Calculate the pressure exerted by 0.250 moles of carbon dioxide in 0.275 litres at `100^@C` and compare this value with that expected for an ideal gas.
(Given : `a = 3.59 L^2 " atm " mol^(-2), b = 0.0427 L mol^(-1)`)

Explanation

To calculate the pressure exerted by 0.250 moles of carbon dioxide in 0.275 liters at 100°C using the Van der Waals equation, we will follow these steps:

### Step-by-Step Solution

1. **Convert Temperature to Kelvin**:
The temperature in Celsius is given as 100°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 100 + 273.15 = 373.15 \, K \]

2. **Identify Given Values**:
- Number of moles (n) = 0.250 moles
- Volume (V) = 0.275 liters
- Van der Waals constants:
- \( a = 3.59 \, L^2 \, atm \, mol^{-2} \)
- \( b = 0.0427 \, L \, mol^{-1} \)
- Gas constant (R) = 0.0821 \( L \, atm \, K^{-1} \, mol^{-1} \)

3. **Use the Van der Waals Equation**:
The Van der Waals equation is given by:
\[ \left(P + \frac{a n^2}{V^2}\right)(V - n b) = nRT \]
Rearranging for pressure (P):
\[ P = \frac{nRT}{V - n b} - \frac{a n^2}{V^2} \]

4. **Calculate \( V - n b \)**:
First, calculate \( n b \):
\[ n b = 0.250 \, mol \times 0.0427 \, L \, mol^{-1} = 0.010675 \, L \]
Now calculate \( V - n b \):
\[ V - n b = 0.275 \, L - 0.010675 \, L = 0.264325 \, L \]

5. **Calculate \( \frac{a n^2}{V^2} \)**:
First, calculate \( n^2 \):
\[ n^2 = (0.250)^2 = 0.0625 \, mol^2 \]
Now calculate \( V^2 \):
\[ V^2 = (0.275)^2 = 0.075625 \, L^2 \]
Now calculate \( \frac{a n^2}{V^2} \):
\[ \frac{a n^2}{V^2} = \frac{3.59 \, L^2 \, atm \, mol^{-2} \times 0.0625 \, mol^2}{0.075625 \, L^2} = \frac{0.224375 \, atm}{0.075625} \approx 2.96 \, atm \]

6. **Calculate Pressure (P)**:
Now substitute the values into the pressure equation:
\[ P = \frac{0.250 \, mol \times 0.0821 \, L \, atm \, K^{-1} \, mol^{-1} \times 373.15 \, K}{0.264325 \, L} - 2.96 \, atm \]
Calculate \( nRT \):
\[ nRT = 0.250 \times 0.0821 \times 373.15 \approx 7.655 \, L \, atm \]
Now calculate \( P \):
\[ P = \frac{7.655 \, L \, atm}{0.264325 \, L} - 2.96 \, atm \approx 28.96 \, atm - 2.96 \, atm \approx 26.00 \, atm \]

7. **Compare with Ideal Gas Pressure**:
For an ideal gas, we use the ideal gas equation \( PV = nRT \):
\[ P = \frac{nRT}{V} = \frac{0.250 \times 0.0821 \times 373.15}{0.275} \]
Calculate:
\[ P \approx \frac{7.655}{0.275} \approx 27.86 \, atm \]

### Final Results
- Pressure exerted by CO2 (real gas) = 26.00 atm
- Pressure expected for ideal gas = 27.86 atm