GAS LAWS
Measureable Properties of Gases
quantity of gas
n
number of molecules, moles
pressure
p
mm Hg (at 0ºC), torr, atmospheres, Pascals
temperature
T
degrees Celsius, Kelvin
volume
V
liters, milliliters
5A-1 (of 13)
QUANTITY OF GAS – A measure of the number of molecules (or atoms)
in a sample
AVOGADRO’S LAW – The volume of a gas is directly
proportional to the quantity of gas, provided the pressure
and temperature of gas do not change
V  n
5A-2 (of 13)
V = kn
PRESSURE – Force per unit area
Due to the constant bombardment of
the inside walls of the container by the
gas molecules
STANDARD PRESSURE – Normal
atmospheric pressure
760.0 mm Hg (at 0ºC)
760.0 torr
1.000 atm
101,325 Pa
5A-3 (of 13)
BOYLE’S LAW – The volume of a gas is inversely
proportional to its pressure, provided the temperature and
quantity of gas do not change
V  1
V = k
p
p
___
5A-4 (of 13)
___
pV = k
Atmospheric pressure is measured with a BAROMETER
Fatm = FHg
FHg = mHgg
mHg = VHgDHg = hHgAHgDHg
Patm =
Fatm
_____
=
AHg
Aatm
=
hHgAHgDHgg
_______________
AHg
=
5A-5 (of 13)
FHg
_____
hHgDHgg
Gas pressure is measured with a MANOMETER
Close-Ended Manometer
Open-Ended Manometer
125 mm Hg more than 0
125 mm Hg more than 765 mm Hg
125 mm Hg
890. mm Hg
5A-6 (of 13)
If a gas can support a column of Hg that weighs 10.0 g, then it will also
support 10.0 g H2O
10.0 g Hg
x 1 mL Hg
= 0.7353 mL Hg
(DHg = 13.6 g/mL)
10.0 g H2O x 1 mL H2O = 10.00 mL H2O
(DH2O = 1.00 g/mL)
_____________
13.6 g Hg
_____________
1.00 g H2O
10.00 mL H2O = 13.6 mL H2O/mL Hg
__________________
0.7353 mL Hg
In a glass tube of constant diameter:
13.6 mm H2O = 1 mm Hg
ratio of the mercury’s density to water’s density
5A-7 (of 13)
54 mm H2O more than 0
54 mm H2O x
1 mm Hg
_________________
13.6 mm H2O
27 mm H2O less than 765 mm Hg
27 mm H2O x
1 mm Hg
_________________
13.6 mm H2O
= 4.0 mm Hg more than 0
= 2.0 mm Hg less than 765 mm Hg
4.0 mm Hg
763 mm Hg
5A-8 (of 13)
GRAPHING EXPERIMENTAL DATA
Scientific relationships are expressed in linear form because they are
easier to read than curves
y = mx + b
variable plotted on x-axis
variable plotted on y-axis
5A-9 (of 13)
V = k
___
p
Relationship:
Inverse
Graph:
Hyperbola
5A-10 (of 13)
To make V and p graph as a line:
(1) Plot the relationship in logarithmic form
log V = log k
___
p
log V = log k - log p
log V = -log p + log k
log V = (-1)log p + log k
y
=
5A-11 (of 13)
m x
+
b
To make V and p graph as a line:
(2) Plot the reciprocal of one variable
V = k
___
p
V = k
1
___
+ 0
p
y = m x
5A-12 (of 13)
+ b
To make V and p graph as a line:
(3) Plot the product of the variables
vs. one of the variables
What does the product of the
variables equal?
V = k
___
p
pV = k
5A-13 (of 13)
TEMPERATURE – A measure of the average kinetic energy of molecules
The lowest possible temperature is the temperature where molecular
motion stops
-273.2C
ABSOLUTE SCALE – One with 0 as the lowest possible value
KELVIN SCALE (K) – A temperature scale where 0 is the lowest possible
temperature
K = Cº + 273.2
ABSOLUTE ZERO – The lowest possible temperature
STANDARD TEMPERATURE – The normal freezing point of water
0.0ºC
273.2 K
5B-1 (of 12)
CHARLES’ LAW – The volume of a gas is directly
proportional to its temperature, provided the pressure and
quantity of gas do not change
V  T
V = kT
T must be measured in Kelvin
so that when T = 0, V = 0
5B-2 (of 12)
THE COMBINED GAS LAW
V  n
Avogadro’s Law
V  1
Boyle’s Law
___
5B-3 (of 12)
____
p
or
p
V  T
V  nT
Charles’ Law
pV  nT
IDEAL GAS LAW EQUATION
pV = nRT
p = pressure (atm)
V
n
T
R
= volume (L)
= quantity (mol)
= temperature (K)
= Universal Gas Constant (0.08206 Latm/molK)
(8.314 J/molK)
5B-4 (of 12)
All relationships can be derived from pV = nRT
p and T
n and T
pV
k = nR
k kT
pV
kk = n R
kT
p =
k = nT
kT
directly proportional
5B-6 (of 12)
inversely proportional
Calculate the quantity of gas in a flask with a capacity of 500. mL when the
pressure is 570. torr at 22ºC.
pV = nRT
pV = n
____
RT
570. torr x
1 atm
____________
= 0.7500 atm
760.0 torr
(0.7500 atm)(0.500 L)
______________________________________
(0.08206 Latm/molK)(295.2 K)
5B-7 (of 12)
= 0.0155 mol
Because moles = mass  molar mass
n = m
___
m
pV = m
n RT
___
m
5B-8 (of 12)
Calculate the density of nitrogen gas at 25.0ºC and 1.00 atm pressure.
pV = mRT
_______
m
D = m
___
V
pm = m
____
___
RT
V
(1.00 atm)(28.02 g/mol)
______________________________________
(0.08206 Latm/molK)(298.2 K)
5B-9 (of 12)
= 1.15 g/L
VOLUME CALCULATIONS IN CHEMICAL REACTIONS
Calculate the mass of sodium bicarbonate that reacted with hydrochloric
acid to produce 1.65 L of carbon dioxide gas at 21ºC and 0.967 atm.
NaHCO3
HCl
→
NaCl
+
H2O
+
CO2
xg
1.65 L
1 mol
1 mol
pV = n
______
+
=
RT
(0.967 atm)(1.65 L)
______________________________________
=
0.06609 mol CO2
(0.08206 Latm/molK)(294.2 K)
0.06609 mol CO2 x 1 mol NaHCO3 x 84.01 g NaHCO3
5B-10 (of 12)
__________________
____________________
1 mol CO2
1 mol NaHCO3
= 5.55 g NaHCO3
Calculate the volume of oxygen gas at 35ºC and 1.25 atm needed to burn
5.00 g of propane, C3H8.
C3H8
5.00 g
1 mol
5.00 g C3H8 x
+
_____
p
5B-11 (of 12)
3 CO2
+
4 H2O
xL
5 mol
1 mol C3H8
________________
44.11 g C3H8
V = nRT =
→
5 O2
x
5 mol O2
______________
= 0.5668 mol O2
1 mol C3H8
(0.5668 mol)(0.08206 Latm/molK)(308.2 K)
_______________________________________________________
(1.25 atm)
= 11.5 L O2
If air is 20.9% oxygen by volume, calculate the volume of air at 35ºC and
1.25 atm needed to burn 5.00 g of propane, C3H8.
C3H8
5.00 g
1 mol
+
5 O2
→
3 CO2
+
4 H2O
xL
5 mol
100% air = 20.9% O2 by volume
100 L air = 20.9 L O2
11.5 L O2 x 100 L air
____________
20.9 L O2
5B-12 (of 12)
= 55.0 L air
Calculate the molar mass of a gas if 2.20 grams of the gas exerts a
pressure of 1.12 atm at 20.0ºC in a 1.00 liter flask.
pV = mRT
_______
m
m = mRT
_______
pV
(2.20 g)(0.08206 Latm/molK)(293.2 K) = 47.3 g/mol
_______________________________________________
(1.12 atm)(1.00 L)
5B-13 (of 13)
GAS MIXTURES
DALTON’S LAW OF PARTIAL PRESSURES – The total pressure of a
mixture of gases is the sum of the pressures exerted by each gas
pN
pO
pAr
2
2
= 591 torr
= 161 torr
=
8 torr
___________
760. torr
PARTIAL PRESSURE – The pressure a gas would exert if it were alone in
a container
5C-1 (of 15)
Partial pressures are proportional to the quantity of gas in the mixture
MOLE FRACTION (Χ) – The moles of one gas in a mixture divided by the
total moles of gas
The partial pressure of one gas in a mixture (pj) is equal to its mole
fraction (Χj) multiplied by the total pressure of the mixture (P)
pj = ΧjP
5C-2 (of 15)
A gas mixture has 2 moles of O2, 3 moles of N2, and a total pressure of
15 atm. Calculate the mole fractions and partial pressures of each gas.
ΧO2 =
2 mol O2
_____________
=
0.40
=
0.60
5 mol gas
ΧN2 =
3 mol N2
_____________
5 mol gas
p O2 =
(0.40)(15 atm) =
6 atm O2
pN2 =
(0.60)(15 atm) =
9 atm N2
5C-3 (of 15)
A gas mixture 70.% Ne and 30.% He by mass has a pressure of 5.0 atm.
Calculate the partial pressures of each gas.
Assume you have 70. g Ne and 30. g He
70. g Ne
x
1 mol Ne
______________
= 3.47 mol Ne
20.18 g Ne
30. g He
x 1 mol He
_____________
= 7.50 mol He
4.00 g He
pNe =
3.47 mol Ne
__________________
(5.0 atm)
= 1.6 atm Ne
(5.0 atm)
= 3.4 atm He
10.97 mol gas
pHe =
7.50 mol He
__________________
10.97 mol gas
5C-4 (of 15)
COLLECTING GASES IN LAB
To determine the quantity of gas collected in lab, you must measure the
(1) gas volume
(2) gas temperature
(3) gas pressure
pV = n
____
RT
EUDIOMETER – A gas collecting tube
5C-5 (of 15)
Atmospheric pressure:
765 mm Hg
1st – Hg level THE SAME in and out
∴ pressure inside eudiometer = pressure outside
p = 765 mm Hg
5C-6 (of 15)
Atmospheric pressure:
765 mm Hg
2nd – Hg level HIGHER inside
∴ pressure inside eudiometer < pressure outside
p = 765 mm Hg – 13 mm Hg = 752 mm Hg
5C-7 (of 15)
Atmospheric pressure:
765 mm Hg
3rd – Hg level LOWER inside
∴ pressure inside eudiometer > pressure outside
p = 765 mm Hg + 6 mm Hg = 771 mm Hg
5C-8 (of 15)
COLLECTING GASES BY WATER DISPLACEMENT
Collecting O2 gas
When a gas is collected by water displacement, some water evaporates into
the eudiometer
The eudiometer contains both oxygen gas and water vapor
If the pressure in the eudiometer is calculated, the partial pressure of the
water vapor must be subtracted to get the partial pressure of the oxygen
The partial pressure of the water vapor (called WATER’S EQUILIBRIUM
VAPOR PRESSURE, or WATER VAPOR PRESSURE) depends only on
temperature, and can be looked up
5C-9 (of 15)
Atmospheric pressure:
765 mm Hg
Temperature: 25ºC
Water Vapor Pressure at 25ºC =
24 mm Hg
1st – H2O level THE SAME in and out
∴ pO2 + pH2O(g) = pressure outside
pO2 + 24 mm Hg = 765 mm Hg
pO2 = 765 mm Hg – 24 mm Hg = 741 mm Hg
5C-10 (of 15)
Atmospheric pressure:
765 mm Hg
Temperature: 25ºC
Water Vapor Pressure at 25ºC =
24 mm Hg
2nd – H2O level HIGHER inside
∴ pO2 + pH2O(g) < atmospheric pressure
27 mm H2O x
1 mm Hg
_________________
= 2.0 mm Hg
13.6 mm H2O
pO2 + 24 mm Hg = 765 mm Hg – 2.0 mm Hg
pO2 = 765 mm Hg – 2.0 mm Hg – 24 mm Hg = 739 mm Hg
5C-11 (of 15)
Atmospheric Pressure = AP
Water Vapor Pressure = WVP
1st –
AP
3rd –
AP – WVP
2nd –
AP – h
4th –
AP – (h÷13.6) – WVP
5C-12 (of 15)
A sample of zinc metal is reacted with hydrochloric acid,
and 44.2 mL of hydrogen gas are collected by water
displacement.
Water temperature is 22ºC, the water level is 54 mm
higher inside the eudiometer, barometric pressure is 752
mm Hg, and water vapor pressure at 22ºC is 20. mm Hg.
Calculate the mass of zinc in the sample.
54 mm H2O x
1 mm Hg
_________________
= 3.97 mm Hg
13.6 mm H2O
pH2 = 752 mm Hg – 3.97 mm Hg – 20. mm Hg = 728.0 mm Hg
728.0 mm Hg x
1 atm
_______________________________
760.0 mm Hg
5C-13 (of 15)
= 0.9579 atm
A sample of zinc metal is reacted with hydrochloric acid,
and 44.2 mL of hydrogen gas are collected by water
displacement.
Water temperature is 22ºC, the water level is 54 mm
higher inside the eudiometer, barometric pressure is 752
mm Hg, and water vapor pressure at 22ºC is 20. mm Hg.
Calculate the mass of zinc in the sample.
Zn
+
2 HCl
→
ZnCl2
+
H2
xg
44.2 mL
1 mol
1 mol
pV = n =
(0.9579 atm)(0.0442 L)
____
______________________________________
RT
(0.08206 Latm/molK)(295.2 K)
5C-14 (of 15)
= 0.001748 mol H2
A sample of zinc metal is reacted with hydrochloric acid,
and 44.2 mL of hydrogen gas are collected by water
displacement.
Water temperature is 22ºC, the water level is 54 mm
higher inside the eudiometer, barometric pressure is 752
mm Hg, and water vapor pressure at 22ºC is 20. mm Hg.
Calculate the mass of zinc in the sample.
Zn
+
2 HCl
→
ZnCl2
+
H2
xg
44.2 mL
1 mol
1 mol
0.001748 mol H2 x 1 mol Zn x 65.38 g Zn = 0.114 g Zn
5C-15 (of 15)
___________
______________
1 mol H2
1 mol Zn
KINETIC ENERGY (EK) – The energy of motion
EK = ½mv2
m = mass (of all the gas molecules)
v = velocity (average velocity of all the gas molecules)
Mathematically for gases:
EK = ½mv2 = ³/2pV
(v2 = the average of the squared velocities)
For 1 mole of a gas, pV = (1)RT, so:
½mv2 = ³/2RT
For 1 mole of a gas, m = m, so:
½mv2 = ³/2RT
5D-1 (of 6)
½mv2 = ³/2RT
TEMPERATURE – A measure of the average kinetic energy of molecules
Molecules at the same temperature have the same average kinetic energy
v2
=
3RT
_____
m
vrms =
3RT
½
_____
m
ROOT-MEAN-SQUARE VELOCITY (vrms) – The average velocity of the
molecules in a gas sample
5D-2 (of 6)
Calculate the root-mean-square velocity of oxygen molecules at 25ºC.
vrms
=
3RT
½
_____
m
=
3(8.314 kgm2/s2molK)(298.2 K)
_______________________________________
(0.03200 kg/mol)
=
5D-3 (of 6)
482 m/s
½
The vrms is the average velocity of a group of molecules at a specific
temperature
At 198 K:
At 298 K:
vrms at 298 K
vrms at 198 K
MAXWELL-BOLTZMAN DISTRIBUTION – The range of velocities of the
molecules in a sample
5D-4 (of 6)
2 gas samples at the same temperature have equal average kinetic energies
½m1v12 = ½m2v22
This relationship is called GRAHAM’S LAW
5D-5 (of 6)
Oxygen gas effuses into a vacuum at a rate of 51 mL/min, while an
unknown gas effuses at a rate of 35 mL/min. Calculate the molar mass of
the unknown gas.
½mo2vo22 = ½mxvx2
mo2vo22 =
mxvx2
mo2vo22 =
mx
_________
vx2
5D-6 (of 6)
=
(32.00 g/mol)(51 mL/min)2
__________________________________
(35 mL/min)2
=
68 g/mol
IDEAL GAS – A hypothetical gas that exactly obey the gas laws
An ideal gas is described by the following assumptions of the kinetic
molecular theory:
1 – the gas particles have no attractive forces between them
2 – the gas particles are point masses that are in constant motion
5E-1 (of 14)
Real gases do not exactly obey the gas laws
Real gases deviate from ideal behavior because
1 – molecules attract each other
this causes the measured pressure
of the real gas to be less than the
pressure of an ideal gas
2 – molecules take up space
this causes the volume available to
each real gas molecule to be less
than the measured volume of the
container
5E-2 (of 14)
The VAN DER WAALS EQUATION corrects for these deviations
p
+
a n
___
2
(V - nb)
=
nRT
V
van der Waals constants:
a – correction for molecular attraction
the “a term” increases the measured pressure of the real gas (p) so that
it equals the pressure of an ideal gas
b – correction for molecular size
the “b term” decreases the volume of the container (V) so that it equals
the actual volume available to each gas molecule
5E-3 (of 14)
Calculate the pressure of 1.000 moles of ammonia gas at 0.0ºC in a 22.42
liter container.
pV = nRT
p
= nRT
______
V
5E-4 (of 14)
=
(1.000 mol)(0.08206 Latm/molK)(273.2 K)
____________________________________________________
(22.42 L)
= 1.000 atm
Calculate the pressure of 1.000 moles of ammonia gas at 0.0ºC in a 22.42
liter container.
For ammonia, a = 4.170 L2atm/mol2, b = 0.03707L/mol.
p
+
a n
2
___
(V - nb)
=
nRT
V
p
+
a n
___
2
=
V
p
nRT
__________
V - nb
=
nRT
__________
V - nb
5E-5 (of 14)
-
a n
___
V
2
Calculate the pressure of 1.000 moles of ammonia gas at 0.0ºC in a 22.42
liter container.
For ammonia, a = 4.170 L2atm/mol2, b = 0.03707L/mol.
(1.000 mol)(0.08206 Latm/molK)(273.2 K) ____________________________________________________
4.170 L2atm/mol2 1.000 mol
______________
22.42 L – (1.000 mol)(0.03707 L/mol)
-
1.0016 atm
p
0.0082959 atm
=
0.993 atm
=
nRT
__________
V - nb
5E-6 (of 14)
22.42 L
-
a n
___
V
2
2
ATMOSPHERIC CHEMISTRY
1 – Combustion of Hydrocarbons from Automobiles
2C8H18 + 25O2 → 16CO2 + 18H2O
However, at high temperatures
nitrogen gas is combusted as well
N2 + O2 → 2NO
2NO + O2 → 2NO2
(orange gas)
And, with light
NO2 → 2NO + O
O + O2 → O3
(irritating)
This is PHOTOCHEMICAL SMOG
5E-7 (of 14)
ATMOSPHERIC CHEMSITRY
2 – Combustion of Coal from Power Plants
C + O2 → CO2
However, coal contains a significant
amount of sulfur
S8 + 8O2 → 8SO2
2SO2 + O2 → 2SO3
SO3 + H2O → H2SO4
CaCO3 is now added to stacks of coal-burning plants
CaCO3 → CaO + CO2
CaO + SO2 → CaSO3 (s)
This is called SCRUBBING
5E-8 (of 14)
ATMOSPHERIC CHEMSITRY
3 – Stratospheric Ozone
Ozone absorbs UV radiation
O3 + UV → O2 + O
O2 + O → O3 + IR
5E-9 (of 14)
ATMOSPHERIC CHEMSITRY
3 – Stratospheric Ozone
1974 SHERWOOD ROWLAND
Showed that chlorofluorocarbons (CFC’s) from air
conditioners, refrigerators, and spray cans destroy
stratospheric ozone
CF2Cl2 + UV → CF2Cl + Cl
Cl + O3 + UV → O2 + ClO
ClO + O → O2 + Cl
x 1,000,000
Cl radicals accumulate on the surfaces of ice clouds over Antarctica
during the dark months
5E-10 (of 14)
ATMOSPHERIC CHEMSITRY
3 – Stratospheric Ozone
Each spring, sunlight releases the Cl radicals, and a thinning of the
ozone layer occurs over Antarctica
5E-11 (of 14)
ATMOSPHERIC CHEMSITRY
4 – Greenhouse Gases
Gases that absorb IR radiation (heat) from the sun, trapping it in the
atmosphere
H2O, CH4, CO2
An increase in greenhouse gases could result in glacier melting, sealevel rise, and more powerful storms
5E-12 (of 14)
ATMOSPHERIC CHEMSITRY
4 – Greenhouse Gases
Models show that if the US follows the world’s most ambitious
climate policy to cut CO2 emissions, the net cost is roughly $250
billion a year for the rest of the century ($833 per person per year),
reducing sea-level rise 9 millimeters by 2100
5E-13 (of 14)
ATMOSPHERIC CHEMSITRY
4 – Greenhouse Gases
The concern for New York City: the 3.3% chance each year (entirely
without global warming) that a Category 3 hurricane will hit, causing
sea surges of up to 7.5 meters (putting Kennedy Airport under 6
meters of water)
Much of the risk could be managed by erecting seawalls, building
storm doors for the Subway, and simple fixes like porous pavements,
at a cost of around $100 million a year
To reduce hurricane damage, should we focus primarily on a very
cheap solution that would enable us to handle storm surges much
better within a few years, or on an expensive solution that would
require almost a hundred years to avoid 9 millimeters of 7.5-meter
surges?
5E-14 (of 14)