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Kinetic Theory of Gases2


Kinetic Theory of Gases
Chapter Outline 1.1 Introduction 1.2 Pressure of an ideal gas 1.3 Temperature and Energy 1.4 Distributions, Mean values, and Distribution Functions 1.5 The Maxwell Distribution of Speeds 1.6 Energy Distributions 1.7 Collisions: Mean Free Path and Collision Number 1.8 Summary

1.1 Introdution
The overall objective of this chapter is to understand macroscopic properties such as pressure and temperature on a microscopic level. We will find that the pressure of an ideal gas can be understood by applying Newton’s law to the microscopic motion of the molecules making up the gas and that a comparison between the Newtonian prediction and the ideal gas law can provide a function that describes the distribution of molecular velocities. This distribution function can in turn be used to learn about the frequency of molecular collisions. Since molecules can react only as fast as they collide with one another, the collision frequency provides an upper limit on the reaction rate.

1.2 Pressure of an ideal gas
Premise:

Pressure exerted by a gas on the wall of a container is due to collisions of molecules with the wall.
Consider molecules: Mass: m Vx: in the x direction Area A: located in the z-y plane The time of striking: t

The number of molecules per unit volume: n
The number of molecules colliding with the wall: (nAVxt)/2

The force on the wall due to the collision of a molecules with the wall is given by Newton’s law:
F=ma=mdv/dt=d(mv)/dt F△t= △( mv) If a molecule rebounds elastically (without losing energy) when it hits the wall, its momentum is changed from +mvx to –mvx, so that the total momentum change is △( mv) =2mvx. Consequently, F△t=2mvx for one molecular collision, and Ft=(nAvxt)(2mvx)/2 for the total number of collisions. Finally, We obtain p=nmvx2. The total velocity of an individual molecule is most likely expressed by; V2=Vx2+Vy2+Vz2 or <V2>=<Vx2>+<Vy2>+<Vz2> We obtain: P=(nm<v2>)/3 n=nNA/V pV=(nNAm<v2>)/3 <E>=(m<v2>)/2 pV=(2nNA<E>)/3

1.3 Temperature and Energy
Two types of molecules: Type one: <E1> T1 Type two: <E2> T2 If <E1> > <E2>, then T1 > T2 If <E1> = <E2>, then T1 = T2 Consequently, there must be a correspondence between <E> and T so that the latter is some function of former: T=T(<E>). We can get the function from the equation pV=(2nNA<E>)/3 and the ideal gas law, and the result is: T=PV/(nR)=2NA<E>/(3R)

<E>=3kT/2=m<v2>/2

<v2>=3kT/m

1.4 Distributions, Mean Values, and Distribution Functions The ordinary way to get the mean values: <S>=(S1+S2+S3+S4+S5)/NT=(ΣSi)/NT

If the number of objection is too large, we can use another method:
<S>=(ΣSjNj)/NT Nj requires that ΣNj=NT Pj=Nj/NT The another way to write the equation of mean values <S>=ΣSjPj Furthermore, we can generalize the equation to provide a method for finding the average of any quantity <Q>=ΣPjQj

The method can be extended to calculate more complicated averages. Let f(Q) be some arbitrary function of the observation Qj. Then the average value of the function f(Q) is given by <f(Q)> = ΣPjf(Qj) If the quantity of S is not an one-point accuracy, but to an accuracy of dS, where dS is a very small fraction of a point. Let P(S)dS be the probability that a quantity will fall in the range between S and S+dS, and let dS become infinitesimally small. The fundamental theorems of calculus tell us that we can convert the sum in equation to the integral <S> = ∫P(S)SdS

or, more generally for any observable quantity,
<Q>= ∫P(Q)QdQ Note that normalization of the probability requires ∫P(Q)dQ =1

1.5 The Maxwell Distribution of Speeds
There are four main points in the derivation:

1.In each direction, the velocity distribution must be an even function of v.
2.The velocity distribution in any particular direction is independent from and uncorrelated with the distribution in orthogonal directions. 3.The average of the square of the velocity <v2> obtained using the distribution function should agree with the value required by the ideal gas law: <v2>=3kT/m 4.The three-dimensional velocity distribution depends only on the magnitude of v and not on the direction.

1.5.1 The velocity distribution must be an even function of v Consider the velocities vx of molecules contained in a box. The number of molecules moving in the positive x direction must be equal to the number of molecules moving in the negative x direction. This conclusion is easily seen by examining the consequences of the contrary assumption. If the number of molecules moving in each direction were not the same, then the pressure on one side of the box would be greater than on the other. Aside from violating experimental evidence that the pressure is the same wherever it is measured in a closed system, our common observation is that the box does not spontaneously move in either the positive or negative x direction, as would be likely if the pressure were substantially different. We conclude that the distribution function for the velocity in the x direction, or more generally in any arbitrary direction, must be symmetric. We can ensure that F(vx) be an even function by requiring that the distribution function depend on the square of the velocity: F(vx)=f(vx2).

1.5.2 The Velocity Distribution are independent and uncorrelated Using the analogy of tossing coins to explain the relationship between the distribution of x-axis velocities and y- or z-axis velocities, and the result is: they are independent and uncorrected, and we can write that F(vx,vy,vz) = F(vx)F(vy)F(vz) Since exp(a+b+c)=eaebec, so we can write F(vx)=f(vx2)=Kexp(+kvx2) The constant K can be determined from normalization : ∫F(vx)dvx=1

1=k∫exp(-kvx2)dvx=k(π/k)1/2
K=(k/π)1/2

1.5.3 <v2> should agree with the ideal gas law The constant k is determined by requiring <v2> to be equal to 3kT/m: <vX2>=∫vx2F(vx)dvx= (k/π)1/2 ∫vx2exp(-kvx2)dvx <v2>= (k/π)1/2 (π/k3) ? = 1/(2k) As a consequence, the average of the square of the total speed, <v2>=<vx2>+<vy2>+<vz2>, is simply <v2>=3/(2k)= 3kT/(m) or k=m/(2kT) The complete one-dimensional distribution function is thus: F(vx)dvx=(m/2πkT)1/2 exp(-mvx2/(2kT))dvx The complete three-dimensional distribution function is thus: F(v x,vy,vz) dvxdvydvz=(m/2πkT)3/2 exp(-mvx2/(2kT))dvxdvydvz

1.5.4 The distribution depends only on the speed We note that the equation of distribution depends on v2 and not the direction property of v, in other word, the function depends on the speed and not on the velocity Root-mean-squared speed: Crms=<v2>1/2 =(3kT/m)1/2 Mean speed: <v>=(8kT/πm)1/2 Most probable speed: C*=(2kT/m)1/2

1.5.5 Experimental Measurement of the Maxwell Distribution og Speeds

1.6 Energy Distribution E=(mv2)1/2 dE=mvdv=(2mE)1/2 We can convert velocities to energies to obtain: G(E)dE=4π(2E/m)(m/(2πkT))3/2exp(-E/kT)/(2mE)1/2dE

=2 π(1/πkT)3/2 E1/2 exp(-E/kT)De
We know that: <E>=3kT/2 The average U of n moles of a monatomic gas: U=(3nNAkT)/2=(3nRT)/2 Heat capacity at constant volume: Cv=(3nR)/2

1.7 Collision: Mean Free Path and Collision Number Simple consideration suggest that the collision rate between molecules should be proportional to the relative speed of the molecules, to their size, and to the number of possible collision pairs. Relative velocity between molecules of type1 and type 2: <vr> The distance of a line perpendicular to the each of the initial velocities of two colliding molecules: b The radii of type 1: r1 The radii of type 2: r2 Premise: a “collision” will occur if the two molecules approach one another so that their centers are within the distance bmax=r1+r2

b

bmax

b=0

A collision will occur if b of two molcules is less than bmax

2
1

A=πb2max

<Vr>△t Consider a molecule of type 1 moving through a gas with a speed equal to the average magnitude of the relative velocity<Vr>. Any molecule of type 2 located in a cylinder of volume πb2max<Vr> △t will then be struck in the time △t.If the density of molecules of type 2 is n*2, then the number of collisions one molecule of type 1 will experience with molecules of type 2 per unit time is Z2= πb2max<Vr> n*2 If type 2 and type 1 are the same:

Z1= πb2max<Vr> n*1

The total number of collision of molecules of type1 with those of type2 per unit time and per unit volume is:

Z12= πb2max<Vr> n*1 n*2
The number of collisions of one type of molecule per unit time per unit volume: Z11=(Z1n*1)/2= (πb2max<Vr> (n*1)2 )/2


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