Kinetic Theory of Gases (1)

Transcription

1 Advanced Thermodynamics (M ) Chapter 11 Kinetic Theory of Gases (1) Min Soo Kim Seoul National University

2 11.1 Basic Assumption Basic assumptions of the kinetic theory 1) Large number of molecules N A = molecules per kilomole 2) Identical molecules which behave like hard spheres 3) No intermolecular forces except when in collision 2/20

3 11.1 Basic Assumption Basic assumptions of the kinetic theory 4) Collisions are perfectly elastic 5) Uniform distribution throughout the container n = N V dn = ndv Average # of molecules per unit volume 3/20

4 11.1 Basic Assumption 6) Equal probability on the direction of molecular velocity average # of intersections of velocity vectors per unit area N 4πr 2 r sin θ dφ r d θ the # of intersections in da r N 4πr 2 da r 2 sin θ dθdϕ θ ϕ = d 2 N θϕ = N sin θ dθdϕ 4π, d 2 n θϕ = n sin θ dθdϕ 4π # of molecules having velocities in a direction (θ< <θ+dθ) (ϕ< <ϕ+dϕ) 4/20

5 11.1 Basic Assumption 7) Magnitude of molecular velocity : 0 ~ c (speed of light) dn v : # of molecules with specified speed (v< <v+dv) 5/20

6 11.1 Basic Assumption Let dn v as # of molecules with specified speed (v< <v+dv) 0 dnv = N Mean speed is v ҧ = 1 N vdnv 0 Mean square speed is v 2 = 1 N 0 v 2 dn v Square root of v 2 is called the root mean square or rms speed: v rms = v 2 The n-th moment of distribution is defined as v n = 1 N 0 v n dn v 6/20

7 11.2 Molecular Flux # of molecules of a gas that strike a surface per unit area and unit time Molecules coming from particular direction θ, ϕ with specified speed v in time dt θϕv collision θ< <θ+dθ ϕ< <ϕ+dϕ v< <v+dv # of θϕv collisions with da dt = θϕv molecules in = θϕ molecules with speed v Fig. Slant cylinder geometry used to calculate the number of molecules that strike the area da in time dt. 7/20

8 11.2 Molecular Flux How many molecules in dn v : # density between speed (v< <v+dv) d 3 n θϕv = dn v da A = dn v sin θdθdϕ 4π dv = da (vdt cosθ ) Volume of cylinder # of θϕv molecules in the cylinder d 3 n θϕv dv = da (vdt cosθ )dn v sin θdθdϕ 4π # of collisions per unit area and time d 3 n θϕv dv da dt = 1 4π vdn vsinθcosθdθdϕ 8/20

9 11.2 Molecular Flux Total # of collisions per unit area and time d 3 n θϕv dv da dt 0 = 2π 0 π/2 sinθcosθdθdϕ 1 4π vdn v = 1 4 vdn v Total # of collisions per unit area and time by molecules having all speed 1 vdn 0 4 v = 1 nvҧ 4 Cf. average speed v ҧ = σ തv N = σ N iv i N = σ n iv i σ n i vdn v = n 9/20

10 11.3 Gas Pressure and Ideal Gas Law Gas pressure in Kinetic theory Gas pressure is interpreted as impulse flux of particles striking a surface mv 1 mv 2 θ θ 10/20

11 11.3 Gas Pressure and Ideal Gas Law Perfect elastic v = v Average force exerted by molecules F = d(mv) dt = m Ԧa + mሶ Ԧv Momentum change of one molecule (normal component only) mvcosθ mvcosθ = 2mvcosθ v sin θ # of θϕv collisions for da, dt v v cos θ 1 4π vdn vsinθcosθdθdϕ θ da θ ϕ 11/20

12 11.3 Gas Pressure and Ideal Gas Law Change in momentum due to θϕv collisions in time dt 2mvcosθ 1 4π vdn vsinθcosθdθdϕ = 1 2π mv2 dn v sinθcos 2 θdθdϕdadt Change in momentum in all v collisions 0 < θ π 2, 0 < ϕ 2π 0 π 2 0 2π 1 2π mv2 dn v sinθcos 2 θdθdϕ dadt = 1 3 mv2 dn v dadt 12/20

13 11.3 Gas Pressure and Ideal Gas Law Change in momentum from collisions of molecules at all speed 1 3 m v2 dn v = d ԦF dt Average pressure തP = d ԦF da 1 m 3 v2 dn v = 1 3 mnv2 cf. v 2 = σ v2 N = v2 dn v n 13/20

14 11.3 Gas Pressure and Ideal Gas Law n = N V, P = 1 3 N V mv2 PV = 1 3 Nmv2 cf. EOS of an ideal gas PV = n തRT = mrt = N N A തRT = NkT N A : Avogadro s number : number/kmole k: Boltzmann constant k B : Boltzmann constant : k B = ത R N A = J/K PV = 1 3 Nmv2 = NkT 1 2 mv2 = 3 2 kt 14/20

15 11.4 Equipartition of Energy Equipartition of energy Because of even distribution of velocity of particles, v 2 = v x 2 + v y 2 + v z 2, v x 2 = v y 2 = v z 2 = 1 3 v2 1 2 mv x 2 = 1 2 kt It can be interpreted that a degree of freedom allocate energy of 1/2 kt 15/20

16 11.4 Equipartition of Energy തE = തE x + തE y + തE z energy = 1 2 mv x mv y mv z 2 തE x = kt 2, തE y = kt 2, തE z = kt 2 16/20

17 11.5 Specific Heat δq = du + δw U = f 1 2 NkT f: number of degrees of freedom u = U N = f 1 2 ത RkT c v = u ቁ = f 1 R ത T v 2 (n തR = Nk) c P = h = f R ത + തR = (f+2) T p 2 2 തR 17/20

18 11.5 Specific Heat Monatomic gas 1 2 mv x 2, 1 2 mv y 2, 1 2 mv z 2 3 c p = c v 3 = 1.67 Diatomic gas 1 2 mv x 2, 1 2 mv y 2, 1 2 mv z 2 Translational 1 2 Iw x 2, 1 2 Iw y 2, 1 2 Iw z 2 Rotational negligible kx2, 1 m ሶ 2 x2 no y,z vibration Vibrational c p = = 1.4 c v 5 Near room temperature, rotational or vibrational DOF are excited, but not both. 18/20

19 11.5 Specific Heat Triatomic gas CO 2 translational 3 rotational 2 vibrational 4 9 c p = c v 7 = 1.28 Vibration modes of CO 2 Stretch Bending 19/20

20 11.5 Specific Heat kt 2 (kinetic) x,y,z direction Solid kt 2 (potential) U = 3NkT c v = 3R(Dulong-Petit Law) 20/20