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Zumdahls Chapter 5 Gases Contents Importance of Gases Gas Pressure Kinetic Theory of Gases Gas Laws Boyle: PV constant Charles: V / T constant Avogadro:

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Presentation on theme: "Zumdahls Chapter 5 Gases Contents Importance of Gases Gas Pressure Kinetic Theory of Gases Gas Laws Boyle: PV constant Charles: V / T constant Avogadro:"— Presentation transcript:


2 Zumdahls Chapter 5 Gases

3 Contents 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 = n R T Gas Stoichiometry Partial Pressures and Mole Fractions, X i Effusion Diffusion Our Atmosphere Ideal gas + g + condensible, heated from the bottom Real Gases

4 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 (G molar ) of solution components!

5 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.

6 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 (m 2 ). Measured with manometers as Pascals = 1 N m –2 = 1 J m –3

7 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. Whats wrong with this picture?

8 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 waters constant density, but airs density varies with pressure hence altitude.

9 Air Pressure With gravity, the change, dP, with altitude, dh, varies with the instantaneous density, = m / V. well 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 = P 0 e – 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.

10 Pressure Rules While it may be instinctively satisfying that varies linearly with P, it would be nice to prove it. Well need Boyles and Avogadros Laws to confirm the atmospheric pressure profile. Theyll need to turn g off and rely on the inherent expansion of gases. AND well have to understand Kinetic Theory.

11 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

12 KINETIC THEORY The 800 lb gorilla of free molecular motion, and roaring success of Bernoulli, Maxwell, and Herepath. EQUIPARTITION THEOREM 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. Translations perfect! Importance: kinetic energy is fixed at fixed T.

13 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.

14 Boyles Law: PV fixed (iff n,T also) P 1 V 1 = P 2 V 2 = PV as long as n and T 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. + p 0 – p0– p0 p = –2p 0 f +2p 0 and conservation requires

15 Boyles Geometry I Regardless of the volume change, each collision transfers the same impulse to the walls. But if the dimensions double, theres 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? A 2 = 4A 1 Is P 2 therefore ¼P 1 ? But V 2 = 8V 1, so P 2 must be 1/8P 1 !?!

16 Boyles 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! P 2 = 1/8 P 1 when V 2 = 8 V 1. Boyle is right!

17 Charless Law: V/T fixed (iff n,P too) Kinetic Theory helps here. Imagine a fixed volume heated such that T 2 = 8 T 1 That means K.E. 2 = 8 K.E. 1 or v 2 2 = 8 v 1 2 More to the point, v 2 = 8 ½ v 1 (if v is a speed), so wall collisions are 8 ½ times more frequent. And molecules have 8 ½ the momentum when they hit. Therefore, P 2 = 8 ½ 8 ½ P 1 = 8 P 1. Want P fixed? Watch how to do it.

18 Charless Law (Geometry) What weve shown is that P/T is fixed when n and V are fixed. Another expression of Charless Law. But if we simply apply our understanding of Boyle to this understanding of (modified) Charles … Keeping the high T 2 fixed, we can expand V to 8V 1 which will lower the P 2 from 8 P 1 to exactly P 1. Thus, T 2 = 8 T 1 implies V 2 = 8 V 1 at fixed P. P, V, T8P, V, 8TP, 8V, 8T

19 Avogadros 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 P back to its original value, Boyle says to double V instead. So Avogadro is right, if Boyle is right. And Boyle is right.

20 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 = kN Av where k is Boltzmanns Constant.

21 PV = n R T 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 / R T = P ( M / R T ) It really is proportional to P for an Ideal Gas. Returning to the Barometric Formula: dP = – g dh = – P g ( M / R T ) dh now gives P = P 0 e – ( Mgh / R T ) ( assuming fixed T which really isnt the case)

22 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 / R T must decrease. Gas molecules leave the balloon! And decreases. hot V is the weight of air left. If = cold – hot, ( )V is the lifting power of the balloon (air mass gone).

23 Prosaic Problems Concentration of O 2 in air. [O 2 ] = n O2 / V = P O2 / R T Need P and T; say STP: 0°C, 1 atm. P O2 = 0.21 atm Must use absolute T, so the R T = 22.4 L / mol 0.0821 atm L/mol K (273 K) [O 2 ] = 0.21 atm/22.4 L/mol [O 2 ] = 0.0094 M Volume of H 2 possible at STP from 10 g Al? Assume excess acid. 3 H + + Al Al 3+ + 1.5 H 2 n H2 = 1.5 n Al n Al = 0.37 mol 10 g (1 mol/27 g) n H2 = 0.55 mol V = n R T/P = 12 L

24 Gas Stoichiometry Last example was one such; finding gas volume since thats 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: V A = n A R T / P A, but unless A is pure, P A P total even though V A = V total. So n P at fixed V too.

25 Daltons 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 dont see one another; so theyd contribute independently to the total pressure. Makes sense. P = P A + P B + P C + … ( Daltons Law; fixed V ) Note the similarity with Avogadros Law which states that at the same pressures, V = V A + V B + V C + …

26 Partial Pressures and Mole Fractions P = P A + P B + P C + … n ( R T/V) = n A ( R T/V) + n B ( R T/V) + n C ( R T/V) + … So n = n A + n B + n C + … (surprise surprise) Now divide both sides by n, the total number of moles of gas 1 = X A + X B + X C + … mole fractions sum to 1. 1 = P A /P + P B /P + P C /P + … Hence X A = P A /P for gases.

27 Grahams Law of Diffusion 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 gass equilibrium. Square of effusion rate is inversely proportional to ( proportional to M.)

28 Kinetic Theory and Rates Presumption behind rate M –½ is comparison of rates at same T and same P. Fixed T implies same K.E. = ½ m v 2 regardless of the identity of the gas molecules! Thus m A v A 2 = m B v B 2 or v A / v B = ( m B / m A ) ½ = ( M B / M A ) ½ 235 UF 6 diffuses (352/349) ½ = 1.004 faster than 238 UF 6

29 Airs Composition as Mole Fraction Dry Atmosphere; XAXA 0.7803 N2N2 0.2099 O2O2 0.0094 Ar 0.0003 CO 2 0.0001 H2 H2 ! Avg MW = 0.02897 kg/mol Mass, 5.2 10 18 kg Standard P, 1 bar = 10 5 Pa 100% Humid Atmosphere At 40°C, P H2O = 55.3 torr 1 = 1 mm Hg 1 bar = 750 torr P H2O = 0.0737 bar 0.9263 0.7803=0.7228 N2N2 0.1944 O2O2 0.0087 Ar, etc. Avg. MW = 0.02816 kg/mol

30 Consequences of M air Humid air may feel heavier but its 3% lighter than dry air. That means a column of it has lower P. The barometer is lower where its stormy, higher where its 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.

31 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!

32 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; its not a problem until … At about 30° N, the ground (and its air) slows too much, and dry tropopause winds whip down, making deserts.

33 Height of a Uniform Dry Atmosphere P 0 = 1 atm = 1.01325 10 5 Pa = 1.01325 10 5 N/m 2 Force on every m 2 is F = M air g = 1.01325 10 5 N N = J m –1 = kg m 2 s –2 m –1 = kg m s –2 in SI M air = F / g = 1.01325 10 5 N / 9.80665 m s –2 M air = 1.03323 10 4 kg = V = Ah ; A = 1 m 2 h = M air / A = ( M air / A ) ( R T / M air ) / P h = 8721 m = 8.721 km = 5.420 mi (at 25°C) g on g off

34 Real Gas: Volume Effect 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 molecules molar condensed volume, ~ b ; we have to exclude n b from V. That gives us the ideal volume the gas is free to use. So a better gas equation is: P ( V – n b ) = n R T Waters exptl. b ~ 30.5 ml, while its liquid molar volume is 18.0 ml. V b 0.1%

35 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.

36 Real Gases: Pressure Effect So the measured P actual is less than the Ideal P. To use P actual 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 n 2 / V 2. Pressure-corrected, its ( P + a n 2 / V 2 ) V = n R T

37 van der Waals Equation ( P + a n 2 / V 2 ) ( V – n b ) = n R T a and b are empirical parameters. Ammonia has a large a value of 4.17 atm L 2 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.

38 Other Non-Idealities Even if PV = n R T, 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: N 2 O 4 2 NO 2 K 300 K = 11 If you evolve 1 mol of N 2 O 4 at 1 atm & 300 K, whats V ? N 2 O 4 isnt a dimer, but HCO 2 H can dimerize a bit. Gases with strong hydrogen-bonds mess with n.

39 NO X Volume Problem N 2 O 4 2 NO 2 or A 2 B 1 mol of N 2 O 4 evolved at 300 K into P total = 1 atm K = (P B ) 2 / P A = 11 K = P (X B ) 2 / X A K / P = X B 2 / X A = 11 / 1 (X B ) 2 / ( 1 – X B ) = 11 X B 2 + 11 X B – 11 = 0 X B = 0.9226; X A = 0.0774 n B = 2 ( 1 – n A ) n = n A + n B = 2 – n A 1 = ( 2 / n ) – ( n A / n ) 1 = ( 2 / n ) – X A n = 2 / ( 1 + X A ) n = 1.856 V = n R T / P = 45.69 L

40 Hydrostatic Pressure Mercury is ~13.6 times as dense as water. Thus, 1 atm = 0.76 m Hg 13.6 = 10.3 m H 2 O 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 Worlds 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 waters vapor pressure?

41 Influence of Vapor Pressure @ 25°C Hg 760 Hg( g ) 2 m Hg 736 H 2 O( g ) 24 mm Hg 701 ethanol( g ) 59 mm Hg 222 ether( g ) 538 mm Acetone 231 mm Methanol 127 mm Propanal 317 mm

42 Gedanken Experiment Gedanken is German for thought. Einstein loved them. P ether + P acetone = 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 Daltons Law? Has it gone bankrupt? See Chapter 11. (§4)

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