Calculate the compressibility factor for `CO_(2)` if one mole of it occupies `0.4` litre at `300K` and `40 atm`. Comment on the result:

Options

A. `0.40, CO_(2)` is more compressible than ideal gas
B. `0.65, CO_(2)` is more compressible than ideal gas
C. `0.55, CO_(2)` is more compressible than ideal gas
D. `0.62, CO_(2)` is more compressible than ideal gas

Answer

To calculate the compressibility factor (Z) for CO₂, we will use the formula:

\[ Z = \frac{PV}{RT} \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify the given values:**
- Pressure, \( P = 40 \, \text{atm} \)
- Volume, \( V = 0.4 \, \text{L} \)
- Temperature, \( T = 300 \, \text{K} \)
- Ideal gas constant, \( R = 0.0821 \, \text{L·atm/(K·mol)} \)

2. **Substitute the values into the formula:**
\[ Z = \frac{PV}{RT} = \frac{(40 \, \text{atm}) \times (0.4 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (300 \, \text{K})} \]

3. **Calculate the numerator:**
\[ PV = 40 \times 0.4 = 16 \, \text{atm·L} \]

4. **Calculate the denominator:**
\[ RT = 0.0821 \times 300 = 24.63 \, \text{L·atm/(mol)} \]

5. **Divide the numerator by the denominator:**
\[ Z = \frac{16}{24.63} \approx 0.649 \]

6. **Round the result:**
\[ Z \approx 0.65 \]

### Conclusion:
The compressibility factor \( Z \) for CO₂ is approximately 0.65.

### Comment on the Result:
Since the compressibility factor \( Z \) is less than 1, it indicates that CO₂ is more compressible than an ideal gas under the given conditions. This suggests that CO₂ experiences stronger intermolecular forces compared to an ideal gas.