Gases
Physical
Properties
Kinetic Molecular Theory
 Particles
•
•
•
•
•
in an ideal gas…
have no volume
have elastic collisions
are in constant, random, straight-line motion
don’t attract or repel each other
average kinetic energy is directly
proportional to the absolute temperature
 To
change pressure you can:
• Change temperature
• Change volume
• Change amount
Characteristics of Gases
 Gases
expand to fill any container
• random motion, no attraction
 Gases
are fluid (like liquids)
• no attraction
 Gases
have very low densities
• no volume = lots of empty space
Characteristics of Gases
 Gases
can be compressed
• no volume = lots of empty space
 Gases
undergo diffusion & effusion
• random motion
State Changes
Describing Gases
 Gases
can be described by their:
• Temperature
•K
• Pressure
• atm
• Volume
•L
• Number of molecules/moles • #
Temperature
 Temperature:
Every time temperature is
used in a gas law equation it must be
stated in Kelvin (an absolute temperature)
ºF
-459
ºC
-273
K
0
C  59 F  32
32
212
0
100
273
373
K = ºC + 273
Pressure
Pressure of a gas is the force of its particles
exerted over a unit area (number of collisions
against a container’s walls)
force
pressure 
area
Which shoes create the most pressure?
Pressure
 Pressure
may have different units:
torr, millimeters of mercury (mm Hg),
atmospheres (atm), Pascals (Pa), or
kilopascals (kPa).

KEY EQUIVALENT UNITS
• 760 torr
• 760 mm Hg
• 1 atm
• 101,325 Pa
• 101.325 kPa (kilopascal)
• 14.7 psi
STP
STP
Standard Temperature & Pressure
0°C
273 K
-OR-
1 atm
101.325 kPa
The volume of 1 mole of gas at STP is 22.4 L
Pressure Problem 1
 The
average pressure in Denver,
Colorado, is 0.830 atm. Express
this in kPa.
0.830 atm 101.325 kPa= 84.1
1 atm
kPa
Pressure Problem 2
 Convert
a pressure of 1.75 atm
to psi.
1.75 atm
14.7 psi
1 atm
= 25.7 psi
The Behavior of Gases
The Gas
Laws
Boyle’s Law
 Investigated
P
the
relationship between
pressure and volume on a
gas.
 He found that pressure and
volume of a gas are
inversely related
• at constant mass & temp
V
Boyle’s Law
 As
pressure increases
 Volume
decreases
Boyles LAW:
P1V1 = k
P
V
Boyle’s Law
Gas Law Problems
A
gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1 = P2V2
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
Charles’ Law
Jacques Charles investigated the
relationship of temperature & volume
in a gas. He found that the volume of
a gas and the Kelvin temperature of a
gas are directly related.
V
T
V1
k
T1
Charles’ Law
 As
volume increases,
temperature increases
 Charles
V
T
Law:
V1
k
T1
Charles’ Law
Gas Law Problems
gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
A
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
Gas Law Problems
 Example:
The volume of a sample of gas is
43.5 L at STP. What will the volume of the gas
be if the temperature is increased to 35.4°C?
CHARLES’ LAW
GIVEN: T V
V1 = 43.5 L
T1 = 0°C = 273K
V2 = ?
T2= 35.4°C= 308.4K
WORK:
V1T2 = V2T1
(43.5 L)(308.4 K)=V2(273 K)
V2 = 49.1 L
Gay-Lussac’s Law
Since temperature and volume are
related, it would make sense that
temperature and pressure are
related. The equation for this will
resemble Charle’s law:
P1
k
T1
P
T
Gay-Lussac’s Law
 The
pressure and absolute
temperature (K) of a gas
are directly related
• at constant mass &
volume
P1
k
T1
P
T
Gas Law Problems
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P T WORK:
P1 = 765 torr
P1T2 = P2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K)
P2 = 560. torr
T2 = 217 K = -56°C
T2 = ?
Combined Gas Law
P
V
PV
PV = k
T
P 1V 1 P 2V 2
=
T1
T2
P 1 V 1T 2 = P 2V 2 T 1
Gas Law Problems
gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
A
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
T2 = 273 K
The First 4 Gas Laws
 The
Gas Laws Table
Gases
II. Ideal Gas
Law
Ideal Gas Law and Gas
Stoichiometry
Ideal Gas Law
Part 1
Avogadro’s Law
The
volume of a gas increases or
reduces, as its number of moles
is being increased or decreased.
The volume of an enclosed gas is
directly proportional to its number
of moles.
Avogadro’s Principle

Equal volumes of gases contain
equal numbers of moles
• at constant temp & pressure
• true for any gas
• n = number of moles
V1
k
n1
V
n
Ideal Gas Law
Merge the Combined Gas Law with Avogadro’s Principle:
PV
V
=R
k
nT
T
n
UNIVERSAL GAS
CONSTANT
R=0.08206 Latm/molK
R=8.315 dm3kPa/molK
Ideal Gas Law
PV=nRT
UNIVERSAL GAS
CONSTANT
R=0.08206 Latm/molK
R=8.315 dm3kPa/molK
Ideal Gas Law Problems
 Calculate
the pressure in atmospheres
of 0.412 mol of Heat 16°C & occupying
3.25 L.
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.08206Latm/molK
Ideal Gas Law Problems
 Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
WORK:
V=?
85 g 1 mol O2 = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
O2
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
V = 64 dm3
Ideal Gas Law and Molar Mass
 The
Ideal Gas Law can be used to calculate
the mass or the molar mass of a gas sample
if the mass of the sample is known.
 number
of molesgas =
 substituting


mass gas or ngas = mgas_
Molar Massgas
MMgas
this for ngas in PV = nRT
P V = mgas__ R T
MMgas
yields
Applications of Ideal Gas Law
 Calculate
the grams of N2 present in a
0.60 L sample kept at 1.00 atm pressure
and a temperature of 22.0oC.
GIVEN:
WORK:
P = 1.00 atm
T = 22°C = 295. K
V = .600 L
R = 0.08206
Latm/molK
MM = 28.01 g
mN2 = ?
MM P V = mgas R T
(28.01)(1.0)(.60)=m (.0821) (295)
g/mol atm L
Latm/molK K
mN2 = .69 g of N2
Applications of Ideal Gas Law
 Can
be used to calculate the molar
mass of a gas and the density
grams of a gas
mass
m
n  moles of a gas 


molar mass
molar mass MM
 Substitute
this into ideal gas law
nRT
m( RT )
P

V
MM (V )
 And
m/V = d in g/L, so
dRT
P
MM
or
dRT
MM 
P
Applications of Ideal Gas Law
 The
density of a gas was measured at 1.50
atm and 27°C and found to be 1.95 g/L.
Calculate the molar mass of the gas.
GIVEN:
WORK:
P = 1.50 atm
MM = dRT/P
T = 27°C = 300. K MM=(1.95)(0.08206)(300.)/1.50
g/L Latm/molK K
atm
d = 1.95 g/L
R = 0.08206 Latm/molK MM = 32.0 g/mol
MM = ?
Applications of Ideal Gas Law
Calculate
the density of carbon
dioxide gas at 25°C and 750. torr.
GIVEN:
WORK:
750 torr 1 atm = .987 atm
d = ? g/L CO2
760 torr
T = 25°C = 298 K
MM = dRT/P →d = MM P/RT
P = 750. torr
= .987 atm d=(44.01 g/mol)(.987 atm)
R = 0.08206 Latm/molK
(0.08206 Latm/molK )(298K)
MM = 44.01 g/mol
d = 1.78 g/L CO2
Part 2
Gas
Stoichiometry
* Stoichiometry Steps Review *
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio
ratio -molesmoles
moles
•• Mole
moles
• Molar mass moles  grams
• Molarity moles  liters soln
• Molar volume - moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm
Pressure
Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
MOLES
6.02 
1023
particles/mol
NUMBER
OF
PARTICLES
Gas Stoichiometry
of one Gas  Liters of another Gas:
• Avogadro’s Principle
• Coefficients give mole ratios and volume
ratios
 Moles (or grams) of A  Liters of B:
• STP – use 22.4 L/mol
• Non-STP – use ideal gas law & stoich
 Non-STP
• Given liters of gas?
 start with ideal gas law
• Looking for liters of gas?
 start with stoichiometry conversion
 Liters
Gas Stoichiometry Problem – STP
many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
 How
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L
O2
1 mol
O2
2 mol
KClO3
122.55
g KClO3
22.4 L
O2
3 mol
O2
1 mol
KClO3
= 32.8 g
KClO3
Gas Stoichiometry Problem – Non-STP
 What
volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g

CaO
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g 1 mol
CaCO3 CaCO3
1 mol
CO2
100.09g 1 mol
CaCO3 CaCO3
+
CO2
?L
non-STP
P = 103 kPa
V = ? NEXT 
n=?
= 0.0525
mol CO2
R = 8.315
3kPa/molK
dmthe
Plug this into
Ideal Gas
Law for n to find liters
T = 298K
Gas Stoichiometry Problem – Non-STP
 What
volume of CO2 forms from
5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 0.0525 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(0.0525mol)(8.315dm3kPa/molK)
(298K)
V = 1.26 dm3 CO2
Gas Stoichiometry Problem
 How
many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2
15.0 L
non-STP

2 Al2O3
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 
n = 0.597 mol O2
Gas Stoichiometry Problem
 How
many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
3 O2 
15.0L
Use stoich to convert moles
of O to grams Al O
non-STP
0.597 2 mol 101.96 g
mol O2 Al2O3
Al2O3
4 Al
2
2
+
2 Al2O3
?g
3
3 mol O2
1 mol
Al2O3
= 40.6 g Al2O3
Ch. 14 - Gases
III. Two More
Laws
Dalton’s Law & Graham’s Law
A. Dalton’s Law
 The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases
Ptotal = P1 + P2 + ...Pn
A. Dalton’s Law
A
sample of air has a total pressure of
1.00 atm. The mixture contains only
CO2, N2, and O2. If PCO2 = 0.12 atm and
PN2 = 0.70, what is the partial pressure
of O2?
GIVEN:
PO2 = ?
Ptotal = 1.00 atm
PCO2 = 0.12 atm
PN2 = 0.70 atm
WORK:
Ptotal = PCO2 + PN2 + PO2
1.00 atm = 0.12 atm + .70
atm + PO2
PO2 = 0.18 atm
A. Dalton’s Law

Mole Fraction: the ratio of the number of moles
of a given component in a mixture to the total
number of moles in the mixture, represented
by chi (  )


n1
nTOTAL
n1

n1  n2  n3  ...
Because n is directly proportional to P
according to the ideal gas law, we can derive
χA 
nA
n TOTAL

PA
PTOTAL
A. Dalton’s Law
 The
mole fraction of nitrogen in the
air is 0.7808. Calculate the partial
pressure of N2 in air when the
atmospheric pressure is 760. torr.
GIVEN:
χN2 = 0.7808
Ptotal = 760. torr
PN2 = ?
WORK:
χN2
= PN2 /PTOTAL
0.7808 = (PN2)/(760. torr)
PN2 = 593 torr
A. Dalton’s Law
•When H2 gas is collected
by water displacement,
the gas in the collection
bottle is actually a mixture
of H2 and water vapor
•Water exerts a pressure known as water-vapor
pressure
•So, to determine total pressure of gas and water
vapor inside a container, make total pressure inside
bottle = atmosphere and use this formula:
Ptotal = Pgas + PH2O
A. Dalton’s Law
 Hydrogen
gas is collected over water at
22.5°C. Find the pressure of the dry gas
if the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
WORK:
PH2 = ?
Ptotal = PH2 + PH2O
Ptotal = 94.4 kPa
94.4 kPa = PH2 + 2.72 kPa
PH2O = 20.4 mm Hg P = 91.7 kPa
H2
= 2.72
kPa
Look up
water-vapor pressure
Sig Figs: Round to lowest
for 22.5°C, convert to kPa
number of decimal places.
B. Graham’s Law
 Diffusion
• Spreading of gas molecules
throughout a container until
evenly distributed
 Effusion
• Passing of gas molecules
through a tiny opening in a
container
B. Graham’s Law
 Speed
of diffusion/effusion
• Kinetic energy is determined by
the temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
 Larger m  smaller v
KE =
2
½mv
B. Graham’s Law
 Graham’s
Law
• Rate of diffusion of a gas is
inversely related to the square root
of its molar mass.
• The equation shows the ratio of
Gas A’s speed to Gas B’s speed.
rate A
rate B

MM B
MM
A
B. Graham’s Law
 Determine
the relative rate of diffusion
for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate means find the ratio “vA/vB”.
vA

vB
v Kr

v Br2
m Br2
m Kr

mB
mA
159.80 g/mol
 1.381
83.80 g/mol
Kr diffuses 1.381 times faster than Br2.
B. Graham’s Law

vA

vB
A molecule of oxygen gas has an average
speed of 12.3 m/s at a given temp and pressure.
What is the average speed of hydrogen
molecules at the same conditions?
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
 3.980
vH2  49.0 m/s
B. Graham’s Law

An unknown gas diffuses 4.0 times faster than
O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA

vB
mB
mA
vA

v O2
mO2
mA

32.00
g/mol
 4.0 

m AA

32.00 g/mol
16 
mA
Square both
sides to get rid
of the square
root sign.




32.00 g/mol
 2.0 g/mol
mA 
16
2