Consider a container that contains \(1.00\) mole of \(\mathrm{CO}_{2}(g)\) at \(298 \mathrm{~K}\). (a) What does the ideal gas law predict the pressure to be in atm? (b) What does the van der Waals equation predict the pressure to be? (c) What is the percent difference of the van der Waals pressure from the ideal pressure? (d) Suppose you increased the temperature to \(1000 \mathrm{~K}\). Would you expect the percent difference to increase or decrease compared to your answer in (c)? Explain.

The Ideal Gas Law formula is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We are given n = 1.00 mole, R = 0.0821 L atm K⁻¹ mol⁻¹ (from the units of R, the pressure will be in atm), and T = 298 K. In order to find the pressure, we need the volume of the container. We can find this information from the van der Waals equation, which is given in part (b) of the question.

The van der Waals equation for CO2 gas is: \((P + a\frac{n^2}{V^2})(V-nb) = nRT\), where a and b are the van der Waals constants for CO2. We are given a = 3.59 L² atm mol⁻² and b = 0.0427 L mol⁻¹. Using the given values, we can solve for the volume V. When the van der Waals equation is applied to the Ideal Gas Law to find the volume, we can consider it as: \(V_\text{ideal}-nb = V_\text{real}\) \(V_\text{ideal} = V_\text{real}+ nb\) In this case, the van der Waals equation for CO2 becomes: \((P + \frac{an^2}{V_\text{ideal}})(V_\text{real}) = nRT\) Since we are talking about 1 mole of CO2 at 298K and 1 atm, we can assume that the gas behaves like an ideal gas. In that case, V_ideal = V_real and the equation becomes: \(V = \frac{nRT}{P}\) Now we have the volume, V, which can be used in the Ideal Gas Law equation in Step 1.

Now we can plug in the values for n, R, and T, to find the pressure: \(P = \frac{nRT}{V} = \frac{(1.00 \, \text{mol})(0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1})(298 \, \text{K})}{V}\) Let this pressure be denoted as P_ideal.

Now we will use the van der Waals equation with the values of a, b, n, R, T, and V to find the pressure, P_vdW: \((P_\text{vdW} + a\frac{n^2}{V^2})(V-nb) = nRT\) Let this pressure be denoted as P_vdW.

Now we can calculate the percent difference between P_ideal and P_vdW: \(\text{Percent Difference} = \frac{| P_\text{vdW} - P_\text{ideal} |}{P_\text{ideal}} \times 100\%\)

To answer part (d), we need to understand the behavior of gases at high temperatures. When the temperature increases, the gas molecules have more kinetic energy and move faster. This has two effects: 1. Increased collisions between gas molecules and the container walls, leading to a higher pressure that can be approximated by the Ideal Gas Law. 2. The intermolecular forces between gas molecules, represented by the van der Waals equation, become less significant due to molecular vibrations at high temperatures. We need to determine if the percent difference between the Ideal Gas Law pressure and the van der Waals pressure increases or decreases as the temperature increases.