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:

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

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 \( 0.04 \, \text{L} \text{mol}^{-1} \))
- \( R \) = gas constant (given as \( 0.083 \, \text{bar dm}^3 \text{mol}^{-1} \text{K}^{-1} \))
- \( T \) = temperature (given as \( 300 \, \text{K} \))
- \( V \) = volume of the gas (given as \( 0.15 \, \text{L} \))

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

We start with the equation:

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

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

First, calculate \( \frac{3.64}{(0.15)^2} \):

\[ (0.15)^2 = 0.0225 \]

Now substitute:

\[ \frac{3.64}{0.0225} \approx 162.22 \]

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

Next, calculate \( 0.15 - 0.04 \):

\[ 0.15 - 0.04 = 0.11 \]

### Step 4: Calculate \( RT \)

Now calculate \( RT \):

\[ 0.083 \times 300 = 24.9 \]

### Step 5: Substitute back into the equation

Now substitute these values back into the equation:

\[ \left(P + 162.22\right)(0.11) = 24.9 \]

### Step 6: Expand and simplify

Expanding this gives:

\[ 0.11P + 17.8442 = 24.9 \]

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

Now isolate \( P \):

\[ 0.11P = 24.9 - 17.8442 \]

Calculating the right side:

\[ 0.11P = 7.0558 \]

Finally, divide by \( 0.11 \):

\[ P = \frac{7.0558}{0.11} \approx 64.58 \, \text{bar} \]

### Final Answer

The pressure exerted by the gas is approximately \( 64.58 \, \text{bar} \).

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