Question

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.

Short answer

Based on the given information and calculations, we find that: (a) The ideal pressure is calculated using the Ideal Gas Law and the provided values. (b) The van der Waals pressure is determined using the van der Waals equation and known constants for CO2. (c) The percent difference between the two pressures is calculated using the formula for percent difference. (d) As temperature increases, the intermolecular forces become less significant due to molecular vibrations, which leads to a decrease in the percent difference between the Ideal Gas Law pressure and the van der Waals pressure.

Step-by-step solution

Use the Ideal Gas Law to find the pressure

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.

Find the volume, using the van der Waals equation

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.

Calculate the pressure using the Ideal Gas Law

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.

Calculate the pressure using the van der Waals equation

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.

Calculate the percent difference between the two pressures

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

Analyze the effect of increasing the temperature

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.

Key concepts

van der Waals equation

The van der Waals equation is a modification of the ideal gas law that accounts for the size of gas molecules and intermolecular forces. This equation is particularly important for describing the behavior of real gases, especially at high pressures and low temperatures, where gas molecules interact more significantly with each other. The van der Waals equation is written as: \[(P + a\frac{n^2}{V^2})(V-nb) = nRT\]Where:

• \(P\) is the pressure of the gas.
• \(n\) is the number of moles of gas.
• \(V\) is the volume.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(a\) and \(b\) are the van der Waals constants for a specific gas, which correct for intermolecular attractions and the volume occupied by gas molecules, respectively.

Using the specified constants provided in the exercise (\(a = 3.59 \, \text{L}^2 \, \text{atm} \, \text{mol}^{-2}\) and \(b = 0.0427 \, \text{L} \, \text{mol}^{-1}\)), we can calculate the real pressure of carbon dioxide to understand its deviation from ideal behavior in real-world conditions.

percent difference

Calculating the percent difference between two values is a useful way of understanding how much they differ in relative terms. In chemistry, we often use this calculation to compare theoretical predictions with experimental or more refined calculations. In the context of the exercise, we are comparing the pressures calculated from the ideal gas law and the van der Waals equation. The formula for percent difference is given by:\[\text{Percent Difference} = \frac{| P_{\text{vdW}} - P_{\text{ideal}} |}{P_{\text{ideal}}} \times 100\%\]Where:

• \(P_{\text{vdW}}\) is the pressure calculated using the van der Waals equation.
• \(P_{\text{ideal}}\) is the pressure calculated assuming the gas behaves as an ideal gas.

A low percent difference suggests that the ideal gas law is a reasonable approximation for the gas under the given conditions. Conversely, a high percent difference indicates that real gas behavior significantly deviates from the ideal assumption, and the van der Waals equation provides a better description.

pressure calculation

Pressure calculations involve understanding the relationships between gas properties such as volume, temperature, and the number of molecules present. When dealing with gases, we often rely on the ideal gas law: \[ PV = nRT \]This equation allows us to predict the pressure \(P\) given the volume \(V\), temperature \(T\), and number of moles \(n\). The constant \(R\) is the universal gas constant, typically valued at \(0.0821 \, \text{L atm K}^{-1} \, \text{mol}^{-1}\).Often in problem-solving, it's necessary to find the volume using other equations such as the van der Waals equation or directly compute pressure values. In our exercise, you needed to find the volume using the corrected van der Waals equation first before applying the ideal gas law. This shows how understanding corrections for real gases helps improve initial calculations. Having both the ideal and van der Waals pressures, calculating the percent difference further characterizes how the real gas behavior varies from ideal predictions.

ideal gas behavior

Ideal gas behavior is a hypothetical scenario where gases perfectly follow the simple relationship given by the ideal gas law. This law assumes that:

• Gas molecules do not attract or repel each other.
• Gas molecules occupy no volume.
• Collisions between gas molecules and container walls are perfectly elastic.

While no gas truly behaves ideally, many gases approximate this behavior well, especially under conditions of low pressure and high temperature. At higher temperatures, gases behave more like ideal gases because:- The kinetic energy surpasses intermolecular forces.- Molecular collisions occur with greater frequency but are less impacted by intermolecular forces.These principles were highlighted in the exercise when discussing the potential changes at \(1000 \, \text{K}\). Understanding when gases deviate from ideality helps in predicting their behavior more accurately and choosing when models like the van der Waals provide necessary corrections.