Zumdahls Chapter 5

1
Zumdahls Chapter 5

• Gases

2
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 nRT

• Gas Stoichiometry
• Partial Pressures and Mole Fractions, Xi
• Effusion
• Diffusion
• Our Atmosphere
• Ideal gas g condensible, heated from the
  bottom
• Real Gases

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-jo
  wl) 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 m2
  1 J m3

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

8
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
  Fmg,
• 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.
• 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.

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.
  Translations 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
Boyles Law PV fixed (iff n,T also)

• P1V1 P2V2 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.

p0
?p 2p0
f ? 2p0
and conservation requires
p0
14
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?

A2 4A1 Is P2 therefore ¼P1?
But V2 8V1, so P2 must be 1/8P1!?!
15
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! P2 1/8 P1 when V2 8 V1. Boyle is right!

?
16
Charless 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 8½ v1 (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 ? 8½ P1 8 P1.
• Want P fixed? Watch how to do it.

17
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 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
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.

?
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 mol1 K1
• If PV is in atm L, then R 0.08206 atm L mol1
  K1
• In fact, R kNAv where k is Boltzmanns 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 isnt 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
Prosaic Problems

• Concentration of O2 in air.
• O2 nO2 / V PO2 / RT
• Need P and T say STP 0C, 1 atm. PO2 0.21 atm
• Must use absolute T, 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

23
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
• VA nA RT / PA, but unless A is pure, PA ?
  Ptotal even though VA Vtotal. So n ? P at
  fixed V too.

24
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 PA PB PC ( Daltons Law fixed V )
• Note the similarity with Avogadros 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
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.)

27
Kinetic Theory and Rates

• Presumption behind rate ? M½ is comparison of
  rates at same T and same P.
• 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
Airs Composition as Mole Fraction

• 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 40C, PH2O 55.3 torr
• 1 torr 1 mm Hg
• 1 bar 750 torr
• ? PH2O 0.0737 bar
• 0.9263?0.78030.7228 N2
• 0.1944 O2
• 0.0087 Ar, etc.
• Avg. MW 0.02816 kg/mol

29
Consequences of Mair

• 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.

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 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.

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 Mair g 1.01325?105 N
• N J m1 kg m2 s2 m1 kg m s2
  in SI
• Mair F / g 1.01325?105 N / 9.80665 m s2
• 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 25C)

g off
g on
33
Real Gas Volume Effect
V
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 molecules
  molar condensed volume, b we have to exclude
  nb from V.
• That gives us the ideal volume the gas is free
  to use.
• So a better gas equation is P ( V nb ) n
  RT
• Waters exptl. b 30.5 ml, while its 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
  aX2 or a( n / V )2 or a n2 / V 2.
• Pressure-corrected, its ( P a n2 / V 2 ) V n
  RT

36
van der Waals Equation

• ( P a n2 / V 2 ) ( V n b ) n RT
• 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,
  whats V ?
• N2O4 isnt a dimer, but HCO2H can dimerize a bit.
• Gases with strong hydrogen-bonds mess with n.

38
NOX Volume Problem

• N2O4 ? 2 NO2 or 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

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
• 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?

40
Influence of Vapor Pressure _at_ 25C
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 Daltons Law?
• Has it gone bankrupt? See Chapter 11. (4)