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?

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: \(1050 \frac{\text{miles}}{\text{hour}} \cdot \frac{1609.34 \text{ meters}}{1 \text{ mile}} \cdot \frac{1 \text{ hour}}{3600 \text{ seconds}}\)

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.

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.

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.