GASES
Chapter 12
The Atmosphere
The atmosphere is a gaseous solution of
nitrogen, N2, and oxygen, O2.
The atmosphere both supports life and acts
as a waste receptacle for the exhaust
gases that accompany many industrial
processes leading to many types of
pollution, including smog and acid rain.
A Gas
-
Uniformly fills any container.
-
Mixes completely with any other gas
-
Exerts pressure on its surroundings.
-
Is easily compressed.
05_47
Vacuum
h = 760mm Hg
for standard
atmosphere
Simple barometer invented by Evangelista Torricelli.
A barometer is used to measure atmospheric pressure.
Atmospheric Pressure
Atmospheric pressure results from the mass
of the air being pulled toward the center
of the earth by gravity--the weight of the
air.
At sea level, atmospheric pressure is
760mm.
Atmospheric pressure varies with altitude-at Breckenridge, CO (elevation 9600 ft),
the pressure is 520 mm.
Figure 12.1: The pressure exerted by the gases in the
atmosphere can demonstrated by boiling water in a can
Pressure
-
is equal to force/unit area
-
SI units = Newton/meter2 = 1 Pascal (Pa)
-
1 standard atmosphere = 101,325 Pa
-
1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr
Atmospheric
pressure (Patm)
05_48
Atmospheric
pressure (Patm)
h
Gas
pressure (Pgas)
less than
atmospheric
pressure
(Pgas) = (Patm) - h
(a)
h
Gas
pressure (Pgas)
greater than
atmospheric
pressure
(Pgas) = (Patm) + h
(b)
A simple manometer is a device for measuring
the pressure of a gas in a container.
Pressure Unit Conversions
The pressure of a tire is measured to be 28
psi. What would the pressure be in
atmospheres, torr, and pascals.
(28 psi)(1.000 atm/14.69 psi) = 1.9 atm
(28 psi)(1.000 atm/14.69 psi)(760.0
torr/1.000atm) = 1.4 x 103 torr
(28 psi)(1.000 atm/14.69 psi)(101,325
Pa/1.000 atm) = 1.9 x 105 Pa
05_1541
Pext
Pext
Volume is decreased
Volume of a gas decreases as pressure increases
at constant temperature
05_50
40
slope = k
V(in3)
P (in Hg)
100
50
20
P
P
2
0
0
20
40
60
0
0.01
0.02
0.03
1/P (in Hg)
V
2V
V(in3)
(a)
P vs V
(b)
V vs 1/P
BOYLE’S LAW DATA
Figure 12.6: Illustration of Boyle’s law
Boyle’s
*
Law
Robert Boyle -- Irish scientist (1627-1691)
(Pressure)( Volume) = Constant (T = constant)
P1V1 = P2V2
V  1/P
(T = constant)
(T = constant)
(*Holds precisely only at very low pressures.)
The temperature and amount of gas must
remain unchanged.
A gas that strictly obeys
Boyle’s Law is called an
ideal gas.
Boyle’s Law Calculations
A 1.5-L sample of gaseous CCl2F2 has a pressure
of 56 torr. If the pressure is changed to 150
torr, will the volume of the gas increase or
decrease? What will the new volume be?
Decrease
P1 = 56 torr
P2 = 150 torr
V1 = 1.5 L
V2 = ?
V1P1 = V2P2
V2 = V1P1/P2
V2 = (1.5 L)(56 torr)/(150 torr)
V2 = 0.56 L
Boyle’s Law Calculations
In an automobile engine the initial cylinder
volume is 0.725 L. After the piston moves up,
the volume is 0.075 L. The mixture is 1.00
atm, what is the final pressure?
P1 = 1.00 atm
P2 = ?
V1 = 0.725 L
V1P1 = V2P2
P2 = V1P1/V2
P2 = (0.725 L)(1.00 atm)/(0.075 L)
P2 = 9.7 atm
V2 = 0.075 L
Is this answer reasonable?
He
05_53
6
5
CH4
V(L)
4
3
H2O
2
H2
1
N2O
-300
-200
-273.2 ºC
-100
0
100
200
300
T(ºC)
Plot of V vs. T(oC) for several gases
05_1543
Pext
Pext
Energy (heat) added
Volume of a gas increases as heat is added
when pressure is held constant.
Charles’s Law
Jacques Charles -- French physicist (17461823)
The volume of a gas is directly proportional to
temperature, and extrapolates to zero at zero
Kelvin.
V = bT
(P = constant)
y = mx + b
b = a proportionality constant
Charles’s Law
V1
V2

T1
T2
( P  constant)
Temperature must be in Kelvin!!
Charles’s Law Calculations
A 2.0-L sample of air is collected at 298 K and
then cooled to 278 K. The pressure is held
constant. Will the volume increase or
decrease? What will the new volume be?
Decrease
V1 = 2.0 L
V2 = ?
T1 = 298 K
T2 = 278 K
V1/V2 = T1/T2
V2 = V1 T2/T1
V2 = (2.0 L)(278 K)/(298 K)
V2 = 1.9 L
Charles’s Law Calculations
Consider a gas with a a volume of 0.675 L at 35 oC and
1 atm pressure. What is the temperature (in Co) of
the gas when its volume is 0.535 L at 1 atm
pressure?
V1/V2 = T1/T2
V1 = 0.675 L
T2 = T1 V2/V1
T2 = (308 K)(0.535 L)/(0.675 L)
V2 = 0.535 L
T2 = 244 K -273
T1 = 35 oC + 273 = 308 K T2 = - 29 oC
T2 = ?
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Pext
Pext
Gas cylinder
Moles of gas
increases
Pext
Increase volume to
return to original
pressure
At constant temperature and pressure, increasing
the moles of a gas increases its volume.
Figure 12.11: The total pressure of a mixture of gases
depends on the number of moles of gas particles
(atoms or molecules) present
Avogadro’s Law
For a gas at constant temperature and
pressure, the volume is directly proportional
to the number of moles of gas (at low
pressures).
V = an
y = mx + b
a = proportionality constant
V = volume of the gas
n = number of moles of gas
AVOGADRO’S LAW
V1 n1

V2 n 2
Temperature and pressure must remain
constant!!
AVOGADRO’S LAW
A 12.2 L sample containing 0.50 mol of oxygen gas,
O2, at a pressure of 1.00 atm and a temperature of
25 oC is converted to ozone, O3, at the same
temperature and pressure, what will be the volume
of the ozone? 3 O2(g) ---> 2 O3(g)
(0.50 mol O2)(2 mol O3/3 mol O2) = 0.33 mol O3
V1 = 12.2 L
V2 = ?
n1 = 0.50 mol
n2 = 0.33 mol
V1/V2 = n1/n2
V2 = V1 n2/n1
V2 = (12.2 L)(0.33 mol)/(0.50 mol)
V2 = 8.1 L
COMBINED GAS LAW
What will be the new volume of a gas under the
following conditions?
V1 = 3.48 L
V2 = ?
P1 = 0.454 atm
P2 = 0.616 atm
T1 = - 15 oC + 273
= 258 K
T2 = 36 oC + 273
T2= 309 K
V1/ V2 = P2 T1/ P1 T2
V2 = V1P1T2/P2T1
V2 = (309 K)(0.454 atm)(3.48 L)
(258 K)(0.616 atm)
V2 = 3.07 L
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Pext
Pext
Temperature is increased
Pressure exerted by a gas increases as temperature
increases provided volume remains constant.
If the volume of a gas is held
constant, then V1 / V2 = 1.
Therefore:
P1 / P2 = T1 / T2
Pressure and Kelvin temperature are directly
proportional.
IDEAL GAS
1. Molecules are infinitely far apart.
2. Zero attractive forces exist between the
molecules.
3. Molecules are infinitely small--zero
molecular volume.
What is an example of an ideal gas?
REAL GAS
1. Molecules are relatively far apart
compared to their size.
2. Very small attractive forces exist between
molecules.
3. The volume of the molecule is small
compared to the distance between
molecules.
What is an example of a real gas?
Real vs. Ideal Gases
When p  1 atm and T  O oC,
real gases approximate ideal
gases.
Ideal Gas Law
-
An equation of state for a gas.
-
“state” is the condition of the gas at a
given time.
PV = nRT
Ideal Gas Law
PV = nRT
R = proportionality constant
= 0.08206 L atm  mol
P = pressure in atm
V = volume in liters
n = moles
T = temperature in Kelvins
Holds closely at P  1 atm
Ideal Gas Law Calculations
A sample of hydrogen gas, H2, has a volume of 8.56 L at a
temperature of O oC and a pressure of 1.5 atm.
Calculate the number of moles of hydrogen present.
p = 1.5 atm
pV
=
nRT
V = 8.56 L
n = pV/RT
R = 0.08206 Latm/molK n = (1.5 atm)(8.56L)
(0.08206 Latm/molK)(273K)
n=?
n = 0.57 mol
T = O oC + 273
T = 273K
Ideal Gas Law Calculations
A 1.5 mol sample of radon gas has a volume of 21.0 L at
33 oC. What is the pressure of the gas?
p=?
pV = nRT
p = nRT/V
V = 21.0 L
p = (1.5mol)(0.08206Latm/molK)(306K)
n = 1.5 mol
(21.0L)
T = 33 oC + 273 p = 1.8 atm
T = 306 K
R = 0.08206 Latm/molK
Dalton’s Law of
Partial Pressures
For a mixture of gases in a container,
the total pressure exerted is the
sum of the partial pressures of the
gases present.
PTotal = P1 + P2 + P3 + . . .
Dalton’s Law of Partial
Pressures Calculations
A mixture of nitrogen gas at a pressure of
1.25 atm, oxygen at 2.55 atm, and
carbon dioxide at .33 atm would have
what total pressure?
PTotal = P1 + P2 + P3
PTotal = 1.25 atm + 2.55 atm + .33 atm
Ptotal = 4.13 atm
Water Vapor Pressure
2KClO3(s) ----> 2KCl(s) + 3O2(g)
When a sample of potassium chlorate is decomposed and
the oxygen produced collected by water
displacement, the oxygen has a volume of 0.650 L at
a temperature of 22 oC. The combined pressure of
the oxygen and water vapor is 754 torr (water vapor
pressure at 22 oC is 21 torr). How many moles of
oxygen are produced?
Pox = Ptotal - PHOH
Pox = 754 torr - 21 torr
pox = 733 torr
Water Vapor Pressure
Continued
p = (733 torr)(1 atm/760 torr)
p = 0.964 atm
V = 0.650 L
n=?
T = 22 oC + 273
pV = nRT
n = pV/RT
n = (0.964 atm)(0.650 L)
(0.08206 Latm/molK)(295K)
n = 2.59 x 10-2 mol
T = 295 K
R = 0.08206 Latm/molK
Kinetic Molecular Theory
1. Volume of individual particles is  zero.
2. Collisions of particles with container
walls cause pressure exerted by gas.
3. Particles exert no forces on each other.
4. Average kinetic energy  Kelvin
temperature of a gas.
5. Gases consist of tiny particles (atoms or
molecules.
Relative number of O2 molecules
with given velocity
05_58
0
4 x 102
8 x102
Molecular velocity (m/s)
Plot of relative number of oxygen molecules with
a given velocity at STP (Boltzmann Distribution).
Relative number of N2 molecules
with given velocity
05_59
273 K
1273 K
2273 K
0
1000
2000
Velocity (m/s)
3000
Plot of relative number of nitrogen molecules with
a given velocity at three different temperatures.
Kinetic Molecular Theory
As the temperature of a gas increases, the
velocity of the molecules increases.
Molecules at a higher temperature have
greater KE and hit the walls of the
container harder and more often exerting
more pressure. If the container is not
rigid, the container and volume of the
gas will expand.
Standard Temperature
and Pressure
“STP”
P = 1 atmosphere
T = C
The molar volume of an ideal gas is 22.42
liters at STP
Molar Volume
pV = nRT
V = nRT/p
V = (1.00 mol)(0.08206 Latm/molK)(273K)
(1.00 atm)
V = 22.4 L
Figure 12.12: The production of oxygen by thermal
decomposition of KClO3
GAS STOICHIOMETRY
Not at STP
Calculate the volume of oxygen gas produced at 1.00
atm and 25 oC by the complete decomposition of
10.5 g of potassium chlorate.
2KClO3(s) ----> 2KCl(s) + 3O2(g)
(10.5g KClO3)(1mol/122.6 g)(3 mol O2/2mol KClO3)
= 1.28 x 10-1 mol O2
Gas Stoichiometry
Not at STP (Continued)
p = 1.00 atm
V=?
n = 1.28 x 10-1 mol
R = 0.08206 Latm/molK
T = 25 oC + 273 = 298 K
pV = nRT
V = nRT/p
V = (1.28 x 10-1mol)(0.08206Latm/molK)(298K)
(1.00 atm)
V = 3.13 L O2
Gases at STP
A sample of nitrogen gas has a volume of 1.75 L at
STP. How many moles of N2 are present?
(1.75L N2)(1.000 mol/22.4 L) = 7.81 x 10-2 mol N2
Gas Stoichiometry
at STP
Quicklime, CaO, is produced by heating calcium
carbonate, CaCO3. Calculate the volume of CO2
produced at STP from the decomposition of 152 g of
CaCO3. CaCO3(s) ---> CaO(s) + CO2(g)
(152g CaCO3)(1 mol/100.1g)(1mol CO2/1mol CaCO3)
(22.4L/1mol) = 34.1L CO2
Note: This method only works when the gas is
at STP!!!!!
Chemistry is a gas!!