Question

Use the van der Waal's equation to determine the pressure of 2.00 moles of helium in a \(5.00 \mathrm{~L}\) balloon at \(300.00 \mathrm{~K}\). How does this value compare to what you would obtain with the ideal gas law?

Short answer

Van der Waals pressure is 9.938 atm, slightly higher than 9.852 atm from ideal gas law.

Step-by-step solution

Understand van der Waals Equation

The van der Waals equation is a modification of the ideal gas law that accounts for the volume occupied by gas particles and the attraction between them. It is given by:\[\left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT\]Where:- \(P\) is the pressure of the gas,- \(V\) is the volume,- \(n\) is the number of moles,- \(R\) is the ideal gas constant (0.0821 L atm/mol K),- \(T\) is the temperature in Kelvin,- \(a\) and \(b\) are van der Waals constants that are specific to each gas (for helium, \(a = 0.0341\, \mathrm{L^2 atm/mol^2}\) and \(b = 0.0237\, \mathrm{L/mol}\)).

Substitute the Known Values in van der Waals Equation

We need to substitute the given values into the van der Waals equation:- \(n = 2.00\, \mathrm{mol}\)- \(V = 5.00\, \mathrm{L}\)- \(T = 300.00\, \mathrm{K}\)- \(R = 0.0821\, \mathrm{L atm/mol K}\)- \(a = 0.0341\, \mathrm{L^2 atm/mol^2}\), \(b = 0.0237\, \mathrm{L/mol}\)The equation becomes:\[\left( P + \frac{0.0341 \times (2.00)^2}{(5.00)^2} \right) (5.00 - 2.00 \times 0.0237) = 2.00 \times 0.0821 \times 300.00\]

Simplify and Solve for Pressure

First, calculate the terms inside the equation:\[\frac{0.0341 \times 4}{25} = 0.005464\]\[5.00 - 2.00 \times 0.0237 = 4.9526\]Substitute back into the equation:\[\left( P + 0.005464 \right) (4.9526) = 49.26\]Solve for \(P\):\[P + 0.005464 = \frac{49.26}{4.9526}\]\[P + 0.005464 = 9.944\]\[P = 9.944 - 0.005464 \approx 9.938\, \mathrm{atm}\]

Calculate Pressure Using the Ideal Gas Law

Use the ideal gas law \( PV = nRT \) for calculation:\[P = \frac{nRT}{V} = \frac{2.00 \times 0.0821 \times 300.00}{5.00} = \frac{49.26}{5.00} = 9.852\, \mathrm{atm}\]

Compare the Results

Compare the pressures obtained from both methods:- Van der Waals Pressure: \(9.938\, \mathrm{atm}\)- Ideal Gas Pressure: \(9.852\, \mathrm{atm}\)The pressure calculated using the van der Waals equation is slightly higher due to adjustments for molecular interactions and volume.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to describe the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of moles of a gas sample. Mathematically, it is represented as: \[PV = nRT\]In this equation, - \(P\) stands for the pressure of the gas,- \(V\) is its volume,- \(n\) represents the number of moles,- \(R\) is the ideal gas constant, and- \(T\) is the temperature in Kelvin.In ideal conditions, gases are assumed to have perfectly elastic collisions and no intermolecular forces. Real gases, however, slightly deviate from this behavior. That’s where modifications like the van der Waals equation come in.
The Ideal Gas Law is simple and useful for predictions when gas behavior closely approximates ideal conditions. However, for substances like helium, especially under extreme conditions, adjustments are necessary.

Pressure Calculation

Pressure calculation in gases involves determining the force that the gas molecules exert on the walls of the container. It is commonly measured in atmospheres (atm), pascals (Pa), or torr. Real gases can have minor deviations in calculated pressure when using different equations due to interactions between molecules and intrinsic properties of the gas. In the exercise, pressure was calculated using both the Ideal Gas Law and the van der Waals equation.

• The Ideal Gas Law produced a pressure of \(9.852\, \mathrm{atm}\).
• The van der Waals equation provided a value of \(9.938\, \mathrm{atm}\).

The small difference highlights the importance of accounting for molecular size and intermolecular forces with the van der Waals modifications when working with real gases.

Helium Properties

Helium is a noble gas known for its stability and low reactivity. It is the second lightest and second most abundant element in the universe. Due to its small atomic size and weak intermolecular forces, helium typically behaves closely to an ideal gas under a wide range of conditions. However, small deviations can occur at high pressures or low temperatures.In terms of gas laws: - Helium has a specific van der Waals constant \(a = 0.0341\, \mathrm{L^2 \, atm/mol^2}\) that accounts for its weak interatomic attractions.- Its \(b\) constant, \(0.0237\, \mathrm{L/mol}\), corrects for the volume occupied by helium atoms themselves.These constants help in refining calculations to more accurately reflect the behavior of helium under non-ideal conditions.

Gas Constants

Gas constants are critical in equations that model the behavior of gases. The ideal gas constant \(R\) is universally accepted as \(0.0821\, \mathrm{L \, atm/mol \, K}\) for use in various gas law calculations. This constant serves as a bridge tying together the gas properties in different conditions.
In van der Waals and other adjusted gas laws, additional specific constants \(a\) and \(b\) are employed. These reflect unique characteristics of each gas, such as molecular size and the intensity of intermolecular forces. For helium:

• \(a = 0.0341\, \mathrm{L^2 \, atm/mol^2}\) considers intermolecular attractions.
• \(b = 0.0237\, \mathrm{L/mol}\) accounts for atomic volume.

These constants are essential in adjusting theoretical predictions to more accurately represent real-world measurements. Understanding these constants is crucial for precision in gas calculations and explaining deviations from ideal behavior.