10.19.2011 - Lecture 18 Wednesday October 19 2011 Today in...

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Unformatted text preview: Lecture 18 Wednesday, October 19, 2011 Today in Chemistry 260 •  The kinetic theory of gases •  Graham’s Law of Effusion •  Collisions and mean free path This Week in Chemistry 260 •  Reading: (Fri.) 12.1,12.2 •  Kinetic theory of gases •  First law of thermodynamics, heat capacities, thermodynamics of gases •  Problem set 6 due Friday (Oct. 21). Image from wikipedia The van der Waals equation of state The van der Waals equation of state 2\$ ! ! − nb\$ # P + a n V &"V % = nRT " % ( Measured pressure ) Effective volume Measured volume ! \$! − nb\$ # P + a n V &"V % = nRT " % Atoms and molecules 2 are not inRinitely small. At very close distances, there are repulsive interactions between the particles ( ) Measured volume Repulsive interactions (excluded volume) The would- have- been pressure in the absence of attractive interactions Veff = V - nb nb The measured volume is bigger than expected because it is the effective volume + the volume of the atoms or molecules. Measured pressure Constant for a particular gas ( ⎛ྎ n a⎜ྎ ⎝ྎ V Pairs of 2 particles ⎞ྏ interact ⎟ྏ ⎠ྏ # of particles/unit volume (~how close the particles are to each other) The measured pressure is smaller than expected because it is the effective pressure - the inRluence of the attractive forces. Attractive interactions Reduce the rate of collisions with the wall, ) Peff > Pmeasured Veff < Vsystem Z = PVideal / PVreal = V / (V − nb) > 1 P = Peff – a(n/V)2 All particles have long range attractive potentials … Vdd(r), Vind- ind(r), etc.) Constant for a particular gas 2\$ ! ! − nb\$ # P + a n V &"V % = nRT " % ) The would- have- been pressure in the absence of attractive interactions # of particles The van der Waals equation of state 2\$ ! ! − nb\$ # P + a n V &"V % = nRT " % ( Repulsive interactions (excluded volume) Note that at smaller volumes (same # of particles), that the – nb term is more important V = Veff + nb The van der Waals equation of state Effective volume 2.0 Z= PV RT 1.5 1.0 0.5 Z=1 Ideal gas 0 1 Some relevant values: The kinetic theory of gases: Compound a (L2atm/mol2) b (L/mol) He Ne H2 0.03412 0.2107 0.2444 0.02370 0.01709 0.02661 Ar O2 1.345 1.360 0.03219 0.03803 N2 1.390 0.03913 CO CH4 CO2 1.485 2.253 3.592 0.03985 0.04278 0.04267 NH3 4.170 So far we have had a rather phenomenological, macroscopic view of gases … Pressure, Volume, Temperature… 0.03707 The kinetic theory of gases: PV=nRT How do we connect macroscopic to the microscopic? The kinetic theory of gases: a constant Ideal Gas Assumptions 1.  A gas consists of a large number of molecules separated by distances large compared to their size. DeRine a cube box of side L. 2.  The molecules have mass – but no volume (negligibly small) PV=nRT V = L3 DeRine a cube box of side L. n = N/NA Now, what can we say about P and/or T? L 3.  The molecules do not interact, no attraction and only hard sphere repulsion. L L 4.  Molecules are in constant random motion 5.  Collisions between molecules and of molecules with the wall are elastic. That is: the velocity change is only a change in sign. What is Pressure? N molecules in the box What is Pressure? L velocity = vx L 1. What is pressure? - vx Consider one molecule moving in one dimension L v The molecule has x collisions with the walls. L Due to the change in velocity each collision exerts pressure on the wall. Force mass×acceleration P= = Area Area L velocity = vx vx Consider one molecule moving in one dimension L Force mass×acceleration P= = Area Area m ⎛ྎ Δv x ⎞ྏ ⎛ྎ collisions ⎞ྏ = 2 × ⎜ྎ ⎟ྏ ⎟ྏ × ⎜ྎ L ⎝ྎ collision ⎠ྏ ⎝ྎ second ⎠ྏ = m ⎛ྎ 1 v x ⎞ྏ × ( 2v x ) × ⎜ྎ ⎟ྏ L2 ⎝ྎ 2 L ⎠ྏ ⇒P= mv 2 x L3 ⇒ PV = mv 2 x We’re only considering one of the walls of the cube 2 What is Pressure? The kinetic theory of gases: PV=nRT Now consider N molecules moving in one dimension 1 ⇒ PV = N m u 2 3 2 L PV = mv x one molecule Ok – we have P, V and n = N/NA We’ve already accounted for motion in one direction with the factor of ½. L average value of vx over the distribution of all N molecules. u = v = speed L Motion in 3 dimensions. vy The atoms/molecules are moving randomly in all directions (no net Rlow). 2 y v =v =v Δv = 2vx -vx 2 z 1 1 ⇒ PV = N m v 2 = N mu 2 3 3 vy vx changes, vy and vz do not. What is Temperature? 3 RT 2 NA 3 = k BT 2 Etrans 1 1 RT = M u 2 = N A m u 2 3 3 1 ⇒ PV = n M u 2 3 Ideal gas law 1 RT = M u 2 3 ⇒ 3RT M urms = root mean square (rms) speed urms is proportional to T½ & M- ½ average kinetic energy Etrans = What about Temperature – What is Temperature? The Maxwell Distribution of Speeds RT 1 2 2 1 = mu = mu2 NA 3 32 1 3 RT mu2 = 2 2 NA M = mNA vx u 2 = v2 = v2 + v2 + v2 x y z 2 x 1 ⇒ PV = n M u 2 3 PV = Nmv 2 ⇒ PV = Nm v 2 x x To take this further we need a distribution function that describes the likelihood of any individual speed. R – Gas Constant 8.3145 J K-1 mol-1 u = ∫ u f (u) d u R ≡ kB NA u, speed of a particular atom/molecule 3 # M \$ 2 2 − Mu 2 2 RT f (u ) = 4π % & ue ' 2π RT ( kB – Boltzmann Constant 1.381x10-23 J K-1 James Clerk Maxwell (1831- 1879) - or- 3 # m \$ 2 2 − mu 2 2 kBT f (u ) = 4π % & ue ' 2π k BT ( Temperature is essentially a measure of the average kinetic energy of the atoms/molecules in the gas Macro- Micro Relations Graham’s Law of Effusion Collision with a wall: The Maxwell Distribution of Speeds gives us information regarding how speed and temperature are related. EFFUSION Zw ∝ à༎ This form is developed speciRically for IDEAL GASES à༎ Similar expressions exist for other phases ⇒ ump = Zw = Peak of distribution 8RT 2.55 RT ≈ πM M u = ∫ u f (u) d u 3RT M Escape of gas through a small hole Graham’s law [1833]: root mean square speed ⇒ urms = 1N uA 4V Rate of effusion: average speed ⇒u = N uA V u2 = cross sectional area average speed Number density = amount per unit volume maximum or most probable speed 2 RT M Collision rate ∫u 2 f (u ) d u ZA NA = ZB NB MB MA The proportionality constant is ¼. Zw = 1N 4V 8RT A πM Helium balloons deRlate … M is low ⇒ rate is high. Used for purifying 235UF from 238UF 6 6 3 Other collisions d A molecule of diameter d sweeps out a volume V – the volume depends on the time d V = π d 2u t d ncollisions = 2 This can be turned around to give the mean free path – the average distance traveled between collisions. Mean free path N Vcylinder V Rate of collision: l1 ncollisions t N = 2 π d 2u V N 8RT = 2 πd2 V πM N 2 π RT d V M l2 l8 l3 l4 λ =l = l7 l6 Z coll = =4 Recap of kinetic model Collisions will occur with other molecules in this volume l9 l5 u Z coll DIFFUSION V λ= 2 Nπ d 2 D= 3π 3 RT V λu = 16 8 π M Nd 2 Diffusion coefRicient Connects microscopic properties (speed, # collisions, etc.) to macroscopic properties (volume, pressure, temperature, etc.) Based on ideal gas assumptions: •  Molecules very small, don't interact, have random motion, elastic collisions •  Pressure related to speed •  Temperature related to average kinetic energy T= Mingling of different substances, e.g., gases PV=nRT 2 Etrans 3 kB 1 PV = N m u 2 3 Etrans = Pressure and speed of molecules related 3 k BT 2 Temperature is essentially a measure of the average kinetic energy of the atoms/molecules in the gas 4 ...
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