Question

At 20 ∘C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph.

Part A What is the kinetic energy (in J) of an N2 molecule moving at this speed? Express your answer using four significant figures.

Part B What is the total kinetic energy of 1 mol of N2 molecules moving at this speed? Express your answer using four significant figures.

Answer 1

Answer: According to the question , Here  Kinetic energy is the energy of motion. It can be calculated by

K.E = 1/2 m v²

Molecular nitrogen (N2) is 28.0 g/mol or 0.028 kg/mol. Using Avogadro's number, you can calculate the mass of a single molecule from this.

hence mass = 0.028 / 6.023 * 10²³ ] = 4.6488 * 10^-26

now the velocity = 1050 meter per h

= 0.29 meter per second

Hence K.E = 1.95 48 * 10^-27

part B ] total kinetic energy = K.e * avogadro number

1.9548 * 10^-27 * 6.022 * 10²³

= 1.17718 * 10^-3 J

Hence it is all about the given question , Thank you :)