Using van der Waals' equation, find the constant 'a' (in atm `L^(2)mol^(-2)`) when two moles of a gas confined in 4 L flask exerts a pressure of 11.0 atmospheres at a temperature of 300 K. The value of b is 0.05 L `mol^(-1)` .(R = 0.082 atm.L/K mol)

Options

A. 2.62
B. 2.64
C. 6.24
D. 6.46

Explanation

To solve the problem using the van der Waals equation, we will follow these steps:

### Step 1: Write down the van der Waals equation.
The van der Waals equation is given by:
\[ \left( P + \frac{a n^2}{V^2} \right) (V - n b) = nRT \]
Where:
- \( P \) = pressure (atm)
- \( V \) = volume (L)
- \( n \) = number of moles (mol)
- \( R \) = ideal gas constant (atm·L/(K·mol))
- \( T \) = temperature (K)
- \( a \) = van der Waals constant (atm·L²/mol²)
- \( b \) = van der Waals constant (L/mol)

### Step 2: Substitute the known values into the equation.
Given:
- \( P = 11.0 \) atm
- \( V = 4.0 \) L
- \( n = 2.0 \) mol
- \( T = 300 \) K
- \( R = 0.0821 \) atm·L/(K·mol)
- \( b = 0.05 \) L/mol

Substituting these values into the van der Waals equation:
\[ \left( 11.0 + \frac{a (2.0)^2}{(4.0)^2} \right) (4.0 - 2.0 \times 0.05) = 2.0 \times 0.0821 \times 300 \]

### Step 3: Simplify the equation.
Calculate \( V - n b \):
\[ 4.0 - 2.0 \times 0.05 = 4.0 - 0.1 = 3.9 \, \text{L} \]

Now substitute this back into the equation:
\[ \left( 11.0 + \frac{4a}{16} \right) (3.9) = 2.0 \times 0.0821 \times 300 \]

Calculate the right side:
\[ 2.0 \times 0.0821 \times 300 = 49.26 \, \text{atm·L} \]

### Step 4: Expand and rearrange the equation.
Now we have:
\[ (11.0 + \frac{a}{4}) \times 3.9 = 49.26 \]

Expanding this gives:
\[ 11.0 \times 3.9 + \frac{3.9a}{4} = 49.26 \]

Calculating \( 11.0 \times 3.9 \):
\[ 42.9 + \frac{3.9a}{4} = 49.26 \]

### Step 5: Isolate the term involving \( a \).
Subtract \( 42.9 \) from both sides:
\[ \frac{3.9a}{4} = 49.26 - 42.9 \]
\[ \frac{3.9a}{4} = 6.36 \]

### Step 6: Solve for \( a \).
Multiply both sides by \( 4 \):
\[ 3.9a = 25.44 \]
Now divide by \( 3.9 \):
\[ a = \frac{25.44}{3.9} \approx 6.54 \, \text{atm·L}^2/\text{mol}^2 \]

### Final Answer:
The value of the constant \( a \) is approximately \( 6.54 \, \text{atm·L}^2/\text{mol}^2 \).

---