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


Gas Stoichiometry
 Partial Pressures and Mole
Fractions, Xi
 Effusion
 Diffusion
 Our Atmosphere


Ideal Gas Law: PV = nRT

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.
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.
 Pressure
 Gravitational
 What’s
force influences P.
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
 And
includes g and the proportionality between P and .
atmospheric P falls off exponentially with
altitude, being only ~1/3 atm on top of Everest.
Pressure Rules
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.
 While
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 = 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
and conservation requires
– p0
f  +2p0
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 = 1/8 P1 when V2 = 8 V1. Boyle is right!

Charles’s Law: V/T fixed (iff n,P too)
 Kinetic Theory
 Imagine
helps here.
a fixed volume heated such that T2 = 8 T1
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.
 That

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,
 And Boyle is right.
if 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
 dP
P
to the Barometric Formula:
= –  g dh = – P g ( M / RT ) dh now gives
= 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 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
= PA + PB + PC + …
 n (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V) + …
 So n = nA + nB + nC + … (surprise surprise)
P
 Now
divide both sides by n, the total number of moles of gas
= XA + XB + XC + … mole fractions sum to 1.
 1 = PA/P + PB/P + PC/P + …
 Hence XA = PA/P for gases.
1
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 Rates
behind “rate  M–½” is comparison of
rates at same T and same P.
 Presumption
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
 Fixed

235UF
6
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, 5.2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
Consequences of Mair
air may feel heavier but it’s 3% lighter than dry
air. That means a column of it has lower P.
 Humid
 The
barometer is lower where it’s stormy, higher
where it’s dry. Winds blow from high P to low P.
  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.
 Since
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
 But
retaliates by firing back.
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.

 Air
Thus the shape of hurricanes (whose upper air is rained out).
“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
= 1 atm = 1.01325105 Pa = 1.01325105 N/m2
 Force on every m2 is F = Mair g = 1.01325105 N
 P0

N = J m–1 = kg m2 s–2 m–1 = kg m s–2 in SI
= 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)
 Mair
g on
g off
0.1%
Real Gas: Volume Effect
b
V
 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.
of V is always taken up with a molecule’s molar
condensed volume, ~ b; we have to exclude nb from V.
 Part

 So
That gives us the “ideal volume” the gas is free to use.
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.
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
/
)(V–n
 a and b are empirical parameters.
2
n
2
V
b ) = n RT
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!
 Ammonia
 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:
 2 NO2 K 300 K = 11
 If you evolve 1 mol of N2O4 at 1 atm & 300 K, what’s V ?
 N2O4
 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 + 1.024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?

Influence of Vapor Pressure @ 25°C
Hg(g)
760 
2 m
H2O(g)
736 
24 mm
ethanol(g)
701 
59 mm
ether(g)
Acetone 231 mm
538 mm
Methanol 127 mm
Propanal 317 mm
222 
Hg
Hg
Hg
Hg
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)