The van der Waal's constants a & b of `CO_(2)` are `3.64 L^(2) mol^(-2)` bar & `0.04 L mol^(-1)` respectively. The value of `R` is `0.083 bar dm^(3)mol^(-1)K^(-1)`. If one mole of `CO_(2)` is confined to a volume of `0.15L at 300 K`, then the pressure (in bar) eaxerted by the gas is :

Options

A. `11 bar`
B. `64.58 bar`
C. `174.50 bar`
D. `135.78 bar`

Explanation

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

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

Where:
- \( P \) = pressure of the gas
- \( a \) = Van der Waals constant for attraction (given as \( 3.64 \, \text{L}^2 \text{mol}^{-2} \text{bar} \))
- \( b \) = Van der Waals constant for volume occupied by gas molecules (given as \( 0.04 \, \text{L mol}^{-1} \))
- \( V \) = volume of the gas (given as \( 0.15 \, \text{L} \))
- \( n \) = number of moles of gas (given as \( 1 \, \text{mol} \))
- \( R \) = universal gas constant (given as \( 0.083 \, \text{bar dm}^3 \text{mol}^{-1} \text{K}^{-1} \))
- \( T \) = temperature (given as \( 300 \, \text{K} \))

### Step 1: Substitute the known values into the Van der Waals equation.

We start with the equation:

\[ \left( P + \frac{3.64}{(0.15)^2} \right) (0.15 - 0.04) = 1 \times 0.083 \times 300 \]

### Step 2: Calculate the term \( \frac{a}{V^2} \).

Calculating \( \frac{3.64}{(0.15)^2} \):

\[ (0.15)^2 = 0.0225 \]

\[ \frac{3.64}{0.0225} = 161.78 \, \text{bar} \]

### Step 3: Calculate \( V - b \).

Calculating \( 0.15 - 0.04 \):

\[ 0.15 - 0.04 = 0.11 \, \text{L} \]

### Step 4: Calculate \( nRT \).

Calculating \( 1 \times 0.083 \times 300 \):

\[ 0.083 \times 300 = 24.9 \, \text{bar} \]

### Step 5: Substitute the calculated values back into the equation.

Now substituting back into the equation:

\[ \left( P + 161.78 \right) (0.11) = 24.9 \]

### Step 6: Expand and rearrange the equation to solve for \( P \).

Expanding the left side:

\[ 0.11P + 17.7958 = 24.9 \]

Rearranging gives:

\[ 0.11P = 24.9 - 17.7958 \]

Calculating the right side:

\[ 0.11P = 7.1042 \]

### Step 7: Solve for \( P \).

Now, divide by \( 0.11 \):

\[ P = \frac{7.1042}{0.11} = 64.58 \, \text{bar} \]

### Final Answer:

Thus, the pressure exerted by the gas is approximately:

\[ P \approx 64.58 \, \text{bar} \]