Heat Capacity Summary for Ideal Gases:Heat Capacity Summary for Ideal Gases:CCvv = (3/2) R, KE change only.= (3/2) R, KE change only.Note, Note, CCvv independent of T.independent of T.CCpp = (3/2) R + R, KE change + work. = (3/2) R + R, KE change + work.Also IndependentAlso Independentof Tof TCCpp//CCvv = [(5/2)R]/[(3/2)R] = 5/3= [(5/2)R]/[(3/2)R] = 5/3CCpp//CCvv = 1.67= 1.67Find for monatomic ideal gases such as He, Find for monatomic ideal gases such as He, XeXe, , ArAr, Kr, , Kr, NeNe CCpp//CCvv = 1.67 = 1.67For For diatomics diatomics and and polyatomics polyatomics find Cfind Cpp//CCvv < 1.67!< 1.67!Since work argument aboveSince work argument aboveP(VP(V22 - V - V11) = RT is simple and holds for all gases,) = RT is simple and holds for all gases,This suggests KE > (3/2)RT for This suggests KE > (3/2)RT for diatomicsdiatomics,,This would make CThis would make Cpp//CCvv < 1.67 < 1.67EquipartitionEquipartition Theorem: Theorem: This is a very general law which This is a very general law which states that for a molecule or atom:states that for a molecule or atom:KE = (1/2)KE = (1/2)kTkT (or 1/2 RT on a mole basis) (or 1/2 RT on a mole basis) perper degree of freedomdegree of freedom..A possible solution:A possible solution:A degree of freedom is a coordinate needed to describeA degree of freedom is a coordinate needed to describe position of a molecule in space. position of a molecule in space.ThusThusKE = 312kTÊ Ë ˆ ¯ =32kTas for a monatomic gasas for a monatomic gasA diatomic molecule is a line (2 points connected by aA diatomic molecule is a line (2 points connected by a chemical bond). It requires 5 coordinates to describe its chemical bond). It requires 5 coordinates to describe its position: x, y, z, position: x, y, z, qq, , jjKE = 512kTÊ Ë ˆ ¯ =52kT(x,y,z)jjqExample: A point has 3 degrees of freedom because Example: A point has 3 degrees of freedom because it requires three coordinates to describe its it requires three coordinates to describe its position: (x, y, z).position: (x, y, z).ZZXXYY(Extra KE comes from Rotation of(Extra KE comes from Rotation ofdiatomic molecule!)diatomic molecule!)Bonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus * Bonus * Bonus * BonusCollision Frequency and Mean Free PathCollision Frequency and Mean Free PathReal gases consist of particles of finite size that bumpReal gases consist of particles of finite size that bump into each other.into each other.Let gas molecules be spheres of radius Let gas molecules be spheres of radius ss or diameter 2 or diameter 2 ss = = rr..Focus on one molecule (say a Focus on one molecule (say a redred one) flying through a one) flying through a background of other molecules (say blue ones). background of other molecules (say blue ones). ¨r¨rÆÆMake the simplifying assumption that only theMake the simplifying assumption that only the redred one is moving. (Will fix later.) one is moving. (Will fix later.)LL = c= c¥¥1s1sV/sec = {V/sec = {ππ rr22}[c]}[c]= {A} = {A} ¥¥ [[ LL / t / t ]]ØØ HitHitMissMissÆÆ↑↑ HitHitHitHitØØ ↑↑ MissMissThe red molecule sweeps out a cylinder of volumeThe red molecule sweeps out a cylinder of volumeprpr22cc in one second. It will collide with any molecules in one second. It will collide with any moleculeswhose centerswhose centers lie within the cylinder. Note that lie within the cylinder. Note thatthe (collision) cylinder radius is the diameter the (collision) cylinder radius is the diameter rr of ofthe molecule NOT its radius the molecule NOT its radius ss!!Note: Note: A= A= pp rr22Gas Kinetic Collision CylinderGas Kinetic Collision CylinderrrrrrrØØ↑↑rr=2=2ss2 2 ss ssrr=2=2ssRed Molecule R sweeps out a Red Molecule R sweeps out a Cylinder of volume Cylinder of volume ππrr22c per c per second (c = speed).second (c = speed).If another molecule has some part in this volume, If another molecule has some part in this volume, VVcc = = pp((rr))2 2 c, c, it will suffer a collision with the red molecule.it will suffer a collision with the red molecule.Gas KineticGas KineticCollisionCollisionCylinderCylinderOn average the collision frequency z will be:On average the collision frequency z will be:z = [volume swept out per second] z = [volume swept out per second] ¥¥ [molecules per unit volume] [molecules per unit volume]z =pr2c ¥NVÊ Ë ˆ ¯ Make the simplifying assumption that only theMake the simplifying assumption that only the redred one is moving. (Will fix later.) one is moving. (Will fix later.)Mean Free Path Mean Free Path ≡≡ average distance traveled between collisions: average distance traveled between collisions:c =distancesec,z =collisionssecll = mean free path, = mean free path,l=czSome typical numbers: STP: 6.02 Some typical numbers: STP: 6.02 ¥¥ 10 102323 molecules / 22.4 liters molecules / 22.4 litersNV=2.69 2.69 ¥¥ 10 101919 molecule / cm molecule / cm33; ; rr ≈≈ 3.5 3.5 ÅÅ;;ππ rr22 = 38.5 = 38.5 ¥¥ 10 10 -16 -16 cm cm2 2 ; c = 4 ; c = 4 ¥¥ 10 1044 cm / sec. cm / sec.ll = 1/[ = 1/[prpr22(N/V)](N/V)]Assumes volume swept out is independent of whetherAssumes volume swept out is independent of whether collisions occur (not a bad assumption in most cases) collisions occur (not a bad assumption in most cases)z z = = prpr2 2 c (N/V)c (N/V)ll >> >> rr, [, [rr ≈≈ 3.5 3.5¥¥1010-8-8 cm], consistent with our initial assumption cm], consistent with our initial assumptionfrom the Kinetic Theory of gases!from the Kinetic Theory of gases!P = 1 P = 1 atmatmÆÆz = 4.14 z = 4.14 ¥¥ 10 1099 sec sec -1-1(P = 1 (P = 1 atmatm))P = 1 P = 1 TorrTorrÆÆz = 5.45 z = 5.45 ¥¥ 10 1066 sec sec -1 -1 (P = 1/760 (P = 1/760 atmatm))lP = 1atm( )=4 ¥ 104cm / sec4.14 ¥ 109collisions / sec= 9.7 ¥10-6cm / collisionlP = 1Torr( )=4 ¥ 104cm / sec5.45 ¥ 106collisions / sec-1= 7.3 ¥ 10-3cm / collision(P = 1 (P = 1 atmatm))(P = 1/760 (P = 1/760 atmatm))Distribution of Molecular SpeedsDistribution of Molecular SpeedsReal gases do not have a single fixed speed. Rather moleculesReal gases do not have a single fixed speed. Rather molecules have speeds that vary giving a have speeds that vary giving a speed distributionspeed distribution..This distribution can be measured in a laboratory (done at This distribution can be measured in a laboratory (done at Columbia by Columbia by Polykarp KuschPolykarp Kusch) or derived from theoretical ) or derived from theoretical