Question

Problem 5.91 At 20 ∘C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph. Part C 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

The K.E of the moelcule will be

K.E = 0.5 X mass X (speed)²

The molecular mass of nitrogen = 28g / mole

Therefore mass of one molecule = 28 / 6.023 X 10²³ Grams = 4.6488 X 10^-23 grams = 4.6488 X 10^-26 Kilograms

Speed = 1050mph = 0.29 m /s

K.E = 0.5 X 4.6488 X 10^-26 Kilograms X ( 0.29m/s)² = 1.9548 10^-27 Joules

The K.E for one mole of molecules = K.E X avagadro.s number = 1.9548 10^-27 X 6.023 X 10²³ Joules

                                                 = 1.177 X 10^-3 Joules