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Zumdahl’s Chapter 5

Zumdahl’s Chapter 5

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Zumdahl’s Chapter 5

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  1. Zumdahl’s Chapter 5 Gases

  2. Importance of Gases Gas Pressure Kinetic Theory of Gases Gas Laws Boyle: PV constant Charles: V / T constant Avogadro: V /n constant Ideal Gas Law: PV = nRT Gas Stoichiometry Partial Pressures and Mole Fractions, Xi Effusion Diffusion Our Atmosphere Ideal gas + g + condensible, heated from the bottom Real Gases Contents

  3. The Significance of Gases • Gases are elementary phases. • Neither condensed (hence low intermolecular forces) • Nor electrified (as would be plasmas) • Equation of State (n,P,V,T) extremely simple. • Vapor pressures betray the equilibrium balance in solutions and tell us of chemical potential (Gmolar) of solution components!

  4. Relation to Other Phases • Gases share the fluidity of liquids & plasmas but not their nonideal high intermolecular interactions. • Gases share the simplicity of geometry (none) with solids (perfectly regular). • Gases share an equilibrium with all of their condensed phases, and their pressure comments upon the shift of that equilibrium.

  5. Gas Pressure • Gases naturally expand to fill all of their container. • Liquids fill only the lower (gravitational) volume equal to their fixed (molecular-cheek-by-jowl) volume. • Fluids (gases and liquids) exert equal pressure (force) in all directions. • Pressure, P, is the (expansive) force (Newtons) applied per unit area (m2). • Measured with manometers as Pascals = 1 N m–2 = 1 J m–3

  6. Isotropic Fluid Pressure • Auto repair hoists work by isotropic oil pressure. • Pressure on one arm of a fluid-filled “U” is transmitted by the fluid to the other arm, raising the car. • But P equivalence is not just up-down. • A pinhole anywhere leaks. • Gas too is a fluid with isotropic P. • Gravitational force influences P. • What’s wrong with this picture?

  7. Pressure and Gravity • While isotropic at every point, P increases linearly with depth in the sea. • Shallow objects must support only shallow columns of water above them. • Deep objects must bear the weight of the deep columns of water above them. • The linearity follows from water’s constant density, but air’s density varies with pressure hence altitude.

  8. Air Pressure • With gravity, the change, dP, with altitude, dh, varies with the instantaneous density,  = m /V. •  we’ll find is proportional to P. And since F=mg, • dF = gdm or dP = dF/A = gdm/A = gd(h) = – gdh • dP/dh = – aP or dP/P = dlnP = – a dh and P = P0e– ah • a includes g and the proportionality between P and . • And atmospheric P falls off exponentially with altitude, being only ~1/3 atm on top of Everest.

  9. Pressure Rules • While it may be instinctively satisfying that  varies linearly with P, it would be nice to prove it. • We’ll need Boyle’s and Avogadro’s Laws to confirm the atmospheric pressure profile. • They’ll need to turn g off and rely on the inherent expansion of gases. • AND we’ll have to understand Kinetic Theory.

  10. Forces and Molecular Forces • Force = mass times acceleration, like mg • Gravitational force is continuous, but the force of gas pressure is discrete. • The pummeling of molecular collisions may be relentless but it is discontinuous. • F = ma = m dv/dt = d(mv)/dt = dp/dt • “p” = momentum, so F = the rate of momentum change

  11. KINETIC THEORY • The 800 lb gorilla of free molecular motion, and roaring success of Bernoulli, Maxwell, and Herepath. • EQUIPARTITION THEOREM • Every “mode of motion” has average thermal energy of ½kT per molecular motion or ½RT for a mole of them. • It works only for continuous energies; it fails if quantum level energy spacings approach ½kT. Translation’s perfect! • Importance: kinetic energy is fixed at fixed T.

  12. Prerequisites for K.T. of Gases • Molecules might as well be mass points, so distant are they from one another in gases. ID irrelevant. • Those distances imply negligible intermolecule forces, so presume them to be zero.  KE fixed… • Until they hit the walls, and those are the only collisions that count.  dp/dt on walls gives P. • Kinetic Energy, KE, directly proportional to T.

  13. Boyle’s Law: PV fixed (iff n,T also) • P1V1 = P2V2 = PVas long asnandT unchanged! • Invariant T means that average Kinetic Energy remains the same; so we expect the same molecular momenta, p • That means that collisions between the molecules and the wall transfer the same average force, f. + p0 p = –2p0 and conservation requires f +2p0 –p0

  14. Boyle’s Geometry I • Regardless of the volume change, each collision transfers the same impulse to the walls. • But if the dimensions double, there’s more wall, and P is force per unit area of wall! • Doubled dimensions means 4 as much wall; thus P should drop to ¼ its original value? A2 = 4A1 Is P2 therefore ¼P1? But V2 = 8V1, so P2 must be 1/8P1!?!

  15. Boyle’s Geometry II • Ahhh … but we forgot that the molecules have twice as far to fly to get to a wall! • That makes those collisions only ½ as frequent! • The total surface experiences only ½ as many impulses per unit time, so there are ½ as many collisions spread over 4 the area. • Yes! P2 = 1/8P1 when V2 = 8 V1. Boyle is right! 

  16. Charles’s Law: V/T fixed (iff n,P too) • Kinetic Theory helps here. • Imagine a fixed volume heated such that T2 = 8 T1 • That means K.E.2 = 8 K.E.1 or v22 = 8 v12 • More to the point, v2 = 8½v1 (if v is a speed), so wall collisions are 8½ times more frequent. • And molecules have 8½ the momentum when they hit. • Therefore, P2 = 8½  8½P1 = 8 P1. • Want P fixed? Watch how to do it.

  17. Charles’s Law (Geometry) • What we’ve shown is that P/T is fixed when n and V are fixed. Another expression of Charles’s Law. • But if we simply apply our understanding of Boyle to this understanding of (modified) Charles … • Keeping the high T2 fixed, we can expand V to 8V1 which will lower the P2 from 8 P1 to exactly P1. • Thus, T2 = 8 T1 implies V2 = 8 V1 at fixed P. P, V, T 8P, V, 8T P, 8V, 8T

  18. Avogadro’s Law: V/n fixed (iff P,T too) • If we double n, the wall experiences twice the frequency of collisions, but each one has the same force as before. • So P doubles. • To reduce Pback to its original value, Boyle says to double V instead. • So Avogadro is right, if Boyle is right. • And Boyle is right. 

  19. Since Everybody is Right … • What Equation of State embodies Boyle, Charles, and Avogadro all at the same time? • Playing with the algebra, convince yourself that only PV/nT = universal constant works. • Doing any number of gas law experiments reveals that the “Gas Constant,” R = 8.314 J mol–1 K–1 • If PV is in atm L, then R = 0.08206 atm L mol–1 K–1 • In fact, R = kNAv where k is Boltzmann’s Constant.

  20. PV = nRT • From this Ideal Gas equation, much Chemistry flows! • Take density, , for example. •  = m/V = n M /V M is the molar weight of the gas. •  / M = n/V = P/RT •  = P ( M /RT ) It really is proportional to P for an Ideal Gas. • Returning to the Barometric Formula: • dP = –  g dh = – P g ( M /RT ) dh now gives • P = P0 e– ( Mgh /RT )( assuming fixed T which really isn’t the case)

  21. They were All balloonists. • Why do you think Charles was fascinated with the volume of heated air? • When you heat a filled hot air balloon, P and V stay the same, but T increases. How can that be? • Rearrange the i.g. eqn., and n = PV/RT must decrease. • Gas molecules leave the balloon! And  decreases. • hotV is the weight of air left. If  = cold – hot , • ()V is the lifting power of the balloon (air mass gone).

  22. Concentration of O2 in air. [O2] = nO2/V = PO2/RT Need P and T; say STP: 0°C, 1 atm. PO2 = 0.21 atm Must use absoluteT, so the RT = 22.4 L / mol 0.0821 atm L/mol K (273 K) [O2] = 0.21 atm/22.4 L/mol [O2] = 0.0094 M Volume of H2 possible at STP from 10 g Al? Assume excess acid. 3 H+ + Al  Al3+ + 1.5 H2 nH2 = 1.5 nAl nAl = 0.37 mol 10 g (1 mol/27 g) nH2 = 0.55 mol V = nRT/P = 12 L Prosaic Problems

  23. Gas Stoichiometry • Last example was one such; finding gas volume since that’s usually its measure. • While a gas has weight, buoyancy corrections are needed to measure it that way since air as weight too. • So the only new wrinkle added to our usual preoccupation with moles in stoichiometry is: • VA = nART/PA, but unless A is pure, PA  Ptotal even though VA = Vtotal. So n  P at fixed V too.

  24. Dalton’s Law: Partial Pressures • Same guy who postulated atoms as an explanation for combining proportions in molecules went on to explain that partial pressures add to the total P. • Kinetic Theory presumes gas molecules don’t see one another; so they’d contribute independently to the total pressure. Makes sense. • P = PA + PB + PC + … ( Dalton’s Law; fixed V ) • Note the similarity with Avogadro’s Law which states that at the same pressures, V = VA + VB + VC + …

  25. Partial Pressures and Mole Fractions • P = PA + PB + PC + … • n (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V) + … • So n = nA + nB + nC + … (surprise surprise) • Now divide both sides by n, the total number of moles of gas • 1 = XA + XB + XC + … mole fractions sum to 1. • 1 = PA/P + PB/P + PC/P + … • Hence XA = PA/P for gases.

  26. Gas Diffusion Mass transport of molecules from a high concentration region to a low one. Leads to homogeneity. Not instantaneous! Hence molecules must collide and impede one another. Square of diffusion rate is inversely proportional to  Gas Effusion Leakage of molecules from negligible pinhole into a vacuum. Leak must be slow relative to maintenance of the gas’s equilibrium. Square of effusion rate is inversely proportional to  ( proportional to M.) Graham’s Law of Diffusion

  27. Kinetic Theory and Rates • Presumption behind “rate  M–½” is comparison of rates at sameT and sameP. • Fixed T implies same K.E. = ½ m v2 regardless of the identity of the gas molecules! • Thus mA vA2 = mB vB2 or • vA / vB = ( mB / mA )½ = ( MB / MA )½  235UF6 diffuses (352/349)½ = 1.004 faster than 238UF6

  28. Dry Atmosphere; XA 0.7803 N2 0.2099 O2 0.0094 Ar 0.0003 CO2 0.0001 H2 ! Avg MW = 0.02897 kg/mol Mass, 5.21018 kg Standard P, 1 bar = 105 Pa 100% Humid Atmosphere At 40°C, PH2O = 55.3 torr 1 torr = 1 mm Hg 1 bar = 750 torr  PH2O = 0.0737 bar 0.92630.7803=0.7228 N2 0.1944 O2 0.0087 Ar, etc. Avg. MW = 0.02816 kg/mol Air’s Composition as Mole Fraction

  29. Consequences of Mair • Humid air may feel heavier but it’s 3% lighter than dry air. That means a column of it has lower P. • The barometer is lower where it’s stormy, higher where it’s dry. Winds blow from high P to low P. • Since   1 / T, higher T regions have less dense air; so tropics get phenomenal thunderclouds as buoyancy (heat) & incoming wind pile up air to flatiron clouds. • Up to the tropopause where it then spreads horizontally.

  30. Moving Air on a Rotating Earth • Imagine a cannon at the N pole fires a shell at NY that takes an hour to travel. • In that time, the Earth rotates to the next time zone, and the shell hits Chicago instead! • The fusilier thinks his shell curved to the right! • Chicago retaliates by firing back. • But its shell is moving east with the city faster than the ground at higher latitudes. It seems to veer right too!

  31. Coriolis (non)Force • All flying things (in the northern hemisphere) veer right. • Wind approaching a low P region misses the center, veering around to the right in a counterclockwise spiral. • Thus the shape of hurricanes (whose upper air is rained out). • Air “fired” from the tropics moves 1000 mph east. • But so does the ground there; it’s not a problem until … • At about 30° N, the ground (and its air) slows too much, and dry tropopause winds whip down, making deserts.

  32. Height of a Uniform Dry Atmosphere • P0 = 1 atm = 1.01325105 Pa = 1.01325105 N/m2 • Force on every m2 is F = Mairg = 1.01325105 N • N = J m–1 = kg m2 s–2 m–1 = kg m s–2 in SI • Mair = F/g = 1.01325105 N / 9.80665 m s–2 • Mair = 1.03323104 kg = V = Ah ; A = 1 m2 • h = Mair/ A = ( Mair /A )  ( RT/Mair ) /P • h = 8721 m = 8.721 km = 5.420 mi (at 25°C) g off g on

  33. 0.1% V Real Gas: Volume Effect b • Odors do NOT diffuse with the speed of sound; so gas molecules must impede one another by collisions. • Kinetic Theory assumed molecules of zero volume, but that would yield liquids of zero volume as well. No way. • Part of V is always taken up with a molecule’s molar condensed volume, ~ b; we have to excludenb from V. • That gives us the “ideal volume” the gas is free to use. • So a better gas equation is: P ( V – nb ) = nRT • Water’s exptl. b ~ 30.5 ml, while it’s liquid molar volume is 18.0 ml.

  34. Real Gas: Intermolecular Forces • For neutrals, all long-range forces are attractive! • In the bulk of a gas, molecular attractions to nearest neighbors are in all directions; they cancel. • At the wall, such attractions are only from the hemisphere behind; they retard the collider! • He strikes the wall less forcefully than had the intermolecular forces actually been zero.

  35. Real Gases: Pressure Effect • So the measured Pactual is less than the Ideal P. • To use Pactual in the Ideal Gas equation, we must add back that lost molecular momentum. • The strength of intermolecular attraction grows as the square of concentration; so the term is a[X]2 or a( n / V )2 or a n2 / V 2. • Pressure-corrected, it’s ( P + an2/V 2 ) V = n RT

  36. van der Waals’ Equation • ( P + an2/V 2 ) ( V – n b ) = nRT • a and b are empirical parameters. • Ammonia has a large a value of 4.17 atm L2 mol –2 • So at STP, the pressure correction term is 0.0083 atm or almost 1%. Hydrogen bonding has l o n g arms! • Van der Waals’ is an empirical equation and not the only one, but a convenient one for estimates.

  37. Other Non-Idealities • Even if PV = nRT, pressures and volumes can be other than elementary. • An obvious source of mischief is uncertainty in n. • Chemical reaction in the gas phase may change n: • N2O4 2 NO2 K 300 K = 11 • If you evolve 1 mol of N2O4 at 1 atm & 300 K, what’s V ? • N2O4 isn’t a dimer, but HCO2H can dimerize a bit. • Gases with strong hydrogen-bonds mess with n.

  38. N2O4 2 NO2or A  2 B 1 mol of N2O4 evolved at 300 K into Ptotal = 1 atm K = (PB)2 / PA = 11 K = P (XB)2 / XA K / P = XB2 / XA = 11 / 1 (XB)2 / ( 1 – XB) = 11 XB2 + 11 XB – 11 = 0 XB = 0.9226; XA = 0.0774 nB = 2 ( 1 – nA ) n = nA + nB = 2 – nA 1 = ( 2 / n ) – ( nA / n ) 1 = ( 2 / n ) – XA n = 2 / ( 1 + XA ) n = 1.856 V = nRT / P = 45.69 L NOX Volume Problem

  39. Hydrostatic Pressure • Mercury is ~13.6 times as dense as water. • Thus, 1 atm = 0.76 m Hg  13.6 = 10.3 m H2O • Pressure increases by 1 atm with each 33 ft of water. • Mariana (deepest) Trench = 11,033 m for what total pressure? • P = 1 + 11,033 m /10.3 m/atm = 1 + 1,071 atm = 1,072 atm • But seawater has  = 1.024 g/cc, so P = 1 + 1.0241,071 = 1,098 atm • World’s Tallest Tree = 376.5 ft • How does it get water from the roots to its topmost leaves? • Pull a vacuum of NEGATIVE 10 atm?!? No … • And what about water’s vapor pressure?

  40. Hg(g) 2 m H2O(g) 24 mm 760  ethanol(g) 59 mm 736  701  ether(g) 538 mm 222  Hg Hg Hg Hg Influence of Vapor Pressure @ 25°C Acetone 231 mm Methanol 127 mm Propanal 317 mm

  41. Gedanken Experiment • Gedanken is German for “thought.” Einstein loved them. • Pether + Pacetone = 538 mm + 231 mm = 769 mm • Does this mean that the total pressure for those two liquids will exceed 1 atm? • If so, how about 1000 liquids with vapor pressures of, say, ½ atm each. Would they exert 500 atm?!? • If not, what happened to Dalton’s Law? • Has it gone bankrupt? See Chapter 11. (§4)