Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible.

Explanation

To calculate the pressure exerted by one mole of CO₂ gas at 273 K using the van der Waals equation, we will follow these steps:

### Step 1: Understand the Van der Waals Equation
The van der Waals equation for one mole of gas is given by:
\[ P + \frac{a}{V^2} = \frac{RT}{V - b} \]
where:
- \( P \) = pressure of the gas
- \( a \) = van der Waals constant (for CO₂, \( a = 3.592 \, \text{dm}^6 \, \text{atm} \, \text{mol}^{-2} \))
- \( V \) = volume of the gas
- \( R \) = universal gas constant (\( R = 0.082 \, \text{atm} \, \text{L} \, \text{K}^{-1} \, \text{mol}^{-1} \))
- \( T \) = temperature in Kelvin (given \( T = 273 \, \text{K} \))
- \( b \) = volume occupied by the gas molecules (assumed negligible, so \( b = 0 \))

### Step 2: Substitute Known Values
Since the volume occupied by CO₂ molecules is negligible, we set \( b = 0 \). The equation simplifies to:
\[ P + \frac{a}{V^2} = \frac{RT}{V} \]

### Step 3: Rearrange the Equation
Rearranging the equation to solve for \( P \):
\[ P = \frac{RT}{V} - \frac{a}{V^2} \]

### Step 4: Use the Ideal Gas Volume
For one mole of an ideal gas at standard temperature and pressure (STP), the volume \( V \) is approximately \( 22.4 \, \text{L} \) (or \( 22.4 \, \text{dm}^3 \)).

### Step 5: Plug in the Values
Now, substitute the values into the equation:
- \( R = 0.082 \, \text{atm} \, \text{L} \, \text{K}^{-1} \, \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \)
- \( V = 22.4 \, \text{L} \)
- \( a = 3.592 \, \text{dm}^6 \, \text{atm} \, \text{mol}^{-2} \)

Substituting these values:
\[ P = \frac{0.082 \times 273}{22.4} - \frac{3.592}{(22.4)^2} \]

### Step 6: Calculate Each Term
1. Calculate \( \frac{0.082 \times 273}{22.4} \):
\[ \frac{0.082 \times 273}{22.4} = \frac{22.406}{22.4} \approx 0.9993 \, \text{atm} \]

2. Calculate \( \frac{3.592}{(22.4)^2} \):
\[ (22.4)^2 = 501.76 \]
\[ \frac{3.592}{501.76} \approx 0.0071 \, \text{atm} \]

### Step 7: Final Calculation
Now substitute these results back into the equation for \( P \):
\[ P = 0.9993 - 0.0071 \]
\[ P \approx 0.9922 \, \text{atm} \]

### Conclusion
The pressure exerted by one mole of CO₂ gas at 273 K is approximately:
\[ P \approx 0.9922 \, \text{atm} \]

---