Chapter 22 Molecular properties of gases

1. Chapter 22 Molecular properties of gases

2. 22-1 The atomic nature of matter (物质的原子本质) J. J. Thomson discovered electrons in 1897. Rutherford discovered the nature of atomic nucleus. He was at his lab at McGill Univ. in 1905.

3. 1. Brownian motion The modern trail to belief in atoms can be said to have started in 1828: the observation of Brownian motion. In 1828 Robert Brown observed through his microscope that tiny grains of pollen suspended in water underwent ceaseless random motion. We now call this phenomenon “Brownian motion”. See动画库\力学夹\4-01布朗运动

4. 2. Properties of the ideal gas (I) The ideal gas consists of particles, which are in random motion and obey Newton’s Laws of motion. These particles are “atoms” or “molecules”. (II) The total number of particles is “large”. The rate at which momentum is delivered to any area A of the container wall is essentially constant. (III) The volume occupied by molecules is a negligibly small fraction of the volume occupied by the gas.

5. (IV) No forces act on a molecule except during a collision. (V) All collisions are elastic and negligible duration. Total kinetic energy of the molecules is a constant, and total potential energy is negligible.

6. We will take the ideal gas as our system. Consider N molecules of an ideal gas confined within a cubical box of edge length L, as in Fig 22-2. 22-2 A molecular view of pressure(压强) How to relate pressure to microscopic quantities? Fig 22-2 y L m L x L z

7. The average impulsive force exerted by the molecule on is (22-4) The total force on by all the gas molecules is the sum of the quantities for all the molecules. Then the pressure on A1 is (22-5)

8. If N is the total number of molecules in container, the total mass is Nm. the density is . (22-6) The quantity in parenthesis is average value of for all the molecules in the container. (22-7)

9. (22-7) For any molecules, , and so Eq(22-7) becomes (22-8) • The result is true even when we consider collisions between molecules. • The result is correct even with consideration of the collisions between molecules and other walls in the box. 3. The result is correct for boxes with any kinds of shape.

10. 4. The “root-mean-square” (均方根)speed of the molecules： (22-9) In Eq(22-8,9), we relate a macroscopic quantity ( the pressure P) to an average value of a microscopic quantity, that is to or .

11. Sample problem 22-1 Calculate the of at , . Under these conditions . Solution:

12. Sample problem 22-2 In Fig 22-2 , L=10cm, P=1atm, T=300K • How many moles of oxygen are in the box? • How many molecules ? Solution: (a) (b) y L m x L L z

13. Problem The mass of the H2 molecule is 3.3*10-24g. If 1.6*1023 hydrogen molecules per second strike 2.0 cm2 of wall at an angle of 600 with the normal when moving with a speed of 1.0*105cm/s, what pressure do they exert on the wall?

14. 22-3 The mean free path 1. Mean free path :The average value of the straight-line distance a molecules travels between collisions. • Which kinds of physical quantities are related to mean free path? (22-3) where P and T are macroscopic quantities pressure and temperature, d is the diameter of a molecule of the gas. See动画库\力学夹\4-11平均自由程 See动画库\力学夹\4-10碰撞频率

15. For air molecules at sea level, . At an altitude of 100 km, .

16. Sample problem 22-4 What are (a) the mean free pathand (b) the average collision rate for nitrogen at T=300k, ? Suppose , . Solution: (a) (b) The average collision rate is

17. N(v) 22-4 The distribution of molecular speeds 1.The Maxwell speed distribution where Nis the total number of molecules; T is temperature, m is the mass of each molecule. N(v) expresses particle number in unit speed range at v. (22-14) Fig 22-6

18. Notes: (a) Avoid the temptation to interpret N(v) as “the number of molecules having a speed v”. • The total number of molecules: (22-15) N is equal to the total area under speed distribution curve in Fig 22-6. (c) The number of moleculeswith speeds in the rangefrom v to v+dv is N(v)dv.

19. N(v) T=80K Fig 22-7 T=300K V(m/s) See动画库\力学夹\4-05麦克斯韦速率分布律 1 兰媚尔实验

20. 2. Consequences of the speed distribution (i) The most probable speed . It is the speed at which N(v) has its maximum value. (22-16) (ii) The average speed (22-17) (22-18)

21. N(v) 400 600 800 Fig 22-6 (iii) The root-mean square speed

22. (iv) Average translational kinetic energy per molecule (22-21) (v) The ideal gas law

23. Scientists contributed to ideal gas law: Boyle(玻意耳,英国），Charles(查理,法国）， Gay-Lussac(盖吕萨克,法国）

24. Sample problem 22-5 The speeds of ten particles in m/s are 0, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0 and 6.0. Find (a) the average speed (b) (c) of these particles Solution: (a)

25. (b) (c) Of the ten particles, three have speeds of 3.0m/s, two have speeds of 4.0m/s, and each of the other has a different speed. Then

26. Sample problem 22-6 A container filled with N molecules of oxygen gas is maintained at 300K . What fraction of the molecules has speeds in the range 599-601 m/s? The molar mass M of oxygen is 32g/mol. The fraction in that interval is , where , . From Solution:

27. 22-5 The distribution of molecular energy 1) Consider a special case that translational kinetic energy is the only form of energy that a molecule can have. 2) Let us consider again the situation of Sample problem 22-6. 3) The number with kinetic energies between E and E+dE is the same as the number with speed between v and v+dv.

28. Mathematically, we express this conclusion as (22-22) (22-23) Since the energy is only kinetic, we must have or , thus (22-24) then (22-25)

29. (22-25) Boltzmann Factor Eq(22-25) is called Maxwell-Boltzmann energy distribution. • Since we have assumed that the molecules can have only Ktrans, this distribution applies only to a monatomic gas. But Boltzmann Factor is always present in N(E) no matter what the form of energy E. (ii) The total number of molecules N is determined from (22-26) (iii) Maxwell-Boltzmann distribution is precisely the same for any gas at a given temperature.

30. Sample problem 22-8 Find (a) The average energy and (b) the most probable energy of a gas in thermal equilibrium at temperature T. Solution: (a) Substituting Eq(22-25), we obtain

31. Using Appendix I (b) Taking the derivative of Eq(22-25), and setting it equal to zero, then

32. *22-6 Equations of state for real gases The equation of statefor an ideal gas holds well enough only for real gases at sufficiently low densities. It does not hold exactly for real gases at any density. Two methods used for real gases: (I) The Virial expansion (维里展开) (II) van der Waals equation of states (proposed in 1873): (22-33) (He received the 1910 Nobel prize for his work.) in which a and b are constants obtained by experiment.

33. 1.The volume correction In section 22-1 we assumed that the volume occupied by the molecules of an ideal gas is negligible. This is not quite true for real gases. Let us regard each molecule of real gas as a hard sphere of diameter d. We can find an approximate value of b

34. 2. The pressure correction For ideal gas, we assumed that the molecules of an ideal gas exert forces on each other only during collisions. That is not quite true for real gases. a) If the molecule is not near the wall of the vessel, it would experience no net force due to the forces exerted on it by the surrounding molecules. b) However, if the molecule is located near the wall of the vessel, as in Fig 22-11. the molecule c would experience a net force of attraction away from the wall.

35. Thus the pressure measured at the wall is somewhat less than what we may call the true pressure. If P in Eq 22-33 is to be the measured pressure, we must increase it by to obtain the “true” pressure. Wall c R R Fig 22-11