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

Short answer

The average speed of N₂ molecules at 20°C is approximately 469.6 m/s. The kinetic energy of a single N₂ molecule moving at this speed is about \(5.12 \times 10^{-21} \, \text{J}\). The total kinetic energy of 1 mole of N₂ molecules moving at this speed is approximately \(3.08 \times 10^3 \, \text{J}\).

Step-by-step solution

Convert the given average speed in m/s

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.

Calculate the kinetic energy of a single N₂ molecule

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.

Calculate the total kinetic energy of 1 mol of N₂ molecules

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.

Key concepts

Molecular Speed Conversion

Understanding how to convert molecular speed between different units can greatly help in solving physics and chemistry problems. Here, we need to transform a velocity from miles per hour (mph) to meters per second (m/s). The conversion involves using the following conversion factors: 1 mile equals 1609.344 meters and 1 hour equals 3600 seconds.
To convert, multiply the original speed in mph (1050 mph in this case) by the factor that converts miles to meters and the factor that converts hours to seconds:

• Multiply the speed by 1609.344 to convert miles to meters.
• Then divide by 3600 to convert hours to seconds.

Utilizing these factors, we find the speed in m/s:
\[ 1050 \times \frac{1609.344}{3600} \approx 469.6 \, \text{m/s} \] Now, our speed is neatly converted to meters per second, making it much easier to calculate related physical properties like kinetic energy.

Molar Mass Calculation

Molar mass is a crucial concept, providing the mass of one mole of a substance. For diatomic nitrogen (\( \mathrm{N}_2 \) ), the molar mass is approximately 28 grams per mole (g/mol). Converting this to kilograms, which is the standard unit for mass in physics calculations, involves a simple division by 1000.
To calculate, first note that:

• 1 g = 0.001 kg

This gives us:\[ 28 \, \text{g/mol} \times 0.001 \approx 0.028 \, \text{kg/mol} \] Using molar mass allows us to compute individual molecular masses, an essential step when calculating properties like kinetic energy.

Avogadro's Number

Avogadro's Number is a fundamental constant in chemistry: the number of atoms, ions, or molecules in one mole of a substance. Its value is approximately \( 6.022 \times 10^{23} \) , and it is indispensable for converting between moles and individual particles.
For example, if we know the molar mass of a substance, we can find the mass of a single molecule. Divide the molar mass (in kg) by Avogadro's Number to find this value. For \( \mathrm{N}_2 \) :

• Molecular mass = \( \frac{0.028 \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \)
• \( \approx 4.65 \times 10^{-26} \text{ kg} \)

This calculated mass is the foundation upon which kinetic energy calculations are built.

Single Molecule Kinetic Energy

The kinetic energy of a single molecule can be determined using the equation: \( KE = \frac{1}{2}mv^2 \). Here, \( m \) is the mass of the molecule and \( v \) is its velocity. With the molecular mass of nitrogen (\( \mathrm{N}_2 \) ) and its speed converted to m/s, we can input these values into the formula.
For nitrogen at 469.6 m/s and a molecular mass of \( 4.65 \times 10^{-26} \text{ kg} \):

• \( KE = \frac{1}{2}(4.65 \times 10^{-26} \, \text{kg})(469.6 \, \text{m/s})^2 \)
• \( \approx 5.12 \times 10^{-21} \, \text{J} \)

This calculation tells us the energy a single molecule carries due to its motion, illustrating how kinetic energy depends on mass and speed.

Mole Total Kinetic Energy

Calculating the total kinetic energy of one mole of molecules involves considering each molecule's kinetic energy and scaling it up using Avogadro's Number. Simply multiply the kinetic energy of one molecule by the number of molecules in a mole (\( 6.022 \times 10^{23} \) ).For nitrogen, where a single molecule has a kinetic energy of \( 5.12 \times 10^{-21} \, \text{J} \):

• Total KE = \( 5.12 \times 10^{-21} \, \text{J} \times 6.022 \times 10^{23} \)
• \( \approx 3.08 \times 10^3 \, \text{J} \)

This method expands the concept of molecular energy to a macro scale, providing insights into the energy contained within a mole of a pure gaseous element.