Question

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a 1.0000-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 Nitrogen gas is approximately 12.28 atm, while using the Van der Waals equation, it is approximately 12.15 atm. The slightly higher value obtained from the Ideal Gas Law is due to it not accounting for the finite size of particles and the intermolecular forces which are considered in the Van der Waals equation. However, the difference in pressure calculated by the two equations is relatively small in this case.

Step-by-step solution

Write down the given information

We are given the following information: - Number of moles (n) = 0.5000 mol - Volume (V) = 1.0000 L - Temperature (T) = 25.0 °C First, let's convert the temperature from °C to Kelvin. T(K) = T(°C) + 273.15 T(K) = 25.0 + 273.15 T(K) = 298.15 K

Calculate pressure using Ideal Gas Law

The Ideal Gas Law can be represented as: \( PV = nRT \) Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant, which is approximately 0.0821 L·atm/(mol·K) for this problem - T is the temperature in Kelvin We have the values for n, V, and T, so we can now solve for P: \( P = \frac{nRT}{V} \) Substitute the given values: \( P = \frac{(0.5000 \,\text{mol})(0.0821 \,\text{L·atm/(mol·K)})(298.15 \,\text{K})}{1.0000 \,\text{L}} \) Calculate the pressure: P ≈ 12.28 atm Using the Ideal Gas Law, the pressure exerted by Nitrogen gas is approximately 12.28 atm.

Calculate pressure using Van der Waals Equation

The Van der Waals equation is a modified form of the Ideal Gas Law that accounts for the finite size of molecules and the attractive forces between them. The equation is: \( \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \) Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant - T is the temperature in Kelvin - a and b are the Van der Waals constants specific to the gas, for N₂: a = 1.390 L²·atm/mol² and b = 0.0391 L/mol We have the values for n, V, T, a, and b, so we can now solve for P: First, let's solve the left side of the equation: \( V - nb = 1.0000 \,\text{L} - (0.5000 \,\text{mol})(0.0391 \,\text{L/mol}) \) \( V - nb = 1.0000 - 0.01955 \) \( V - nb = 0.98045 \,\text{L} \) Now, let's solve the right side: \( \frac{an^2}{V^2} = \frac{(1.390 \,\text{L²·atm/mol²})(0.5000 \,\text{mol})^2}{(1.0000 \,\text{L})^2} \) \( \frac{an^2}{V^2} = 0.3475 \,\text{atm} \) Now, we can re-write the Van der Waals equation as: \( \left(P + 0.3475 \,\text{atm}\right)(0.98045 \,\text{L}) = (0.5000 \,\text{mol})(0.0821 \,\text{L·atm/(mol·K)})(298.15 \,\text{K}) \) Let \(x = P + 0.3475 \,\text{atm}\) Now, the equation is: \( x(0.98045 \,\text{L}) = 12.2489 \,\text{L·atm} \) Now, solve for x: \( x = \frac{12.2489 \,\text{L·atm}}{0.98045 \,\text{L}} \) \( x ≈ 12.50 \,\text{atm} \) Finally, calculate P: \( P = x - 0.3475 \,\text{atm} \) P ≈ 12.15 atm Using the Van der Waals equation, the pressure exerted by Nitrogen gas is approximately 12.15 atm.

Compare the results

- Using the Ideal Gas Law, the pressure exerted by Nitrogen gas is approximately 12.28 atm. - Using the Van der Waals equation, the pressure exerted by Nitrogen gas is approximately 12.15 atm. The pressure calculated using the Ideal Gas Law is slightly higher than the pressure calculated using the Van der Waals equation. This difference can be attributed to the fact that the Ideal Gas Law doesn't account for the finite size of the particles and the intermolecular forces between them, which the Van der Waals equation does. In this case, however, the difference in the pressure calculated by the two equations is relatively small.

Key concepts

Van der Waals equation

The Van der Waals equation is a more realistic approach for calculating the behavior of gases compared to the Ideal Gas Law. It accounts for the fact that gas molecules have a finite size and experience intermolecular forces. While the Ideal Gas Law assumes molecules are point particles with no attraction to each other, the Van der Waals equation introduces two corrective factors to address these limitations.

• a: A constant representing the strength of the attraction between molecules, which leads to a reduction in pressure.
• b: A constant accounting for the volume occupied by the gas molecules themselves, effectively reducing the volume in which the molecules can move.

The Van der Waals equation is expressed as: \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]Where:- P is the pressure of the gas.- V is the volume available to the gas molecules.- n is the number of moles.- R is the ideal gas constant.- T is the temperature in Kelvin.- a and b are Van der Waals constants specific to each gas.Students often see the need for the Van der Waals equation when the Ideal Gas Law fails at high pressures or low temperatures, highlighting its practical importance.

Gas constant

The gas constant, often denoted as R, is a key figure in several equations dealing with gases, including the Ideal Gas Law and the Van der Waals equation. It bridges the units of energy, temperature, and amount of substance, making these equations applicable in different scenarios. For common calculations involving gases at standard conditions, the value of R is typically given as 0.0821 L·atm/(mol·K).

This value can change based on the units used in a problem. For instance:

• 8.314 J/(mol·K) when using SI units, linking energy in joules.
• 62.3637 L·mmHg/(mol·K) for when pressure is measured in millimeters of mercury.

Understanding the gas constant is crucial for correctly using gas laws, as using the right units ensures consistency in calculations and results. Students should be careful to match the value of R with the specific units used for pressure, volume, and temperature in their calculations.

Moles calculation

Moles are a fundamental unit in chemistry used to measure the amount of substance. Knowing the number of moles is essential when applying gas laws, such as the Ideal Gas Law and Van der Waals equation, because these laws require the amount of substance to predict the behavior of gases.

The calculation of moles involves:

• Using the molar mass: To convert between grams and moles, you divide the mass of the substance by its molar mass (g/mol).
• Finding relationships: In reactions or gas behaviors, utilizing stoichiometric relationships to find moles when given different properties.

In the context of our exercise, we had a known value of 0.5000 mol of Nitrogen (\(N_2\)), which we used directly in calculations. Determining the number of moles accurately is essential for ensuring other calculations, such as pressures and volumes, are correct and representative of real conditions.

Temperature conversion

Temperature conversion is often a necessary step in gas calculations because many formulas require temperature to be in Kelvin, the absolute temperature scale. The Ideal Gas Law and Van der Waals equation both rely on absolute temperature to ensure accurate results.

The conversion from Celsius to Kelvin is simple: add 273.15 to the Celsius temperature. For example, in our exercise,
- 25.0°C is converted as follows:
\[T(K) = 25.0 + 273.15 = 298.15 \, \text{K}\]

• This conversion is essential because the kinetic energy of gas particles, which influences other properties like volume and pressure, is directly proportional to absolute temperature.
• Using Kelvin ensures no mathematical issues arise, such as division by zero, which can occur if Celsius or Fahrenheit scales were improperly used.

Being adept in temperature conversion is crucial for any student tackling gas law problems, as it affects the accuracy and feasibility of their calculations.