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Chapter 19 the Kinetic Theory of Gases

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Chapter 19 the Kinetic Theory of Gases Chapter 19 The kinetic theory of gases In this chapter we will introduce the kinetic theory of gases which relates the motion of the constituent atoms to the volume, pressure and temperature of the gas. The following topics will be covered: (( Ludwig Boltzmann )) Ludwig Eduard Boltzmann (February 20, 1844 – September 5, 1906) was an Austrian physicist famous for his founding contributions in the fields of statistical mechanics and statistical thermodynamics. He was one of the most important advocates for atomic theory when that scientific model was still highly controversial. Boltzmann’s equation-curved in stone. This equation was originally formulated by Ludwig Boltzmann between 1872 to 1875, but later put into its current form by Max Planck in about 1900. To quote Planck, "the logarithmic connection between entropy and probability was first stated by L. Boltzmann in his kinetic theory of gases." Reference Carlo Cercignani, Ludwig Boltzmann The man who trusted atoms (Oxford University Press, 1998) E.A. Guggenheim, Boltzmann’s distribution law (North-Holland, 1955). G. Gallavotti, W.L. Reiter, and J. Yngvason edited, Boltzmann’s Legacy (European Mathematical Society, 2008). 1 Molecules colliding with wall We consider molecules that strike a unit area of the wall of a container. Let vx denote the velocity component normal to the plane of the wall. The particles are inside the cube with the volume V. There are N particles. We consider the particle (mass m) with the velocity (vx ,vy ,vz ) . After the reflection at the wall, the velocity of the particles is changed into (vx ,vy ,vz ) .The change of the linear momentum gives a force on the plane normal to the x axis. px 2mvx F1xt As a result of the Newton’s third law (the action-reaction), the force 2mv F F x x 1x t is exerted on the wall (as an impulse), contributing to the pressure. Here we assume that the number of molecules per unit volume (number density), which have velocity componenets of vx – (vx+d vx), vy – (vy+d vy), and vz – (vz+d vz) is given by 1 1 N v)( d 3v N (v , v , v )dv dv dv V V x y z x y z 3 where N v)( d v is the number of molecules whose velocities are vx – (vx+d vx), vy – (vy+d vy), and vz – (vz+d vz. ). N N v)( d 3v The number of particles (velocity v) colliding with the plane is evaluated as follows. The volume of the cylinder is given by Avxt The number of molecules in the cylinder is calculated as 1 (Av t) N()v d 3v x V The force applied along the x axis is 2mv 1 A (Av t) x N()v d 3v 2mv 2 N()v d 3v x t V V x Then the pressure P is defined by F total 1 2Amv 2 P x x N()v d 3v A A V 1 2mv 2 N()v d 3v V x vx 0 where only the particles with vx>0 contribute to the pressure. 2 Thermal equilibrium In thermal equilibrium, the gas is assumed to be homogeneous for each direction. So it is natural to consider that N(v) is dependent only on the magnitude of v. 1 1 PV 2mv 2 N()v d 3v 2mv 2 N()v d 3v mv2 N()v d 3v x 2 x 3 vx 0 2 2 where vx is replaced by v /3. Here we note that average of the total energy is 1 E mv2 N()v d 3v 2 Then we have 2 PV E . 3 This equation is called a Bernoulli’s equation. For an ideal gas, we have the Boyle’s law PV NkBT , leading to the expression for E 3 E N Nk T 2 B Note that is the average energy of each particle. 3 k T 2 B 3 Maxwell velocity distribution f(v) 3.1 Determination of the form N(v) It is assumed that N(v) depends only on the magnitude of v. The velocity distribution for vx is independent of those for vy and vz. Then we have the form of N(v) as 2 N()v G(v ) g(vx )(g vy )(g vz ) (1) with g(0) = , where G and g are functions to be determined. In Eq.(1), we put vy = vz = 0. 2 2 G(vx ) g(vx )g (0)g (0) g( vx ) Then we have 1 g(v ) G(v 2 ) x 2 x Similarly, 1 g(v ) G(v 2 ) y 2 y 1 g(v ) G(v 2 ) z 2 z Then Eq.(1) can be rewritten as 2 2 2 2 G(v ) G(vx vy vz ) g(vx )(g vy )(g vz ) 1 G(v 2 )G(v 2 )G(v 2 ) 6 x y z 2 2 2 For simplicity we put vx = , vy = , and vz = . Then we have 1 G( ) G() G() G() . 6 Taking a derivative of the above equation with respect to and putting = = 0, 1 G'() G( )G '(0)G )0( aG() 6 or G( ) Aea where 1 a G )0(' G )0( . 6 The final form of N(v) is obtained as 2 2 2 2 2 2 2 N()v G(vx v y vz ) Aexp[a( vx v y vz )] Aexp(av ) (Gaussian form) Now we determine the values of A and a. (a) Determination of A N N()v d 3v Aexp[a( v 2 v 2 v 2 )]dv dv dv x y z x y z A[ exp(av 2 )dv ]3 x x A( ) 2/3 a or A ( ) 2/3 N a where exp(av 2 )dv exp(av 2 )dv x x x x a (( Note )) We can also calculate the above integral as follows. N A 4v2 exp(av2 )dv 0 A 4 v2 exp(av2 )dv 2 1 2A a 2/3 2 A( ) 2/3 a (( Mathematica )) ∞ J n_ := x n Exp −a x2 x Simplify , a > 0 & @ D ‡−∞ A E êê @ D = list1 Table @8 n,J@nD<, 8n, 0, 10, 2<D; list1 êê TableForm 0 π a 2 π 2 a3ê2 4 3 π 4 a5ê2 6 15 π 8 a7ê2 8 105 π 16 a9ê2 10 945 π 11 ê2 32 a (b) Determination of a The total energy <E> is given by 1 E m(v 2 v 2 v 2 )N()v dv dv dv 2 x y z x y z 1 m(v 2 v 2 v 2 )Aexp[a( v 2 v 2 v 2 )]dv dv dv 2 x y z x y z x y z 1 mv2 Aexp(av2 4) v2dv 2 1 mA4 v4 exp(av2 )dv 2 0 1 1 mA4 v4 exp(av2 )dv 2 2 1 1 3 mA4 2 2 4a 2/5 3 2/3 mA 4a 2/5 2 where v is the radius of the sphere in the ( vx, vy, vz) and 4v dv is the volume element enclosed by the spherical shell of inner radius v and outer radius v +d v, 2 2 2 2 v vx vy vz 3 2 d v dvxdvydvz 4v dv Then we have 3 2/3 3 mA m E 2/5 2/5 3m 3 4a 4a k T 1 B N A( ) 2/3 4a 2 a a 2/3 From this equation, the value of a is determined as m a 2kBT In summary we have the Maxwell distribution or Maxwell-Boltzmann distribution, m mv 2 N()v N( ) 2/3 exp( ) . 2kBT 2kBT We define n(v) as N()v m mv 2 n()v ( ) 2/3 exp( ) N 2kBT 2kBT with 1 n()v d 3v n( v )4v2dv (1) n(v)dv is the probability of finding particles whose velocities are vx – (vx+d vx), vy – (vy+d vy), and vz – (vz+d vz. ). (2) Note that n(v)4 v2dv is the probability of finding particles whose magnitude of velocities are between v and (v + dv ). Using the Avogadro number NA, where NAm = M (the molar mass) NAkB = R (gas constant), n(v) can be rewritten as 2/3 M Mv2 n()v exp( ) 2RT 2RT The function n(v) has a Gaussian type with a peak at v = 0. Here we introduce the probability f( v ) dv of finding the particle having the velocity between v and v + d v, as f() v dv 4v 2n() v dv 2/3 2 m 2 mv 4v exp( )dv 2kBT 2k BT 2/3 M Mv 2 4v 2 exp( )dv 2RT 2RT where f() v dv 1 0 f(v) is called the Maxwell velocity distribution and is given by 2/3 M Mv2 f() v 4v2 exp( ) 2RT 2RT ((Note)) 1 N()v dv n()v dv 4v2 n() v dv f() v dv N We note that f( v ) has a local maximum at 2k T 2RT RT v B 1.41421 (most probable speed) mp m M M ((Note)) 3/ 2 df ()v M Mv2 Mv2 Mv 4{2vexp( ) v2 exp( )( ) dv 2RT 2RT 2RT RT 3/ 2 M Mv2 Mv2 4 ( )(v 2) 0 2RT 2RT RT The maximum value of f( v ) at v = vmp is 2 2M f max e RT The distribution function is normalized as f() v v v ( )2 exp[1 ( )]2 fmax vmp vmp f fmax 1.0 0.8 0.6 0.4 0.2 v 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 vmp Fig.
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