \( V_2 = 3V \).

### Step 2: Calculate the number of moles of \( H_2 \) and \( N_2 \)
Using the ideal gas equation \( PV = nRT \), we can express the number of moles \( n \) as:
\[ n = \frac{PV}{RT} \]

For \( H_2 \):
\[ n_{H_2} = \frac{P_{H_2} \cdot V_{H_2}}{RT} = \frac{600 \, \text{mm} \cdot 2V}{RT} = \frac{1200V}{RT} \]

For \( N_2 \):
\[ n_{N_2} = \frac{P_{N_2} \cdot V_{N_2}}{RT} = \frac{900 \, \text{mm} \cdot 3V}{RT} = \frac{2700V}{RT} \]

### Step 3: Calculate the total number of moles
The total number of moles when both gases are combined is:
\[ n_{total} = n_{H_2} + n_{N_2} = \frac{1200V}{RT} + \frac{2700V}{RT} = \frac{3900V}{RT} \]

### Step 4: Calculate the final volume
When the two vessels are connected, the total volume \( V_{total} \) is:
\[ V_{total} = V_1 + V_2 = 2V + 3V = 5V \]

### Step 5: Calculate the final pressure using the ideal gas law
Using the ideal gas equation again for the combined system:
\[ P_{final} = \frac{n_{total} \cdot RT}{V_{total}} \]
Substituting the values we found:
\[ P_{final} = \frac{\left(\frac{3900V}{RT}\right) \cdot RT}{5V} \]
The \( RT \) and \( V \) terms cancel out:
\[ P_{final} = \frac{3900}{5} = 780 \, \text{mm} \]

### Final Answer:
The final pressure when the two vessels are connected together is **780 mm**.

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