Question

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a \(10.000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 107 .

Short answer

a. Using the Ideal Gas Law, the calculated pressure is \(P = 124344.7845 \mathrm{~Pa}\). b. Using the Van der Waals equation, the calculated pressure is \(P_{vdW} = 127284.4734 \mathrm{~Pa}\). c. The Ideal Gas Law gives a slightly lower value compared to the Van der Waals equation, as the Ideal Gas Law does not consider the weak forces between particles and their actual size. d. The comparison with Exercise 107 cannot be made due to lack of context or meaning.

Step-by-step solution

Convert Temperature and Volume to SI units

Convert the given temperature from Celsius to Kelvin and the volume from liters to cubic meters: Temperature T = 25.0°C + 273.15 = 298.15 K Volume V = 10.0 L × (1 m³ / 1000 L) = 0.0100 m³

Determine the Gas Constant (R) value

We will use the gas constant R = 8.314 J/(mol.K)

Calculate the Pressure using the Ideal Gas Law

Using the ideal gas law formula PV=nRT, we can calculate the pressure: P = nRT/V P = (0.5 mol × 8.314 J/(mol.K) × 298.15 K) / 0.0100 m³ P = 124344.7845 Pa b. Using the Van der Waals equation

Choose Van der Waals Constants (a, b) for Nitrogen

We need to know the a and b constants for nitrogen to use in the van der Waals equation. These constants are typically given: a = 1.39 \(\frac{\text{Pa} \cdot \text{m}^6}{\text{mol}^2}\) b = 3.87 × 10⁻² \(\frac{\text{m}^3}{\text{mol}}\)

Calculate the Pressure using the Van der Waals equation

Using the Van der Waals equation is a more accurate representation of gas behavior because it accounts for the size of the particles and their interactions. The equation is given by: \( \Bigg[P+\frac{an^2}{V^2}\Bigg]\Bigg(V-{nb}\Bigg) = nRT \) Rearranging the equation for P, we can solve for P_vdW: \( P_{vdW} = \frac{nRT}{V-nb} - \frac{an^2}{V^2} \) Plugging in the values, we calculate the pressure using van der Waals equation: \( P_{vdW} = \frac{0.5 \cdot 8.314 \cdot 298.15}{0.0100 - (0.5)(3.87 \times 10^{-2})} - \frac{1.39 \cdot 0.5^2}{0.0100^2} \) \( P_{vdW} = 127284.4734 \text{ Pa} \) c. Compare the results From Ideal Gas Law: P = 124344.7845 Pa From Van der Waals equation: \( P_{vdW} \) = 127284.4734 Pa Upon comparison, the Ideal Gas Law gives a slightly lower value compared to the Van der Waals equation, which is due to the Ideal Gas Law not considering the weak forces between the particles and their actual size. d. Compare with Exercise 107 This question cannot be answered as Exercise 107 lacks context or meaning.

Key concepts

Ideal Gas Law Calculation

The ideal gas law is an equation of state for a hypothetical ideal gas. It is a good approximation for the behavior of real gases under many conditions, although it has its limitations. It is represented by the equation PV=nRT, where P is the pressure, V is the volume, T is the absolute temperature, n is the amount of substance of gas (in moles), and R is the ideal gas constant.

The solution for calculating the pressure exerted by nitrogen in the problem provided begins with converting the temperature to Kelvin and the volume to cubic meters, which are the SI units compatible with the gas constant in the ideal gas equation. Then, by rearranging the equation to solve for P (pressure), the calculation is straightforward. This law assumes that the gas particles have no volume and do not interact with each other, which is not entirely true for real gases, but works well for many scenarios especially at high temperatures and low pressures.

It's important for students to grasp not only the calculation but also the underlying assumptions and conditions where the ideal gas law applies accurately. This provides a foundation to understand why other more complex equations, like the van der Waals equation, may sometimes be necessary.

Van der Waals Equation Application

The van der Waals equation provides a more accurate description for the behavior of real gases compared to the ideal gas law, especially at high pressures and low temperatures. It adjusts the ideal gas law to account for the volume occupied by gas molecules and the attractive forces between them. The equation is \( \Bigg[P+\frac{an^2}{V^2}\Bigg]\Bigg(V-{nb}\Bigg) = nRT \), where 'a' and 'b' are substance-specific constants which represent the magnitude of intermolecular forces and the volume occupied by the gas molecules, respectively.

In the step-by-step solution, we first need to introduce the constants for nitrogen. The calculations require careful attention to how the equation is manipulated to solve for pressure (P), specifically remembering to include the correction factors for the volume and pressure terms. Understanding the van der Waals equation helps students account for non-ideal behavior in gases and can provide a deeper insight into factors affecting gas pressure in confined systems.

Due to its complexity, the van der Waals equation is best applied in cases where the ideal gas law deviates significantly from observed behavior, and understanding its application reinforces concepts of molecular interaction in gases.

Comparative Gas Pressure Analysis

Comparing the results of the ideal gas law and the van der Waals equation from the exercise reveals that the pressure calculated using the ideal gas law is slightly lower. This difference arises because the ideal gas law does not account for the finite size of gas particles or the intermolecular forces between them, which the van der Waals equation does. Comparative analysis of gas pressures using different gas laws is critical in developing a complete understanding of the behavior of gases under various conditions.

Students should recognize that the ideal gas law serves as a simplified model for gas prediction, useful for many situations but not all. The ability to compare and understand deviations from ideality is pivotal in chemistry, especially in cases requiring high precision, such as in industrial applications involving gases. Analyzing the difference in predicted pressure values helps to appreciate the complexity of real-world gas behavior and the importance of choosing the correct equation for the situation at hand.

While comparative analysis is a powerful tool, it is also crucial to have context, such as Exercise 107 in the problem set, to establish a reference or benchmark for comparison. Without context, comparative analysis loses its significance, as it becomes challenging to gauge the magnitude and implications of the observed differences.