Question

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(1.0000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

Short answer

Using the ideal gas law, the pressure exerted by the nitrogen gas is \(12.25 \: \text{atm}\). When calculated using the van der Waals equation, the pressure is \(12.10 \: \text{atm}\). The difference is due to the ideal gas law making simplifying assumptions, while the van der Waals equation accounts for intermolecular forces and particle volume, providing a more accurate representation of real gases.

Step-by-step solution

Gather information and formulas

First, let's gather the given information: - Number of moles (n) = 0.5 mol N₂ - Volume (V) = 1.0 L - Temperature (T) = 25.0 °C Next, we need the relevant formulas: - Ideal gas law: \(PV = nRT\) - Van der Waals equation: \( \left[P+a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT \) In order to use these formulas, we must remember the gas constant (R) with units that will match our given information. We will use R = 0.0821 L atm / (K mol). Additionally, we need to convert the temperature from Celsius to Kelvin. Lastly, we need the van der Waals constants (a and b) for N₂. For N₂, a = 1.390 L² atm / mol² and b = 0.0391 L / mol.

Convert temperature to Kelvin

To convert the temperature from Celsius to Kelvin, we add 273.15 to the given temperature. T = 25.0 °C + 273.15 = 298.15 K

Calculate pressure using the ideal gas law

Now we can plug in the given information into the ideal gas law equation and solve for pressure (P). \(PV = nRT\) \(P = \frac{nRT}{V}\) P = (0.5 mol)(0.0821 L atm / (K mol))(298.15 K) / (1.0 L) P = 12.25 atm Using the ideal gas law, the pressure exerted by the nitrogen gas is 12.25 atm.

Calculate pressure using the van der Waals equation

Now we will use the van der Waals equation to calculate the pressure: \( \left[P+a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT \) Plugging in the values, we have: \( \left[P+1.390\left(\frac{0.5}{1.0}\right)^2\right](1.0-(0.5)(0.0391)) = (0.5)(0.0821)(298.15) \) After solving and isolating P, we get: P = 12.10 atm Using the van der Waals equation, the pressure exerted by the nitrogen gas is 12.10 atm.

Compare the results

Now, we'll compare the pressures obtained from both methods: Ideal gas law: P = 12.25 atm Van der Waals equation: P = 12.10 atm The pressure calculated using the ideal gas law is slightly higher than the pressure calculated using the van der Waals equation. This is because the ideal gas law assumes no intermolecular forces or volume occupied by the gas particles, whereas the van der Waals equation accounts for these factors. In reality, the pressure of the nitrogen gas would be closer to the value obtained using the van der Waals equation since it provides a more accurate representation of real gases.

Key concepts

Ideal Gas Law

The ideal gas law is a crucial formula in understanding how gases behave under certain conditions of pressure, volume, and temperature. It's mathematically expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) is volume, \( n \) indicates the amount of substance in moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

This equation provides a simplified model by assuming that the gas particles are point particles with no volume and that they exert no force on each other except during elastic collisions. However, the ideal gas law doesn't take into account the size of gas molecules or intermolecular forces, which can lead to inaccuracies when applied to real gases, especially under high pressure or low temperature.

Nevertheless, the ideal gas law serves as an essential first step in understanding more complex behavior in gases, and for many conditions, it's a pretty accurate way to calculate pressure, volume or temperature for a gas.

Gas Constant

The gas constant, denoted as \( R \), is a fundamental parameter in the ideal gas equation. Its value of 0.0821 L atm / (K mol) is used when working with pressure in atmospheres, volume in liters, and quantity in moles.

This constant essentially encapsulates the proportional relationship between the product of pressure and volume, and the product of the mole number and temperature for an ideal gas. It provides a bridge between macroscopic properties of gases and the amount of substance, connecting various units involved in gas calculations.

Remembering the correct value and units for the gas constant is essential, as using the wrong units can lead to incorrect results. The numeric value of \( R \) changes based on the system of units employed in the calculation, making it adaptable to different scientific and engineering contexts.

Intermolecular Forces

Intermolecular forces are the attractive and repulsive forces that occur between molecules. They play a significant role in determining the physical properties of substances, including gases. While the ideal gas law assumes no intermolecular forces, real gases exhibit these forces, which can impact their behavior.

There are several types of intermolecular forces, including dispersion forces, dipole-dipole interactions, and hydrogen bonds. These forces influence how molecules are held together and affect properties such as boiling points, melting points, and vapor pressures.

The presence of intermolecular forces can cause the pressure of a real gas to be lower than what would be predicted by the ideal gas law, as molecules attract each other and reduce the impact they have on the container walls. Understanding intermolecular forces is essential for predicting and explaining the behavior of real gases.

Real Gases

Real gases are gases that do not perfectly follow the ideal gas law due to intermolecular forces and the finite volume occupied by the gas molecules themselves. Under many conditions, real gases behave similarly to ideal gases, but at high pressures and low temperatures, the differences become more noticeable.

The van der Waals equation is a refined version of the ideal gas law that takes these deviations into account. It corrects for the volume occupied by gas molecules (represented by \( b \)) and the attraction between them (represented by \( a \)). The equation is expressed as \( \[P+a\left(\frac{n}{V}\right)^2\] (V-nb) = nRT \).

By using the van der Waals equation, we get a more accurate pressure for real gases as seen in the exercise provided, leading to the conclusion that real gases have slightly less pressure than ideal gases in the same conditions, due to their real-world interactions.