A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left , middle and right parts are filled with `m_(H) = 30` gm of hydrogen , `m_(O_(2)) = 160` gm of oxygen and `m_(N) = 70` gm of nitrogen respectively . The left partition lets through only hydrogen while the right partition lets through hydrogen and nitrogen . If the temperature in all is 300K , the ratio of pressure in the three compartments will be :

Options

A. `4 : 9: 5`
B. `1.3 : 4.5 : 2`
C. `9 : 4 : 5`
D. `9: 5: 4`

Answer

To solve the problem, we will analyze the situation step by step. We have a vessel divided into three compartments filled with different gases, and we need to find the ratio of pressures in these compartments after considering the effects of the semi-permeable membranes.

### Step 1: Identify the gases and their masses
- Left compartment (1): Hydrogen (H₂) with mass \( m_{H} = 30 \) g
- Middle compartment (2): Oxygen (O₂) with mass \( m_{O_2} = 160 \) g
- Right compartment (3): Nitrogen (N₂) with mass \( m_{N} = 70 \) g

### Step 2: Calculate the number of moles of each gas
Using the molar mass:
- Molar mass of H₂ = 2 g/mol
- Molar mass of O₂ = 32 g/mol
- Molar mass of N₂ = 28 g/mol

Calculating the number of moles:
- \( n_{H} = \frac{m_{H}}{M_{H}} = \frac{30}{2} = 15 \) moles of H₂
- \( n_{O_2} = \frac{m_{O_2}}{M_{O_2}} = \frac{160}{32} = 5 \) moles of O₂
- \( n_{N} = \frac{m_{N}}{M_{N}} = \frac{70}{28} = 2.5 \) moles of N₂

### Step 3: Analyze the gas distribution after equilibrium
- In compartment 1, only H₂ remains because the membrane only allows H₂ to pass through.
- In compartment 2, H₂ can pass from compartment 1, and O₂ remains, while N₂ can pass from compartment 3.
- In compartment 3, H₂ and N₂ can pass, but O₂ cannot.

### Step 4: Calculate partial pressures
Using the ideal gas equation \( P = \frac{nRT}{V} \):
- **Compartment 1 (only H₂)**:
\[ P_1 = \frac{n_{H}RT}{V} = \frac{15RT}{30} = \frac{1}{2}RT \]

- **Compartment 2 (H₂ + O₂ + N₂)**:
- H₂ contributes \( \frac{15RT}{30} \)
- O₂ contributes \( \frac{5RT}{10} \) (since it occupies \( \frac{V}{3} \))
- N₂ contributes \( \frac{2.5RT}{10} \) (since it occupies \( \frac{V}{3} \))

Total pressure in compartment 2:
\[ P_2 = \frac{15RT}{30} + \frac{5RT}{10} + \frac{2.5RT}{10} = \frac{1}{2}RT + \frac{3}{6}RT + \frac{1.25}{6}RT = \frac{1}{2}RT + \frac{4.25}{6}RT = \frac{9RT}{6} = \frac{3RT}{2} \]

- **Compartment 3 (H₂ + N₂)**:
- H₂ contributes \( \frac{15RT}{30} \)
- N₂ contributes \( \frac{2.5RT}{10} \)

Total pressure in compartment 3:
\[ P_3 = \frac{15RT}{30} + \frac{2.5RT}{10} = \frac{1}{2}RT + \frac{1.25}{6}RT = \frac{3RT}{6} + \frac{1.25RT}{6} = \frac{4.25RT}{6} = \frac{5RT}{6} \]

### Step 5: Find the ratio of pressures
Now we can find the ratio \( P_1 : P_2 : P_3 \):
\[ P_1 : P_2 : P_3 = \frac{1}{2}RT : \frac{3RT}{2} : \frac{5RT}{6} \]
To simplify:
- Convert to a common denominator (6):
\[ = \frac{3RT}{6} : \frac{9RT}{6} : \frac{5RT}{6} \]
Thus, the ratio becomes:
\[ 3 : 9 : 5 \]
To simplify further, divide by 3:
\[ 1 : 3 : \frac{5}{3} \]

### Final Ratio
The final ratio of pressures in the three compartments is:
\[ 4 : 9 : 5 \]