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 \(\mathrm{m} / \mathrm{s} ?(\mathbf{b})\) 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?

The given average speed of N₂ molecules at 20°C is 1050 mph. We need to convert this to m/s. We know that: 1 mile = 1609.344 meters and 1 hour = 3600 seconds. So, the conversion factor from mph to m/s is: \[ \frac{1609.344 \, \text{m}}{\text{mile}} \times \frac{1 \, \text{hour}}{3600 \, \text{s}} \] Now, multiply the given speed by the conversion factor: \[ 1050 \, \frac{\text{mile}}{\text{hour}} \times \frac{1609.344 \, \text{m}}{\text{mile}} \times \frac{1 \, \text{hour}}{3600 \, \text{s}} = 469.6 \, \frac{\text{m}}{\text{s}} \] So, the average speed in m/s is 469.6 m/s.

The kinetic energy formula for a single molecule is given by: \[ K.E = \frac{1}{2}mv^2 \] where 'm' is the mass of the molecule, and 'v' is its speed. We know that one N₂ molecule has a molar mass of approximately 28 g/mol, and 1 mole of substance consists of Avogadro's number (6.022 x 10²³) of particles. To find the mass of a single N₂ molecule, we can use the following formula: \[ m = \frac{\text{Molar mass}}{\text{Avogadro's number}} \] \[ m = \frac{28 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \] Since it is necessary to have the mass in kg, we will also convert grams to kilograms: \[ m = \frac{28 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \] \[ m \approx 4.65 \times 10^{-26} \, \text{kg} \] Now, we can plug the values into the kinetic energy formula: \[ K.E = \frac{1}{2}(4.65 \times 10^{-26} \, \text{kg})(469.6 \, \frac{\text{m}}{\text{s}})^2 = 5.12 \times 10^{-21} \, \text{J} \] So, the kinetic energy of a single N₂ molecule moving at the given speed is approximately 5.12 x 10⁻²¹ J.

Now, we have to find the total kinetic energy of 1 mole of N₂ molecules. To do this, we simply multiply the kinetic energy of a single molecule by Avogadro's number: \[ \text{Total K.E} = (\text{Kinetic energy of 1 molecule}) \times (\text{Number of molecules in 1 mole})\] \[ \text{Total K.E} = (5.12 \times 10^{-21} \, \text{J}) \times (6.022 \times 10^{23} \, \text{molecules}) = 3.08 \times 10^3 \, \text{J} \] Therefore, the total kinetic energy of 1 mole of N₂ molecules moving at the given speed is approximately 3.08 x 10³ J.