1、Chapter 22 Molecular properties of gasesJ.J.Thomson discovered electrons in 1897.22-1 The atomic nature of matter(物质的原子本质)Rutherford discovered the nature of atomic nucleus.He was at his lab at McGill Univ.in 1905.In 1828 Robert Brown observed through his microscope that tiny grains of pollen suspen
2、ded inwater underwent ceaseless random motion.We now call this phenomenon“Brownian motion”.The modern trail to belief in atoms can be said to have started in 1828:the observation of Brownian motion.1.Brownian motionSee动画库力学夹4-01布朗运动(I)The ideal gas consists of particles,which are in random motion an
3、d obey Newtons 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 occ
4、upied by the gas.2.Properties of the ideal gas(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.22-2 A molecular view of pressure(压强)We will take
5、 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.LLyxzLm1Av2AFig 22-2How to relate pressure to microscopic quantities?,.vPThe average impulsive force exerted by themolecule on is 1ALmvvLmvFxxxx222=(22-4)The total force on
6、 by all the gas molecules is the sum of the quantitiesfor all the molecules.Then the pressure on A1is(22-5)1ALmvx/2)(1.)(12221322212212 +=+=+=xxxxxxvvLmLmvmvLFFLPIf 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
7、 offor all the molecules in the container.(22-7)3/LNm=2xvavxvP)(2=223NvNvLmNPixiixi=)(22213 +=xxvvLmPNvvixiavx/)()(22=For any molecules,andso Eq(22-7)becomes(22-8)1.The result is true even when we consider collisions between molecules.2.The result is correct even with consideration of the collisions
8、 between molecules and other walls in the box.3.The result is correct for boxes with any kinds of shape.2222zyxvvvv+=avavzavyavxvvvv)(31)()()(2222=avvP)(312=avxvP)(2=(22-7)(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
9、.Pvvavrms3)(2=rmsvavv)(24.The“root-mean-square”(均方根)speed of the molecules：Sample problem 22-1Calculate the of at ,.Under these conditions .Solution:rmsvcT0=atmP1=32/1099.8mkgH=smmkgPaPvrms/1840/1099.8)1001.1(33325=2HSample problem 22-2In Fig 22-2,L=10cm,P=1atm,T=300K(a)How many moles of oxygen are
10、in the box?(b)How many molecules?Solution:(a)(b)molkmolkJmPaRTPVn041.0)300)(/31.8()10.0)(1001.1(35=moleculesmolnNNA2223105.2)1002.6)(041.0(=LLyxzLm1AThe mass of the H2molecule is 3.3*10-24g.If 1.6*1023hydrogen molecules per second strike 2.0 cm2of wall at an angle of 600with the normal when moving w
11、ith a speed of 1.0*105cm/s,what pressure do they exert on the wall?Problem 22-3 The mean free path1.Mean free path:The average value of the straight-line distance a molecules travels between collisions.2.Which kinds of physical quantities are related to mean free path?(22-3)where P and T are macrosc
12、opic quantities pressure and temperature,d is the diameter of a molecule of the gas.PdKT22=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For air molecules at sea level,.At an altitude of 100 km,.cm16m710Sample problem 22-4What are(a)the mean free path and(b)the average collision rate for nitrogen at T=300k,?Su
13、ppose ,.Solution:(a)(b)The average collision rate ismPdKT82103.92=mdN101015.32=smvav/478=PaP51001.1=ondcollisionsvZavsec/101.59=22-4 The distribution of molecular speeds1.The Maxwell speed distributionwhere N is the total number of molecules;T is temperature,m is the mass of each molecule.N(v)expres
14、ses particle number in unit speed range at v.KTmvevKTmNvN222/32)2(4)(=N(v)Fig 22-6(22-14)(c)The number of molecules with speeds in the range from v to v+dv is N(v)dv.(b)The total number of molecules:(22-15)N is equal to the total area under speed distribution curve in Fig 22-6.=0)(dvvNNNotes:(a)Avoi
15、d the temptation to interpret N(v)as“the number of molecules having a speed v”.Fig 22-7V(m/s)N(v)T=80KT=300KSee动画库力学夹4-05麦克斯韦速率分布律1 兰媚尔实验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 speedPv0)(=dvvdNMRTmKT
16、vP22=KTmvevKTmNvN222/32)2(4)(=piavvMRTmKTdvvvNNNvv=88)(10(22-17)(22-18)N(v)800Fig 22-6rmsvavvpv400600=022)(1)(dvvNvNvvavrmsavrmsvMRTmKTv=33(iii)The root-mean square speedrmsv(v)The ideal gas lawpMRTvrms33=VnM/=nRTPV=(iv)Average translational kinetic energy per moleculeKTmKTmmvNvvvmvvvmNKrmsNNtrans23)3(2121)(21)(2122222122221=+=+=(22-21)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）Sample problem 22-5 The speeds of ten particles in m/s are 0,1.0,2.0,3.0,3
