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: 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 :)
Get Answers For Free
Most questions answered within 1 hours.