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 10^{23}
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 10^{23} Joules

= 1.177 X 10^{-3} Joules

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