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## Zumdahl’s Chapter 5

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**Zumdahl’s Chapter 5**Gases**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**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!**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.**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**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?**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.**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.**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.**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**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.**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.**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**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!?!**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! **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.****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**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. **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.**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)**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).**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**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.**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 + …**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.**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**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**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, 5.21018 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 0.92630.7803=0.7228 N2 0.1944 O2 0.0087 Ar, etc. Avg. MW = 0.02816 kg/mol Air’s Composition as Mole Fraction**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.**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!**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.**Height of a Uniform Dry Atmosphere**• P0 = 1 atm = 1.01325105 Pa = 1.01325105 N/m2 • Force on every m2 is F = Mairg = 1.01325105 N • N = J m–1 = kg m2 s–2 m–1 = kg m s–2 in SI • Mair = F/g = 1.01325105 N / 9.80665 m s–2 • Mair = 1.03323104 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**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.**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.**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**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.**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.**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**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 + 1.0241,071 = 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?**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**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)