We need to determine the volume $V$ of 1 mole of $CO_2$ gas at $460 \, \text{K}$ and $100 \, \text{atm}$ using the van der Waals equation of state:

$P = \frac{nRT}{V - nb} - \frac{a n^2}{V^2}$

Given Data:

• $P = 100 \, \text{atm}$
• $T = 460 \, \text{K}$
• $n = 1 \, \text{mol}$ (1 mole of gas)
• $R = 82.06 \, \text{cm}^3 \cdot \text{atm} / \text{mol} \cdot \text{K}$
• $a = 3592000 \, \text{atm} \cdot \text{cm}^6 / \text{mol}^2$
• $b = 42.67 \, \text{cm}^3 / \text{mol}$

We will now substitute the known values into the van der Waals equation and solve for $V$. To simplify, the equation becomes:

$100 = \frac{1 \times 82.06 \times 460}{V - 42.67} - \frac{3592000 \times 1^2}{V^2}$

Rearranging this equation gives a cubic form in $V$, which we will solve numerically to find the largest real root. Let me now compute this.The largest real value of the volume $V$ is approximately $324 \, \text{cm}^3$.

Do you need any further details or have any questions? Here are 5 related questions for deeper understanding:

1. How does the ideal gas law compare to the van der Waals equation?
2. What would the volume be if the gas behaved ideally under the same conditions?
3. How does the parameter $a$ in the van der Waals equation account for molecular interactions?
4. How would the volume change if the temperature increased to 500 K?
5. What effect does the parameter $b$ have on the behavior of real gases?

Tip: The van der Waals equation modifies the ideal gas law by correcting for molecular volume and intermolecular forces.