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 average speed 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 \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) molecules moving at this speed?

Short answer

In summary, the average speed of N₂ molecules at room temperature is 469.034 m/s, the kinetic energy of one N₂ molecule moving at this speed is approximately \(1.33 \times 10^{-22} J\), and the total kinetic energy of one mol of N₂ molecules moving at this speed is approximately 23605.40 J.

Step-by-step solution

Convert mph to m/s

To convert the speed from miles per hour (mph) to meters per second (m/s), we'll use the following conversion factors: 1 mile = 1609.34 meters 1 hour = 3600 seconds Now, let's convert 1050 mph to m/s: \(1050 \frac{\text{miles}}{\text{hour}} \cdot \frac{1609.34 \text{ meters}}{1 \text{ mile}} \cdot \frac{1 \text{ hour}}{3600 \text{ seconds}}\)

Simplify the expression

Now, we can simplify the expression by canceling the units and calculating the numerical value: \(1050 \cdot \frac{1609.34}{3600} \frac{\text{meters}}{\text{seconds}} = 469.034 \frac{\text{meters}}{\text{seconds}}\) So, the average speed in meters per second is approximately 469.034 m/s. Next, let's calculate the kinetic energy of one N₂ molecule at this speed.

Calculate the kinetic energy

First, we need to find the mass of one N₂ molecule. The molecular mass of N₂ is 28.02 g/mol, and since there are \(6.022 \times 10^{23}\) molecules in a mol (Avogadro's number), the mass of one molecule can be obtained as follows: \(m = \frac{28.02 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}}\). Now, we need to convert the mass from grams to kilograms by dividing by 1000: \(m = \frac{28.02}{6.022 \times 10^{23} \times 1000} \text{ kg}\) Now, we can use the kinetic energy formula to find the kinetic energy of one N₂ molecule: \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(\frac{28.02}{6.022 \times 10^{23} \times 1000}) \cdot (469.034)^2\) Now let's find the total kinetic energy for one mol of N₂ molecules.

Calculate the total kinetic energy

We are given that there are \(6.022 \times 10^{23}\) molecules in one mol. So, to find the total kinetic energy, we will multiply the kinetic energy of one molecule by the number of molecules in one mol: \(Total \, KE = KE \times 6.022 \times 10^{23}\) \(Total \, KE = (\frac{1}{2}(\frac{28.02}{6.022 \times 10^{23} \times 1000}) \cdot (469.034)^2) \times 6.022 \times 10^{23}\) After calculating, we get: \(Total \, KE \approx 23605.40 J\) In summary, the average speed of N₂ molecules at room temperature is 469.034 m/s, the kinetic energy of one N₂ molecule moving at this speed is approximately \(1.33 \times 10^{-22} J\), and the total kinetic energy of one mol of N₂ molecules moving at this speed is approximately 23605.40 J.

Key concepts

Average Velocity

Average velocity is a term used to describe the speed of an object in a specific direction over a period of time. In this case, we're talking about the average velocity of nitrogen gas (\( ext{N}_2 \)) molecules at room temperature, which is initially given as 1050 miles per hour (mph).
This velocity is important because it gives us an idea of how fast the molecules are moving around in a gas under standard conditions.
To understand average velocity in the context of molecular motion, it's crucial to translate this speed from a macroscopic unit (mph) to a more scientific one (meters per second or m/s). This conversion allows us to make further calculations, such as kinetic energy, using standard scientific notation.
The conversion process involves reducing distances from miles to meters (1 mile = 1609.34 meters) and time from hours to seconds (1 hour = 3600 seconds). By applying these conversions, we find that the average velocity of \( ext{N}_2 \) molecules is approximately 469.034 m/s. This precise measurement helps in quantifying molecular behavior in scientific studies.

Molecular Mass

Molecular mass is the mass of a single molecule of a substance. For \( ext{N}_2 \) (nitrogen gas), the molecular mass is 28.02 grams per mole (g/mol).
Knowing the molecular mass is essential to calculate various properties, such as the kinetic energy of molecules.
When dealing with individual molecules rather than moles, we must convert the molecular mass into kilograms (the SI unit for mass), since energy calculations require mass to be in kilograms. Given that there are \( 6.022 \times 10^{23} \) molecules per mole (Avogadro’s number), the mass of a single \( ext{N}_2 \) molecule can be calculated by dividing the molar mass by Avogadro's number and converting grams to kilograms by dividing by 1000.
This computation allows us to use the molecular mass in kinetic energy equations to find out how much energy is possessed by single molecules moving at set speeds.

Unit Conversion

Unit conversion is a mathematical process used to change one unit of measure to another to facilitate calculations. In this exercise, it plays a pivotal role in translating speed from miles per hour to meters per second.
Proper unit conversion ensures that scientific formulas remain valid, as these formulas typically rely on standardized units.
To convert speeds, multiply the given speed in mph by the conversion factor for miles to meters (1609.34 meters per mile) and then divide by the time conversion factor (3600 seconds per hour). Completing these steps provides a speed in the desired unit of meters per second (m/s), critical for calculating kinetic energy accurately.
Understanding unit conversions allows students to effectively work in various scientific disciplines, ensuring precision across different calculations.

Avogadro's Number

Avogadro's number, \( 6.022 \times 10^{23} \), is a fundamental constant in chemistry that represents the number of atoms, ions, or molecules contained in one mole of a substance.
It's essential for converting between the macroscopic scale of moles and the microscopic scale of individual molecules.
In this exercise, Avogadro’s number is crucial to calculate the kinetic energy per mole of \( ext{N}_2 \) molecules. First, it helps us determine the mass of a single molecule from the molar mass, allowing precise calculations at the molecular level.
Then, it's used again to scale this individual kinetic energy up to the macroscopic level by multiplying it by Avogadro's number, resulting in the total kinetic energy for a whole mole of molecules.
Understanding Avogadro’s number and its applications is key to grasping fundamental concepts in chemistry and physics, linking the microscopic world to the quantities we measure in labs.