Two closed vessel A and B of equal volume of 8.21 L are connected by a narrow tube of negligible volume with open valve. The left hand side container is found to contain 3 mole `CO_(2)` and 2 mole of He at 400 K. What is the partial pressure of He in vessel B at 400 K?

Options

A. 2.4 atm
B. 8 atm
C. 12atm
D. None of these

Explanation

To find the partial pressure of helium in vessel B at 400 K, we can follow these steps:

### Step 1: Determine the total number of moles in vessel A
In vessel A, we have:
- 3 moles of CO₂
- 2 moles of He

Total moles in vessel A (n_total) = 3 moles (CO₂) + 2 moles (He) = 5 moles

### Step 2: Use the Ideal Gas Law to find the total pressure in vessel A
The Ideal Gas Law is given by the equation:
\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

Rearranging the equation to solve for pressure \( P \):
\[ P = \frac{nRT}{V} \]

Substituting the known values:
- \( n = 5 \) moles
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 400 \) K
- \( V = 8.21 \) L

Calculating the total pressure:
\[ P = \frac{5 \times 0.0821 \times 400}{8.21} \]

### Step 3: Calculate the total pressure
Calculating the numerator:
\[ 5 \times 0.0821 \times 400 = 164.2 \]

Now divide by the volume:
\[ P = \frac{164.2}{8.21} \approx 20.0 \text{ atm} \]

### Step 4: Calculate the mole fraction of helium
The mole fraction of helium (\( \chi_{He} \)) is given by:
\[ \chi_{He} = \frac{\text{moles of He}}{\text{total moles}} = \frac{2}{5} \]

### Step 5: Calculate the partial pressure of helium
The partial pressure of helium (\( P_{He} \)) can be calculated using:
\[ P_{He} = \chi_{He} \times P \]

Substituting the values:
\[ P_{He} = \frac{2}{5} \times 20.0 \]

### Step 6: Calculate the partial pressure of helium
Calculating the partial pressure:
\[ P_{He} = \frac{2 \times 20.0}{5} = 8.0 \text{ atm} \]

### Final Answer
The partial pressure of helium in vessel B at 400 K is **8.0 atm**.

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