Chemistry C-2407Chemistry C-2407Course InformationCourse InformationName: Name: Intensive General ChemistryIntensive General Chemistry

Instructor: Instructor: George FlynnGeorge Flynn

Office: Office: 501 Havemeyer Extension501 Havemeyer ExtensionMail Stop 3109, 3000 Broadway,Havemeyer HallMail Stop 3109, 3000 Broadway,Havemeyer HallPhone: Phone: 212 854 4162212 854 4162Email: Email: flynn@chem.columbia.eduflynn@chem.columbia.eduFAX: FAX: 212 854 8336212 854 8336

Office hours: Office hours: TuTh: 2:30-3:00 or by appointmentTuTh: 2:30-3:00 or by appointment

Website Address for course: Website Address for course: http://www.columbia.edu/itc/chemistry/chem-c2407/http://www.columbia.edu/itc/chemistry/chem-c2407/

Chemistry C-2407Chemistry C-2407More Course InformationMore Course Information

Required Recitations: Required Recitations: M, 3-5, 6-8; Tu, 3-5; W, 4-6; F, 10-12, 2-4M, 3-5, 6-8; Tu, 3-5; W, 4-6; F, 10-12, 2-4[Sign up for [Sign up for oneone only!] only!]

These are held in the Chemistry Computer RoomThese are held in the Chemistry Computer Room

Room 211 HavemeyerRoom 211 HavemeyerTelephone Registration: Course # C2409Telephone Registration: Course # C2409

Teaching Assistants: Teaching Assistants: Jennifer InghrimJennifer Inghrim (jai2002@columbia.edu, (jai2002@columbia.edu, mail box #3133, Havemeyer Hall) mail box #3133, Havemeyer Hall) (854-4964);;Office Hour: Wednesdays 10:00-11:00, Room 343 HavemeyerOffice Hour: Wednesdays 10:00-11:00, Room 343 Havemeyerand Sean Moran (sdm2007@columbia.edu, and Sean Moran (sdm2007@columbia.edu, mail box 3139, Havemeyer Hall) (854 8468);mail box 3139, Havemeyer Hall) (854 8468);Office Hour: Tuesdays 2:00-3:00, Room 343 HavemeyerOffice Hour: Tuesdays 2:00-3:00, Room 343 Havemeyer

First Required Recitation: Tomorrow, Wednesday, First Required Recitation: Tomorrow, Wednesday, September 3, 2003--September 3, 2003--Bring a blank, unformatted floppy disk!Bring a blank, unformatted floppy disk!

A Little ObservationA Little Observation

Consider 1 mole of water molecules. HConsider 1 mole of water molecules. H22O has a O has a

molecular weight of 18 gm/mole.molecular weight of 18 gm/mole.

So, 6.02x10So, 6.02x102323 molecules weigh 18 gm molecules weigh 18 gm

But, 18 gm of liquid water occupies 18 ml volume (18 cmBut, 18 gm of liquid water occupies 18 ml volume (18 cm33))

Liquid water is pretty incompressible. So we guess that theLiquid water is pretty incompressible. So we guess that thewater molecules in the liquid are literally in contact with each water molecules in the liquid are literally in contact with each other. No significant space in between.other. No significant space in between.

So the actual volume of one water molecule must be roughlySo the actual volume of one water molecule must be roughly18 ml/6.02x1018 ml/6.02x102323= 3x10= 3x10-23-23 ml ml

Contrast this with the volume occupied by a water molecule inContrast this with the volume occupied by a water molecule inthe gas phase. To find this number, treat water as an ideal gasthe gas phase. To find this number, treat water as an ideal gasat 300 K and use the ideal gas law to compute the volume:at 300 K and use the ideal gas law to compute the volume:

pV=nRT with n=1 mole, p= 1 atm. and R=0.082 pV=nRT with n=1 mole, p= 1 atm. and R=0.082 ll-atm/mole-deg-atm/mole-deg

V=(1)(.082)(300)/(1)= 24.6 litersV=(1)(.082)(300)/(1)= 24.6 liters

Or, the volume occupied per molecule is 24,600ml/6.023x10Or, the volume occupied per molecule is 24,600ml/6.023x102323==4.1x104.1x10-20-20 ml ml

Thus the ratio of the volume occupied by an ideal gas molecule Thus the ratio of the volume occupied by an ideal gas molecule to its actual volume is 24600/18=1400 !!to its actual volume is 24600/18=1400 !!

Compare!Compare!

This leads us to conclude that molecules in an ideal gas mustThis leads us to conclude that molecules in an ideal gas mustbe far apart, “rarely” bumping into each other. The volume be far apart, “rarely” bumping into each other. The volume of an actual molecule is tiny by comparison to the volume of an actual molecule is tiny by comparison to the volume occupied at 1 atmosphere and 300K in the gas phase.occupied at 1 atmosphere and 300K in the gas phase.

The picture we walk away with for the gaseous state of The picture we walk away with for the gaseous state of a molecule in an ideal gas is one of huge empty spacesa molecule in an ideal gas is one of huge empty spacesbetween molecules with “rare” collisions. between molecules with “rare” collisions.

This will allow us to develop a simple This will allow us to develop a simple MODELMODEL of the of thegaseous state which provides remarkable insight intogaseous state which provides remarkable insight intothe properties of molecules and matter.the properties of molecules and matter.

This This MODELMODEL is called the Kinetic Theory of Gases. is called the Kinetic Theory of Gases.

Kinetic Theory PreliminariesKinetic Theory Preliminaries1) Particle velocity v includes both the speed ( c ) of a

particle (cm/s) and its direction. In one dimension:

3) Change in Momentum for an elastic collision: An elastic collision is one where speed is the same before and after the collision:

0

-x +x

0

-x +x

2) Particle Momentum is P = mv (not mc). In one dimension:

m gm at c = 10 cm/sgm at c = 10 cm/sP = - 10P = - 10m gm-cm/s gm-cm/s

m gm at c = 10 cm/s gm at c = 10 cm/sP = + 10P = + 10m gm-cm/s gm-cm/s

v = +10 cm / s, c = 10 cm / sv = +10 cm / s, c = 10 cm / s

v = -10 cm / s, c = 10 cm / sv = -10 cm / s, c = 10 cm / s

-x-x

v =+10cm/sv =+10cm/s

c=10cm/sc=10cm/s

WallWall

+x+x

{v = -10cm/sv = -10cm/sc=10cm/sc=10cm/s

4) Conservation of Momentum: what particle loses,4) Conservation of Momentum: what particle loses, wall must gainwall must gain

∆∆PPwallwall = +2mc = +2mc ∆P∆Pww + ∆P + ∆Pmm = 0 = 0

∆∆P = PP = Pff - P - Pii = -mc - (mc) = -2mc [ = -mc - (mc) = -2mc [∆∆ is the symbol for “change”] is the symbol for “change”]

4LConser vat i onof Moment um: whatpar t i cl el oses, wal lmustgai n

PPfinalfinal = -10m(gm-cm/s) = mv = -mc = -10m(gm-cm/s) = mv = -mc

PPinitialinitial = +10m(gm-cm/s) = mv = mc = +10m(gm-cm/s) = mv = mc

Force is change in momentum with time:

F = dP / dtF = dP / dt

F = ma = m∆v / ∆t = ∆P / ∆tF = ma = m∆v / ∆t = ∆P / ∆t

6) Force = F = ma6) Force = F = ma

a = ∆v / ∆ta = ∆v / ∆t ( a = dv / dt )( a = dv / dt )

5) Acceleration 5) Acceleration aa is is (velocity) / (velocity) / (time). (time). a like v has direction.a like v has direction.

Kinetic Theory of GasesKinetic Theory of GasesAssumptionsAssumptions1) Particles are point mass atoms1) Particles are point mass atoms (volume = 0) (volume = 0)

2) No attractive forces between atoms.2) No attractive forces between atoms. Behave independently Behave independently except for brief moments of collision.except for brief moments of collision.

Model SystemModel SystemA box of volume V with N atoms of mass m all moving withA box of volume V with N atoms of mass m all moving with the same speed c.the same speed c.

We wish to calculate the pressure exerted by the gas on theWe wish to calculate the pressure exerted by the gas on the walls of the box:walls of the box:

V,V, N,N, m,m, cc

Force of atom impinging on wall creates pressure Force of atom impinging on wall creates pressure that we can that we can measure [measure [ppV = nRT].V = nRT].

Typical Path for a gas atom or molecule in a box. Typical Path for a gas atom or molecule in a box.

Pressure Pressure Force / unit area Force / unit areaThus, we need to find force exerted by atoms Thus, we need to find force exerted by atoms on the the wall of the box.on the the wall of the box.

Let’s try to calculate the force exerted by the gas on a segmentLet’s try to calculate the force exerted by the gas on a segment of the box wall having area A.of the box wall having area A.To do this we will make one more simplifying assumption:To do this we will make one more simplifying assumption:

We assume that all atoms move either along the x, y, or z axesWe assume that all atoms move either along the x, y, or z axes but not at any angle to these axes!but not at any angle to these axes! (This is a silly assumption (This is a silly assumptionand, as we shall see later, causes some errors that we must and, as we shall see later, causes some errors that we must correct.)correct.)

F = ma = m( ∆v )/∆tF = ma = m( ∆v )/∆t(∆P)/∆t=(Change in momentum) / (change in time)(∆P)/∆t=(Change in momentum) / (change in time)

Vectors and Vector Components:Vectors and Vector Components:The MovieThe MovieZZ

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