Zumdahl’s Chapter 5

Gases

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

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) = – gdhdP/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 THEORYKINETIC THEORYThe 800 lb gorilla of free molecular motion, and

roaring success of Bernoulli, Maxwell, and Herepath. EQUIPARTITION THEOREMEQUIPARTITION 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 = 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, pThat means that collisions between the molecules and

the wall transfer the same average force, f.

+ p0

– p0

p = –2p0 f +2p0and conservation requires

Boyle’s Geometry IRegardless 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 = 1/8 P1 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 v1

2

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.

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 P back 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 = nRTFrom 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).

Prosaic Problems

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 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

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

Dalton’s Law: Partial PressuresSame 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.

Graham’s 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 gas’s equilibrium.

Square of effusion rate is inversely proportional to

( proportional to M.)

Kinetic Theory and RatesPresumption behind “rate M–½” 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 vA

2 = mB vB2 or

vA / vB = ( mB / mA )½ = ( MB / MA )½ 235UF6 diffuses (352/349)½ = 1.004 faster than 238UF6

Air’s 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, 5.21018 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

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 = Mair g = 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 on

g off

Real Gas: Volume EffectOdors 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 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 Water’s exptl. b ~ 30.5 ml, while it’s liquid molar volume is 18.0 ml.

Vb

0.1%

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 + a n2 / V 2 ) V = n RT

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.

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.

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

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?

Influence of Vapor Pressure @ 25°C

Hg

760 Hg(g)

2 m

Hg

736 H2O(g)

24 mm

Hg

701 ethanol(g)

59 mm

Hg

222

ether(g)

538 mm

Acetone 231 mm

Methanol 127 mm

Propanal 317 mm

Gedanken ExperimentGedanken 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)