Calculate the pressure exerted by 3 mole `CO_2` ​of gas at 273 K, if the van der Waal's constat "a"=3.592`dm^6atm^(-2)` . Assume that the volume occupied by `CO_2` molecules is negligible.

Answer

To calculate the pressure exerted by 3 moles of CO₂ gas at 273 K 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:

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

Where:
- \( P \) = pressure
- \( a \) = Van der Waals constant
- \( n \) = number of moles
- \( V \) = volume
- \( b \) = volume occupied by one mole of gas molecules (which we assume to be negligible here, so \( b = 0 \))
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Simplify the equation
Since the volume occupied by CO₂ molecules is negligible, we can set \( b = 0 \). The equation simplifies to:

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

### Step 3: Rearrange the equation to solve for pressure \( P \)
Rearranging gives:

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

### Step 4: Substitute the known values
Given:
- \( n = 3 \) moles
- \( T = 273 \) K
- \( a = 3.592 \, \text{dm}^6 \text{atm}^{-2} \)
- \( R = 0.0821 \, \text{dm}^3 \text{atm} \text{K}^{-1} \text{mol}^{-1} \)
- At standard temperature and pressure (STP), the volume \( V \) for 1 mole of gas is approximately \( 22.4 \, \text{dm}^3 \). Therefore, for 3 moles, \( V = 3 \times 22.4 = 67.2 \, \text{dm}^3 \).

Substituting these values into the equation:

\[ P = \frac{3 \times 0.0821 \times 273}{67.2} - \frac{3.592 \times 3^2}{67.2^2} \]

### Step 5: Calculate the first term
Calculating the first term:

\[ \frac{3 \times 0.0821 \times 273}{67.2} = \frac{67.2113}{67.2} \approx 1.0017 \, \text{atm} \]

### Step 6: Calculate the second term
Calculating the second term:

\[ \frac{3.592 \times 9}{67.2^2} = \frac{32.328}{4515.84} \approx 0.00715 \, \text{atm} \]

### Step 7: Combine the results to find \( P \)
Now, substituting back into the equation for \( P \):

\[ P \approx 1.0017 - 0.00715 \approx 0.99455 \, \text{atm} \]

### Step 8: Round off the answer
Rounding to three decimal places gives:

\[ P \approx 0.995 \, \text{atm} \]

### Final Answer
The pressure exerted by 3 moles of CO₂ gas at 273 K is approximately **0.995 atm**.

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