Calculate the pressure exerted by one mole `CO_2` ​of gas at 273 K, if the van der Waal's constat "a"=3.502`dm^6atm^(-2)` . Assume that the volume occupied by `CO_2` molecules is negligible.

Options

A. 1.99
B. 0.993
C. 99.3
D. 9.93

Answer

To calculate the pressure exerted by one mole of \( CO_2 \) gas at 273 K using the van der Waals equation, we can follow these steps:

### Step 1: Write down the van der Waals equation for one mole of gas.
The van der Waals equation is given by:

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

For one mole of gas, we can simplify this to:

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

### Step 2: Identify the values needed for the calculation.
- **R** (Ideal gas constant) = 0.0821 atm L K\(^{-1}\) mol\(^{-1}\)
- **T** (Temperature) = 273 K
- **a** (van der Waals constant for \( CO_2 \)) = 3.502 dm\(^6\) atm\(^{-2}\)
- **b** (volume occupied by gas molecules) is negligible, so we can assume \( b = 0 \).
- **V** (Volume at standard temperature and pressure, STP) = 22.4 L

### Step 3: Substitute the values into the equation.
Using the values identified:

\[ P = \frac{(0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1})(273 \, \text{K})}{22.4 \, \text{L}} - \frac{3.502 \, \text{dm}^6 \text{atm}^{-2}}{(22.4 \, \text{L})^2} \]

### Step 4: Calculate the first term.
Calculating the first term:

\[ \frac{(0.0821)(273)}{22.4} \approx \frac{22.4133}{22.4} \approx 1.0006 \, \text{atm} \]

### Step 5: Calculate the second term.
Calculating the second term:

\[ \frac{3.502}{(22.4)^2} = \frac{3.502}{501.76} \approx 0.006975 \, \text{atm} \]

### Step 6: Combine the results to find the pressure.
Now substitute these results back into the equation for pressure:

\[ P = 1.0006 \, \text{atm} - 0.006975 \, \text{atm} \approx 0.9936 \, \text{atm} \]

### Step 7: Round the final answer.
Rounding to three significant figures, we get:

\[ P \approx 0.994 \, \text{atm} \]

### Conclusion
The pressure exerted by one mole of \( CO_2 \) gas at 273 K is approximately **0.994 atm**.