Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 6 (Chp 10):
Gases
John Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Characteristics of Gases
• Unlike liquids and solids, they…
 expand to fill their containers.
(indefinite volume)
 are highly compressible.
 have extremely low densities.
Pressure
• Pressure is the amount
of force applied per area.
22,000 lbs!!!
(per sq. meter)
F
P=
A
Atmospheric Pressure:
weight of air per area
empty space
(a vacuum)
Pressure
Units (at sea level)
1 atm = 760 mmHg
= 760 torr
= 101.3 kPa
STP
(standard T & P)
273 K 1 atm
1N
1 m2
657
1) 657 mmHg to atm 760
2) 830 torr to atm 830
760
3) 0.59 atm to torr
0.59 x 760
h
760 mm
Atmospheric
Pressure
(weight of air)
Kinetic-Molecular Theory
KMT is a model
which explains the
Properties
(P, V, T, n)
and
Behavior
(motion, energy,
speed, collisions)
of gases.
5 Parts of Kinetic-Molecular Theory
1) Gas particles are in constant
random motion.
2) Gas pressure is caused by
collisions with the container walls.
P=
F
A
F
P=
A
F
P=
A
Collisions
are elastic
(no KE lost).
(ideally)
5 Parts of Kinetic-Molecular Theory
3) Attractive forces
(IMAFs) are negligible.
IMAFs
4) Volume of gas particles is negligible,
compared to total volume of container.
Ideally: Vgas = Vcontainer
1.000 L
container
has 1.000 L
of gas
Vgas = Vcontainer – Vparticles
In a 1.000 L
container,
gas only expands
into about
0.999 L of
volume
(negligible)
5 Parts of Kinetic-Molecular Theory
5) Average KE of gas particles is…
…directly proportional to Kelvin Temp.
(K not oC)
KEavg α T
no negative temp’s,
no negative energies,
no negative volumes,
etc.)
video clip
Boyle’s Law (P & V)
P↑,V↓
(inversely proportional)
10 L
5L
1 atm
2 atm
Charles’ Law (V & T)
T↑,V↑
(directly proportional)
30 L
60 L
how absolute zero
was estimated
150 K
300 K
Lussac’s Law (P & T)
T↑,P↑
(directly proportional)
100 kPa
300 K
200 kPa
600 K
500 kPa
Avogadro's Hypothesis
• At the same ___
volumes of
T & ___,
P equal _________
moles (n) (particles)
gas must contain equal _________.
All at:
P = 1 atm
T = 25oC
V = 1.0 L
CO2
He
O2
Avogadro's Law (V & n)
n↑,V↑
(directly proportional)
1 mol add gas
He
2 mol
He
• So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
P  T (Lussac’s law)
V  n (Avogadro’s law)
PV
=R
nT
constant
(all gases
same ratio)
ideal gas constant: R = 0.08206 L∙atm/mol∙K
given on exam
Ideal Gas Law
PV = nRT
NO Units of : mL , mmHg , kPa , grams , °C
PV = nRT
given on
exam
P1 V1
P2 V2
=R=
n1T1 constant n2T2
(initial)
(final)
NOT
given on exam
P1 V1 P2 V2
=
n1T1 n2T2
(changes in P,V,T,n)
(inversely proportional) (directly proportional)
P↑,V↓
T↑,V↑
T↑,P↑
n↑,V↑
Ideal-Gas Changes
PV = nRT
V1
1. The pressure on a 411 mL sample of gas is
decreased from 812 torr to 790 torr. P2
What will be the new volume of the gas?
P1
V2 = ?
P1V1
P2V2
= 422 mL
n1T1
=
n2T2
(812)(411) = (790) V2
(812)(411) = V2
(790)
PV
=R
nT
Ideal-Gas Changes
PV = nRT
V1
2. A 10.0 L sample of a gas is collected at
25oC and then cooled to a new volume of
T1 8.83 L while the pressure remains at 1.20 atm.
What is the final temperature in oC?
V2
oC
=
–10
=
?
263
K
T
P1V1
P2V2
2
n1T1
=
n2T2
(10.0) = (8.83)
(298)
T2
(8.83)(298)
T2 =
(10.0)
T2 (10.0) = (8.83)(298)
PV
=R
nT
Ideal-Gas Changes
PV = nRT
HW p. 432 # 1, 23, 89, 26
V1
n1
3. A 13.1 L sample of 0.502 moles of O2 is held
under conditions of 1.00 atm and 25.0oC.
If all of the O2 is converted to Ozone (O3),
what will be the volume of O3?
V2 = 8.73 L
n2 = ?
3 O2  2 O3
P1V1 P2V2 (13.1) = V2
=
n1T1 n2T2 (0.502) (0.335)
n2
0.502 mol O2 x 2 mol O3 = 0.335
n1
3 mol O2 mol O3
.
PV
=R
nT
Ideal-Gas Equation
PV = nRT
V
P
T
o
1. A 5.00 L He balloon has 1.20 atm at 0.00 C.
How many moles of He gas are in the balloon?
n PV = nRT R = 0.08206
L∙atm∙mol–1∙K–1
(1.20 atm)(5.00 L) = n (0.08206)(273 K)
(1.20)(5.00) = n
(0.08206)(273)
How many
atoms
n = 0.268 mol He
of He?
Ideal-Gas Equation
PV = nRT
HW p. 438 #92, 29, 46, 38, 35
m 2. A sample of aluminum chloride gas weighing
0.0500 g at 350.oC and 760 mmHg of
T
pressure occupies a volume of 19.2 mL.
P
Calculate the Molar Mass of the gas.
V
M=?
PV = nRT R = 0.08206 L∙atm
(1.00 atm)(0.0192 L) = n (0.08206)(623 K) mol∙K
n = 0.000376 mol
n = m (given on
M exam)
__0.0500
g_
M=
so…
0.000376 mol
m
Molar
grams
M=
=
n Mass
mole AlCl3 = 133 g/mol
1 mole
Molar Volume: the volume
______ of ______
STP
of any gas at ____.
PVm = nRT
(1.00 atm) Vm = (1 mol )(0.08206)(273 K)
• The volume of 1 mole of any gas at STP
will be:
L but…
Vm = 22.4
_____
ONLY
at
STP!!!
1 mol
Gas Stoich with Molar Volume
The reactions below occurred at STP.
1. Calculate the mass of NH4CI reacted with
Ca(OH)2 to produce 11.6 L of NH3(g) .
2 NH4Cl + Ca(OH)2  2 NH3 + CaCI2 + 2 H2O
11.6 L NH3 x 1 mol NH3 x 2 mol NH4Cl x 53.49 g NH4Cl =
22.4 L NH3
2 mol NH3
1 mol NH4Cl
27.7 g
2. Calculate the volume of CO2 gas produced
when 9.85 g of BaCO3 is decomposed.
BaCO3(s)  BaO(s) + CO2(g) 1.12 L
9.85 g BaCO3 x 1 mol BaCO3 x 1 mol CO2 x 22.4 L CO2 =
197.34 g BaCO3 1 mol BaCO3 1 mol CO2
NOT at STP use… PV = nRT
3. What volume of O2 gas is produced from
490 g KClO3 at 298 K and 1.06 atm?
KClO3(s)  KClO(s) + O2(g)
1 mol KClO3 x 1 mol O2
490 g KClO3 x
= 4.00 mol O2
122.55 g KClO3 1 mol KClO3
(1.06 atm) V = (4.00 mol )(0.08206)(298 K)
= ____L
92.3 L O22
Molar Mass of KClO3 is 122.55 g/mol
Dalton’s Law of Partial Pressures
Ptotal = PA + PB + PC + …
HW
p. 436
#63
64
66
Mole Fraction (XA)
XA =
moles of A
total moles
WS 6b
#1-4
PA = Ptotal x XA
• The mole fraction (XA) is like a % of total
moles that is A, but without the % or x 100.
equalize water level
inside & outside
=
Ptotal = PH2O + Pgas
• When one collects a gas over water, there is
water vapor mixed in with the gas.
• To find only the pressure of the gas, one
must subtract the water vapor pressure
from the total pressure. P
=P
–P
gas
total
H2O
Ptotal = PH2O + Pgas
PV = nRT
1. Calculate the mass of 0.641 L of H2 gas
collected over water at 21.0oC with a total
pressure of 750. torr.
The vapor pressure of water at 21.0oC is
20.0 torr.
750. = 20.0 + PH2
PH2 = 730. torr
nH2 = 0.0255 mol
mH2 = 0.0514 g
PH2 = 730/760 = 0.961 atm
(0.961)(0.641) = nH2 (0.08206)(294)
Study the models below. What can be said
quantitatively about the molecular speed (v) of a gas in
relation to its molar mass (M) and its temperature (T)?
↑M,↓v
↑T,↑v
KE = ½ mv2
(given on exam)
Compare the molecular speed (v) of these gases:
1) at 25.0 oC
(i) Helium (ii) Oxygen (O2)
1360 m/s
482 m/s
2) at 50.0 oC
(i) Helium (ii) Oxygen (O2)
1420 m/s
502 m/s
3) Does the data support your conclusions from
the models on the previous slide about effects
of T and M on v ?
↑M,↓v
↑T,↑v
WHY?
Distributions of Molecular Speed
Temp (K) & KEavg
are directly proportional
average molecular speed (v)
T α KEavg
(KMT)
KE = ½ mv2
(given on exam)
Therefore:
T & v are
__________
directly
proportional
↑T,↑v
Speed vs. Molar Mass
avg
Gases at the same Temp, have the same KE
_____.
KE = ½ mv2
Ar: M = 40 g/mol
He: M = 4.0 g/mol
KE1 = ½ m1v12
KE2 = ½ m2v22
KEHe = KEAr
(at same T)
½
m
2
v
↑M,↓v
=½m
v
2
↓ M,↑v
M & v are
__________
inversely
proportional
Effusion
escape of
gas particles
through a
tiny hole
KE = ½ mv2
Diffusion
↑T, ↑v
KE = ½ mv
2
↑M,↓v
½m
v
2
HW p.437
#8,74,76a
=½
m
v2
spread of
gas particles
throughout
a space
64 g/mol (SO2)
16 g/mol (CH4)
CH4 is __
2 times
_____
SO2
faster than _____.
Real (non-Ideal) Gases
In the real world, the behavior of gases only
conforms to the ideal-gas equation under
“ideal” conditions.
(PV
usually)
= nRT
(ONLY under ideal conditions)
Ideal: (High T) (Low P) (weaker IMAFs)
WHY? (high KE)(high Vtotal) (negligible)
Non-Ideal: (Low T) (High P) (stronger IMAFs)
(low KE) (low Vtotal) (not negligible)
Ideal Gas vs Non-Ideal Gas
IDEAL
more KE/speed
T: ↑
weaker IMAFs
more avg. dist.
P: ↓
IMAFs
(ideal P)
NON
less KE/speed
T: ↓ stronger IMAFs
P: ↑ less avg. dist.
HW p.437
(ideal V)
#81,82,83
n2a
(P + 2 ) (V − nb ) = nRT
V
“observed” P too low “observed” V too high b/c
size of particles
b/c attractive forces
not negligible
not negligible,
collisions less frequent compared to total volume
Vgas = Vcontainer – Vparticles
and of less force