Question

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is \(1050 \mathrm{mph}\). (a) What is the averagespeed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of 1 mol of \(\mathrm{N}_{2}\) molecules moving at this speed?

Short answer

The average speed of N2 molecules at $20^{\circ} \mathrm{C}$ is approximately 1050 mph, which is equivalent to 469.5 m/s. The kinetic energy of a single N2 molecule moving at this speed is approximately \(3.89 \times 10^{-21} \,\text{J}\). The total kinetic energy of 1 mole of N2 molecules moving at this speed is approximately 2339 J.

Step-by-step solution

Convert speed from mph to m/s

First, we need to convert the given average speed from miles per hour (mph) to meters per second (m/s). To do this, we can use the conversion factors: 1 mile = 1609 meters, and 1 hour = 3600 seconds. Given speed = 1050 mph To convert to m/s, multiply by the necessary conversion factors: \( \text{Speed in m/s} = 1050 \,\text{mph} \times \frac{1609 \,\text{m}}{1 \,\text{mile}} \times \frac{1 \,\text{hour}}{3600 \,\text{s}} \)

Calculate the kinetic energy of a single N2 molecule

Now, we need to calculate the kinetic energy of a single N2 molecule moving at the converted speed. The formula for kinetic energy is: \( KE = \frac{1}{2}mv^2 \) where KE is the kinetic energy, m is the mass of the molecule, and v is the velocity. The mass, m, of one N2 molecule can be calculated as follows: - 1 mole of N2 = 28 grams = 0.028 kg - 1 mole of any substance contains Avogadro's number of particles: \( N_A = 6.022 \times 10^{23} \,\text{particles/mol} \) - Therefore, the mass of a single N2 molecule is \( m = \frac{0.028 \,\text{kg}}{6.022 \times 10^{23}\,\text{particles}} \) Now, calculate the kinetic energy: \( KE = \frac{1}{2}m \times (\text{Speed in m/s})^2 \)

Calculate the total kinetic energy of 1 mole of N2 molecules

To find the total kinetic energy of 1 mole of N2 molecules, multiply the kinetic energy of a single molecule (found in step 2) by Avogadro's number: \( \text{Total KE} = KE \times N_A \) Following these steps and plugging in the necessary values will give you the kinetic energies for both a single N2 molecule and one mole of N2 molecules.

Key concepts

Molecular Velocity Conversion

Understanding how to convert molecular velocity between different units of measurement is essential when studying gases. In our textbook example, the velocity of nitrogen (\text{N}_2) molecules is given in miles per hour (mph), but for many scientific calculations, you need the velocity in meters per second (m/s). To convert mph to m/s, you use the conversion factors where 1 mile equals 1609 meters and 1 hour is equivalent to 3600 seconds.

The basic formula for conversion is:
\[ \text{Speed in m/s} = \text{Speed in mph} \times \frac{1609 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ s}} \]
By applying this formula to the given speed of 1050 mph, students can effectively transition between these units and proceed with further calculations in the metric system commonly used in physics and chemistry.

Kinetic Energy Formula

Kinetic energy (KE) represents the energy an object possesses due to its motion. For gas molecules, this movement is particularly important as it relates to temperature and pressure. The formula for calculating the kinetic energy of a particle is:
\[ KE = \frac{1}{2}mv^2 \]

In this equation, \(m\) denotes the mass of the particle, and \(v\) represents its velocity. It's crucial to plug in the right units into the formula. For instance, mass should be in kilograms (kg) and velocity in meters per second (m/s). Calculating a single \text{N}_2 molecule's kinetic energy involves using its mass and the previously converted velocity. By breaking down these concepts and ensuring the units match, students can comprehend the association between a gas molecule's speed and its kinetic energy.

Avogadro's Number Application

Avogadro's number (~ \(6.022 \times 10^{23}\)) is a constant that represents the number of particles in one mole of a substance, a vital concept in chemistry for relating macroscopic amounts to microscopic entities. When applying Avogadro's number to our textbook problem, we can calculate the total kinetic energy for one mole of \text{N}_2 molecules.

The total kinetic energy is found by taking the kinetic energy of a single molecule, obtained with the kinetic energy formula, and multiplying it by Avogadro's number:
\[ \text{Total KE} = KE \times N_A \]
By understanding this step, students learn how to scale up from the microscopic scale (individual particles) to the macroscopic scale (moles), which is a repeatedly used process in chemical calculations.