1. Physics 101 Lecture 10

2. Atomic Mass Unit (u)

3. Atomic Mass Unit For one element choose  Unit  Reference value Unit is called atomic mass unit (u) Reference element is carbon-12 Atomic mass is defined to be exactly 12 atomic mass units

4. Atomic Mass Unit Periodic Table Atomic masses of elements are listed in the periodic table Masses listed are average values and take into account isotopes of an element that exist naturally

7. Periodic Table Example In the periodic table, the atomic mass of carbon is 12.011  98.93% of naturally occuring carbon is the isotope carbon-12  1.07% of naturally occuring carbon is the isotope carbon-13  Periodic table atomic mass value is  (0.9893)(12) + (0.0107)(13) = 12.011

8. Relationship of AMU to Kg 1 u = 1.6605 x 10 -27 kg

9. Atomic Mass of Molecule The atomic mass of a molecule is the sum of the atomic masses of its atoms –Hydrogen has an atomic mass of 1.00794 –Oxygen has an atomic mass of 15.9994 u –The atomic mass of water molecule is –2(1.00794 u) + 1(15.9994 u) = 18.0152 u

10. Mole

11. Mole or Gram-Mole One mole of a substance contains as many particles (atoms or molecules) as there are atoms in 12 grams of the isotope carbon-12

12. Particles per Mole of Carbon-12 Number of particles, N A, is Avogadro’s number Use carbon-12 as an example 12 g = (12u/particle)(1.6605x10 -24 g/u)(N A particles) N A particles = 12g / [(12u/particle)(1.6605x10 -24 g/u)] N A particles = 6.022 x 10 23 particles

13. Mole of Particles other than C-12 The mole is defined in terms of Carbon-12 Concept of a mole can be applied to any collection of objects One mole contains Avogadro’s number of objects  One mole of atomic sulfur contains 6.022 x 10 23 sulfur atoms  One mole of water contains 6.022 x 10 23 H 2 0 molecules  One mole of golf balls contains 6.022 x 10 23 golf balls

14. Avogadro’s Number Symbol: N A Number of atoms per mole is known as Avogadro’s number N A = 6.022 x 10 23 particles / mol

15. Number of Moles Sample of N Particles n moles = (N particles)/ (N A particles/mole)

16. Number of Moles in Mass m of Particles n moles = (mass in g) / (g/mole)

17. Mass of One Particle M particle = (mass in g / mole) / (N A particles/mole)

18. Mole – Example 1 Ordinary nitrogen gas consists of molecules of N 2 Find the mass, m, of one such molecule The molecular mass is 28 g/mole  m = [28 g/mole] /[ 6.02 x 10 23 particles/mole]  m = 4.7 x 10 -23 g/particle

19. Mole – Example 2 Helium gas consists of separate He atoms There are 2.0 g of He Molecular mass of helium is 4.0 g/mole How many moles of He are there? How many He atoms are there?  moles = (2.0 g) / (4.0 g/mole) = 0.5 mole  particles = (0.5 mole) x (6.02 x 10 23 particles/mol)  particles = 3.0 x 10 23 He atoms

20. Ideal Gases

21. Ideal Gas Law Based on experiments by  Robert Boyle in 1662  Jacques Charles in 1787  Joseph Louis Gay-Lussac in 1802

22. Ideal Gas Law – Form 1 PV = nRT  R = 8.31 J/molK is universal gas constant  P is pressure in pascals  V is volume in m 3  n is number of moles  T is Kelvin temperature

23. Ideal Gas Law – Form 2 PV = NkT  k = 1.38 x 10 -23 J/K is Boltzmann’s constant  P is pressure in pascals  V is volume in m 3  N is number of particles  T is Kelvin temperature

24. Ideal Gas Law - Example Given a mass of oxygen  Volume = 0.0200 m 3  Pressure = atmospheric pressure, 101 x 10 3 Pa  Temperature = 5.0 0 C Determine volume if pressure increased to 108 x 10 3 Pa while temperature is changed to 30 0 C  PV = nRT  (101 x 10 3 )(0.0200) = nR(5 + 273) = nR(278)  nR = (101 x 10 3 )(0.0200)/278 = 7.266  (108 x 10 3 )V = nR(30 + 273) = nR(303)  V = 7.266(303)/108 x 10 3 = 0.0204 m 3

25. Kinetic Theory

26. Kinetic Theory - 1 Ideal gas N particles Cubical container of side L

27. Kinetic Theory - 2 One particle of mass m hits the right wall perpendicularly and rebounds elastically While approaching the wall, the particle has velocity +v and momentum +mv After rebounding, the particle has velocity –v and momentum –mv Particle heads to the left wall, rebounds and returns to the right wall Time t between collisions with the right wall is the round-trip distance 2L divided by the speed of the particle

28. Kinetic Theory - 3 t = 2L/v F = (p f – p i ) / t F = [-mv – (+mv)] / t = -2mv / t F = -2mv / (2L/v) = -mv 2 / L Newton’s third law, force applied to wall by particle is oppositely directed F w = mv 2 / L

29. Kinetic Theory - 4 There are N particles in the gas 1/3 are travelling in the horizontal direction Particles have a statistical distribution of velocities Average of velocity squared is v 2 avg F w = (N/3)(mv 2 avg / L)

30. Kinetic Theory - 5 Pressure is force divided by area P = F w / L 2 =(N/3)(mv 2 avg / L 3 ) Volume is V = L 3 PV = (2N/3)(½mv 2 avg ) ½m(v 2 ) avg = KE avg The ideal gas law is PV = NkT PV = (2N/3)KE avg KE avg = (3/2)kT

31. Kinetic Theory - Example Find average speed of a nitrogen molecule in air at 0 0 C ½ mv 2 avg = (3/2)kT v = sqrt(3kT/m) Atomic mass = 28 g/mol m = (28 g/mol / 6.02 x 10 23 particles/mol) m = 4.65 x 10 -23 g = 4.65 x 10 -26 kg