GASES
Chapter 5
A Gas
-
Uniformly fills any container.
-
Mixes completely with any other gas
-
Exerts pressure on its surroundings.
05_47
Vacuum
h = 760mm Hg
for standard
atmosphere
Simple barometer invented by Evangelista Torricelli
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
Pressure Unit Conversions
The pressure of a tire is measured to be 28
psi. What would the pressure 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
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)
Simple Manometer
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
Boyle’s Law*
(Pressure)( Volume) = Constant (T = constant)
P1V1 = P2V2
V  1/P
(T = constant)
(T = constant)
(*Holds precisely only at very low pressures.)
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?
A gas that strictly obeys
Boyle’s Law is called an
ideal gas.
05_51
Ideal
22.45
Ne
PV (L • atm)
22.40
O2
22.35
CO2
22.30
22.25
0
0.25
0.50
0.75
1.00
P (atm)
Plot of PV vs. P for several gases at pressures
below 1 atm.
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
The volume of a gas is directly proportional
to temperature, and extrapolates to zero at
zero Kelvin.
V = bT (P = constant)
b = a proportionality constant
Charles’s Law
V1
V2

T1
T2
( P  constant)
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 = ?
05_1544
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.
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
a = proportionality constant
V = volume of the gas
n = number of moles of gas
AVOGADRO’S LAW
V1/V2 = n1/n2
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
V1/ V2 = P2 T1/ P1 T2
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
05_1542
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
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
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?
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 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
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
Standard Temperature
and Pressure
“STP”
P = 1 atmosphere
T = C
The molar volume of an ideal gas is 22.42
liters at STP
GAS STOICHIOMETRY
1. Mass-Volume
2. Volume-Volume
Molar Volume
pV = nRT
V = nRT/p
V = (1.00 mol)(0.08206 Latm/molK)(273K)
(1.00 atm)
V = 22.4 L
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!!!!!
Volume-Volume
If 25.0 L of hydrogen reacts with an excess of
nitrogen gas, how much ammonia gas will be
produced? All gases are measured at the same
temperature and pressure.
2N2(g) + 3H2(g) ----> 2NH3(g)
(25.0 L H2)(2 mol NH3/3 mol H2) = 16.7 L NH3
MOLAR MASS OF A GAS
n = m/M
n = number of moles
m = mass
M = molar mass
MOLAR MASS OF A GAS
P = mRT/VM
or
P = DRT/M
therefore:
M = DRT/P
Molar Mass Calculations
M=?
d = 1.95 g/L
T = 27 oC + 273
p = 1.50 atm
T = 300. K
R = 0.08206 Latm/mol K
M = dRT/p
M = (1.95g/L)(0.08206Latm/molK)(300.K)
(1.5atm)
M = 32.0 g/mol
Dalton’s Law of
Partial Pressures
For a mixture of gases in a container,
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
MOLE FRACTION
-- the ratio of the number of moles of a
given component in a mixture to the
total number of moles of the mixture.
1 = n1/ ntotal
1 = V1/ Vtotal
1 = P1 / Ptotal (volume & temperature
constant)
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.
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.
The Meaning of
Temperature
(KE) avg
3

RT
2
Kelvin temperature is an index of the random
motions of gas particles (higher T means greater
motion.)
 rms
3RT

M
Diffusion: describes the mixing of
gases. The rate of diffusion is the
rate of gas mixing.
Effusion: describes the passage of
gas into an evacuated chamber.
05_60
Pinhole
Gas
Vacuum
Effusion of a gas into an evacuated chamber
Effusion:
Rate of effusion for gas 1

Rate of effusion for gas 2
M2
M1
Diffusion:
Distance traveled by gas 1

Distance traveled by gas 2
M2
M1
Real Gases
Must correct ideal gas behavior
when at high pressure (smaller
volume) and low temperature
(attractive forces become important).
05_62
CH4
N2
2.0
H2
PV
nRT
CO2
1.0
Ideal
gas
0
0
200
400
600
800
1000
P(atm)
Plots of PV/nRT vs. P for several gases at 200 K.
Note the significant deviation from ideal behavior.
Real Gases
2
[ Pobs  a (n / V ) ]  V  nb  nRT

corrected pressure
Pideal

corrected volume
Videal
05_68
Molecules of unburned
fuel (petroleum)
0.4
Other
pollutants
0.3
NO2
0.2
O3
NO
6:00
4:00
2:00
Noon
10:00
8:00
0
6:00
0.1
4:00
Concentration (ppm)
0.5
Time of day
Concentration for some smog components
vs. time of day
NO2(g) NO(g) + O(g)
O(g) + O2(g)  O3(g)
NO
+ 1/2 O
 NO
(g)
2(g)
2(g)
__________________________
3/2 O2(g)  O3(g)
What substances represent intermediates?
Which substance represents the catalyst?
Chemistry is a gas!!