If volume occupied by `CO_(2)` molecules is negligible, then calculate pressure `((P)/(5.277))` exerted by one mole of `CO_(2)` gas at `300 K`. `(a=3.592 atm L^(2)mol^(-2))`

Options

A. (a) 7
B. (b) 8
C. (c) 9
D. (d) 3

Explanation

To solve the problem, we will use the Van der Waals equation for real gases, which is given by:

\[ \left( P + \frac{a n^2}{V^2} \right) (V - nB) = nRT \]

Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin
- \( a \) = Van der Waals constant (for attraction between molecules)
- \( B \) = Van der Waals constant (for volume occupied by the gas molecules)

Given:
- \( n = 1 \) mole (since we are considering one mole of \( CO_2 \))
- \( T = 300 \) K
- \( a = 3.592 \, \text{atm L}^2/\text{mol}^2 \)
- Volume occupied by \( CO_2 \) molecules is negligible, so we can assume \( B \approx 0 \).

### Step 1: Simplify the Van der Waals equation
Since \( B \) is negligible, the equation reduces to:

\[ P + \frac{a n^2}{V^2} = \frac{nRT}{V} \]

Substituting \( n = 1 \):

\[ P + \frac{a}{V^2} = \frac{RT}{V} \]

### Step 2: Rearranging the equation
Rearranging gives us:

\[ P = \frac{RT}{V} - \frac{a}{V^2} \]

### Step 3: Formulate a quadratic equation
To find \( P \), we can multiply through by \( V^2 \) to eliminate the denominators:

\[ PV^2 = RTV - a \]

Rearranging gives us:

\[ PV^2 - RTV + a = 0 \]

This is a quadratic equation in terms of \( V \).

### Step 4: Apply the quadratic formula
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = P \)
- \( b = -RT \)
- \( c = a \)

Using the quadratic formula:

\[ V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting the values:

\[ V = \frac{RT \pm \sqrt{(RT)^2 - 4Pa}}{2P} \]

### Step 5: Set the discriminant to zero
For the equation to have one solution (which is the case for a specific pressure and temperature), we set the discriminant to zero:

\[ (RT)^2 - 4Pa = 0 \]

Solving for \( P \):

\[ P = \frac{(RT)^2}{4a} \]

### Step 6: Substitute the known values
Now substituting the known values:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)
- \( a = 3.592 \, \text{atm L}^2/\text{mol}^2 \)

Calculating \( P \):

\[ P = \frac{(0.0821 \times 300)^2}{4 \times 3.592} \]

Calculating \( (0.0821 \times 300)^2 \):

\[ = (24.63)^2 = 606.5769 \]

Now substituting back:

\[ P = \frac{606.5769}{14.368} \approx 42.22 \, \text{atm} \]

### Step 7: Calculate \( \frac{P}{5.277} \)
Now we need to calculate \( \frac{P}{5.277} \):

\[ \frac{42.22}{5.277} \approx 8 \]

Thus, the final answer is:

\[ \boxed{8} \]