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

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.

Homework Answers

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

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