Gases

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Chapter 5

• Gases

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Chapter 5 Gases
5.1 Early Experiments 5.2 The gas laws of
Boyle, Charles, and Avogadro 5.3 The Ideal
Gas Law 5.4 Gas Stoichiometry 5.5
Daltons Law of Partial Pressures 5.6 The
Kinetic molecular Theory of Gases 5.7
Effusion and Diffusion 5.8 Collisions of Gas
Particles with the Container Walls 5.9
Intermolecular Collisions 5.10 Real Gases 5.11
Chemistry in the Atmosphere
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Hurricanes, such as this one off the coast of
Florida, are evidence of the powerful forces
present in the earth's atmosphere.
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Important Characteristics of Gases
1) Gases are highly compressible An external
force compresses the gas sample and decreases
its volume, removing the external force
allows the gas volume to increase. 2) Gases
are thermally expandable When a gas sample
is heated, its volume increases, and when it is
cooled its volume decreases. 3) Gases have
low viscosity Gases flow much easier than
liquids or solids. 4) Most Gases have low
densities Gas densities are on the order of
grams per liter whereas liquids and solids
are grams per cubic cm, 1000 times greater. 5)
Gases are infinitely miscible
Gases mix in any proportion such as in air, a
mixture of many gases.
5

Substances that are Gases under
Normal Conditions
Substance Formula
MM(g/mol)

• Helium He 4.0
• Neon Ne 20.2
• Argon Ar 39.9
• Hydrogen H2 2.0
• Nitrogen N2 28.0
• Nitrogen Monoxide NO
  30.0
• Oxygen O2 32.0
• Hydrogen Chloride HCL 36.5
• Ozone O3 48.0
• Ammonia NH3
  17.0
• Methane CH4 16.0

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Some Important Industrial Gases
Name - Formula Origin and
use
Methane (CH4) Natural
deposits domestic fuel Ammonia (NH3)
From N2 H2 fertilizers,
explosives Chlorine (Cl2)
Electrolysis of seawater bleaching
and
disinfecting Oxygen (O2)
Liquefied air steelmaking Ethylene (C2H4)
High-temperature decomposition of

natural gas plastics
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Pressure of the Atmosphere

• Called Atmospheric pressure, or the force
  exerted upon us by the atmosphere above us.
• A measure of the weight of the atmosphere
  pressing down upon us.
• Measured using a Barometer! - A device that can
  weigh the atmosphere above us!

Force Area
Pressure
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Figure 5.1 A torricellian barometer.
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Construct a Barometer using Water!

• Density of water 1.00 g/cm3
• Density of Mercury 13.6 g/cm3
• Height of water column Hw
• Hw Height of Hg x Density of Mercury
• 
  
• Hw 760 mm Hg x 13.6/1.00 1.03 x 104 mm
• Hw 10.3 m __________ ft

HeightWater HeightMercury
DensityMercury DensityWater

Density of Water
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Common Units of Pressure
Unit Atmospheric Pressure
Scientific Field Used
Pascal (Pa) 1.01325 x 105 Pa
SI unit physics, kilopascal (kPa)
101.325 kPa
chemistry Atmosphere (atm) 1
atm Chemistry Millimet
ers of mercury 760 mmHg
Chemistry, medicine (mmHg)

biology Torr
760 torr
Chemistry Pounds per square inch 14.7
lb/in2 Engineering (psl or
lb/in2) Bar
1.01325 bar Meteorology, chemistry
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Converting Units of Pressure
Problem A chemist collects a sample of Carbon
dioxide from the decomposition of Limestone
(CaCO3) in a closed end manometer, the height of
the mercury is 341.6 mm Hg. Calculate the CO2
pressure in torr, atmospheres, and
kilopascals. Plan The pressure is in mmHg, so we
use the conversion factors from Table 5.2(p.178)
to find the pressure in the other units. Solution
converting from mmHg to torr
1 torr 1 mm Hg
PCO2 (torr) 341.6 mm Hg x
341.6 torr
converting from torr to atm
1 atm 760 torr
PCO2( atm) 341.6 torr x
0.4495 atm
converting from atm to kPa
101.325 kPa 1 atm
PCO2(kPa) 0.4495 atm x
_____________ kPa
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Figure 5.2 A simple manometer, a device for
measuring the pressure of a gas in a container.
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Figure 5.3 A J-tube similar to the one used by
Boyle.
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Boyles Law P - V relationship

• Pressure is inversely proportional to Volume
• P or V
  or PVk
• Change of Conditions Problems
• if n and T are constant !
• P1V1 k P2V2 k
• k k
• Then
• P1V1 P2V2

k V
k P
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Figure 5.4 Plotting Boyles data from Table
5.1.
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Figure 5.5 Plot of PV versus P for several
gases.
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Applying Boyles Law to Gas Problems
Problem A gas sample at a pressure of 1.23 atm
has a volume of 15.8 cm3, what will be the
volume if the pressure is increased to 3.16
atm? Plan We begin by converting the volume that
is in cm3 to ml and then to liters, then we do
the pressure change to obtain the final
volume! Solution
V1 (cm3)
P1 1.23 atm P2 3.16 atm V1 15.8 cm3
V2 unknown T and n remain constant
1cm3 1 mL
V1 (ml)
1000mL 1L
1 mL 1 cm3
1 L 1000mL
V1 15.8 cm3 x x
0.0158 L
V1 (L)
x P1/P2
P1 P2
1.23 atm 3.16 atm
V2 V1 x 0.0158 L x
________ L
V2 (L)
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Boyles Law - A gas bubble in the ocean!
A bubble of gas is released by the submarine
Alvin at a depth of 6000 ft in the ocean, as
part of a research expedition to study under-
water volcanism. Assume that the ocean is
isothermal (the same temperature throughout) ,a
gas bubble is released that had an initial
volume of 1.00 cm3, what size will it be at the
surface at a pressure of 1.00 atm?(We will
assume that the density of sea water is 1.026
g/cm3, and use the mass of Hg in a barometer for
comparison!)
Initial Conditions
Final Conditions
V 1 1.00 cm3
V 2 ?
P 1 ?
P 2 1.00 atm
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Calculation Continued
0.3048 m 1 ft
100 cm 1 m
1.026 g SH2O 1 cm3
Pressure at depth 6 x 103 ft x
x x
Pressure at depth 187,634.88 g pressure from
SH2O
For a Mercury Barometer 760 mm Hg 1.00 atm,
assume that the cross-section of the barometer
column is 1 cm2.
The mass of Mercury in a barometer is
10 mm 1 cm
Area 1 cm2
1.00 cm3 Hg 13.6 g Hg
1.00 atm 760 mm Hg
Pressure x x
x x

187,635 g
Pressure _____ atm Due to the added atmospheric
pressure ___ atm!
V1 x P1 P2
1.00 cm3 x ____ atm 1.00 atm
V2
____ cm3 liters
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Boyles Law Balloon

• A balloon has a volume of 0.55 L at sea level
• (1.0 atm) and is allowed to rise to an altitude
  of 6.5 km, where the pressure is 0.40 atm. Assume
  that the temperature remains constant (which
  obviously is not true). What is the final volume
  of the balloon?
• P1 1.0 atm P2 0.40 atm
• V1 0.55 L V2 ?
• V2 V1 x P1/P2 (0.55 L) x (1.0 atm / 0.40 atm)
• V2 __________ L

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Figure 5.6 PV plot verses P for 1 mole of
ammonia
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Charles Law - V - T- relationship

• Temperature is directly related to volume
• T proportional to Volume T kV
• Change of conditions problem
• Since T/V k or T1 / V1 T2 / V2
  or

Temperatures must be expressed in Kelvin to avoid
negative values!
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Figure 5.7 Plots of V versus T (c) for several
gases.
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Figure 5.8 Plots of V versus T as in Fig. 5.7
except that here the Kelvin scale is used for
temperature.
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Charles Law Problem

• A sample of carbon monoxide, a poisonous gas,
  occupies 3.20 L at 125 oC. Calculate the
  temperature (oC) at which the gas will occupy
  1.54 L if the pressure remains constant.
• V1 3.20 L T1 125oC 398 K
• V2 1.54 L T2 ?
• T2 T1 x ( V2 / V1) T2 398 K x
  ______K
• T2 ___ K oC K - 273.15 ______ - 273
  
• oC ______oC

1.54 L 3.20 L
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Charles Law Problem - I

• A balloon in Antarctica in a building is at room
  temperature ( 75o F ) and has a volume of 20.0 L
  . What will be its volume outside where the
  temperature is -70oF ?
• V1 20.0 L V2 ?
• T1 75o F T2 -70o F
• o C ( o F - 32 ) 5/9
• T1 ( 75 - 32 )5/9 23.9o C
• K 23.9o C 273.15
  __________ K
• T2 ( -70 - 32 ) 5/9 - 56.7o C
• K - 56.7o C 273.15
  ___________ K

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Antarctic Balloon Problem - II

• V1 / T1 V2 / T2 V2 V1 x ( T2 / T1 )
• V2 20.0 L x
• V2 __________ L
• The Balloon shrinks from 20 L to 15 L !!!!!!!
• Just by going outside !!!!!

216.4 K 297.0 K
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Applying the Temperature - Pressure
Relationship (Amontons Law)
Problem A copper tank is compressed to a
pressure of 4.28 atm at a temperature of 0.185
oF. What will be the pressure if the temperature
is raised to 95.6 oC? Plan The volume of the
tank is not changed, and we only have to deal
with the temperature change, and the pressure, so
convert to SI units, and calculate the change
in pressure from the Temp.and Pressure
change. Solution
T1 (0.185 oF - 32.0 oF)x 5/9 -17.68 oC T1
-17.68 oC 273.15 K 255.47 K T2 95.6 oC
273.15 K 368.8 K
368.8 K 255.47 K
P2 4.28 atm x _______ atm
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Avogadros Law - Amount and Volume
The Amount of Gas (Moles) is directly
proportional to the volume of the Gas.
n ? V or
n kV
For a change of conditions problem we have the
initial conditions, and the final conditions, and
we must have the units the same.
n1 initial moles of gas V1 initial
volume of gas n2 final moles of gas V2
final volume of gas
n1 V1
n2 V2
V1 V2

or n1 n2 x
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Avogadros Law Volume and Amount of Gas
Problem Sulfur hexafluoride is a gas used to
trace pollutant plumes in the atmosphere, if the
volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC
is 2.93 m3, what will be the mass of SF6 in a
container whose volume is 543.9 m3 at 1.143 atm
and 28.5 oC? Plan Since the temperature and
pressure are the same it is a V - n problem, so
we can use Avogadros Law to calculate the moles
of the gas, then use the molecular mass to
calculate the mass of the gas. Solution Molar
mass SF6 146.07 g/mol
2.67g SF6 146.07g SF6/mol
0.0183 mol SF6
543.9 m3 2.93 m3
mass SF6 3.40 mol SF6 x 146.07 g SF6 / mol
__________ g SF6
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Volume - Amount of Gas Relationship
Problem A balloon contains 1.14 moles(2.298g H2)
of Hydrogen and has a volume of 28.75 L. What
mass of Hydrogen must be added to the balloon to
increase its volume to 112.46 Liters? Assume T
and P are constant. Plan Volume and amount of
gas are changing with T and P constant, so we
will use Avogadros law, and the change of
conditions form. Solution
V1 28.75 L V2 112.46 L
T constant P constant
n1 1.14 moles H2 n2 1.14 moles ? moles
n1 n2 V1 V2
V2 V1
112.46 L 28.75 L

n2 n1 x 1.14 moles H2 x
mass moles x molecular mass mass 4.46 moles x
2.016 g / mol mass ___________ g H2 gas
n2 4.4593 moles 4.46 moles
added mass 8.99g - 2.30g 6.69g
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Change of Conditions, with no change in the
amount of gas !

• constant Therefore for a
  change
• of
  conditions
• T1
  T2

P x V
T
P1 x V1
P2 x V2

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Change of Conditions Problem -I

• A gas sample in the laboratory has a volume of
  45.9 L at 25 oC and a pressure of 743 mm Hg. If
  the temperature is increased to 155 oC by pumping
  (compressing) the gas to a new volume of 3.10 ml
  what is the pressure?
• P1 743 mm Hg x1 atm/ 760 mm Hg0.978 atm
• P2 ?
• V1 45.9 L V2 3.10 ml 0.00310 L
• T1 25 oC 273 298 K
• T2 155 oC 273 428 K

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Change of Condition Problem I continued
P1 x V1
P2 x V2

• 
  
• 
  

T1
T2
P2 (0.00310 L)
( 0.978 atm) ( 45.9 L)

( 298 K)
( 428 K)
( 428 K) ( 0.978 atm) ( 45.9 L)
________ atm
P2
( 298 K) ( 0.00310 L)
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Change of Conditions Problem II

• A weather balloon is released at the surface of
  the earth. If the volume was 100 m3 at the
  surface ( T 25 oC, P 1 atm ) what
  will its volume be at its peak altitude of 90,000
  ft where the temperature is - 90 oC and the
  pressure is 15 mm Hg ?
• Initial Conditions Final
  Conditions
• V1 100 m3 V2 ?
• T1 25 oC 273.15 T2 -90 oC
  273.15
• 298 K
  183 K
• P1 1.0 atm P2 15 mm Hg
  

760 mm Hg/ atm P2 ___________ atm
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Change of Conditions Problem II continued

• P1 x V1 P2 x V2
  
• V2
  
• 
  
• V2
  
• V2 3117.23 m3 __________ m3 or 31 times
  the
• 
  volume !!!

P1V1T2

T1
T2
T1P2
( 1.0 atm) ( 100 m3) (183 K)
(298 K) (0.0197 atm)

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Change of Conditions Problem III

• How many liters of CO2 are formed at 1.00 atm and
  900 oC if 5.00 L of Propane at 10.0 atm, and 25
  oC is burned in excess air?
• C3H8 (g) 5 O2 (g) 3 CO2 (g) 4 H2O(g)
• 25 oC 273 298 K
• 900 oC 273 1173 K

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Change of Conditions Problem IIIcontinued

• V1 5.00 L V2 ?
• P1 10.0 atm P2 1.00 atm
• T1 298K T2 1173 K
• P1V1/T1 P2V2/T2 V2 V1P1T2/
  P2T1
• V2
  197 L
• VCO2 (197 L C3H8) x (3 L CO2 / 1 L C3H8)
• VCO2 _________ L CO2

( 5.00 L) (10.00 atm) (1173 K)
( 1.00 atm) ( 298 K)
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Standard Temperature and Pressure (STP)
A set of Standard conditions have been chosen to
make it easier to understand the gas laws, and
gas behavior.
Standard Temperature 00 C 273.15 K
Standard Pressure 1 atmosphere 760 mm Mercury
At these standard conditions, if you have 1.0
mole of a gas it will occupy a standard molar
volume.
Standard Molar Volume 22.414 Liters 22.4 L
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Table 5.2 (P 149) Molar Volumes for Various
Gases at 00 and
1 atm
Gas
Molar Volume (L)
Oxygen (O2)
22.397 Nitrogen (N2)
22.402 Hydrogen (H2)
22.433 Helium (He)
22.434 Argon (Ar)

22.397 Carbon dioxide (CO2)
22.260 Ammonia (NH3)
22.079
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IDEAL GASES

• An ideal gas is defined as one for which both the
  volume of molecules and forces between the
  molecules are so small that they have no effect
  on the behavior of the gas.
• The ideal gas equation is
• PVnRT
• R Ideal gas constant
• R 8.314 J / mol K 8.314 J mol-1 K-1
• R 0.08206 l atm mol-1 K-1

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Variations on the Gas Equation

• During chemical and physical processes, any of
  the four variables in the ideal gas equation may
  be fixed.
• Thus, PVnRT can be rearranged for the fixed
  variables
• for a fixed amount at constant temperature
• P V nRT constant Boyles
  Law
• for a fixed amount at constant volume
• P / T nR / V constant Amontons Law
• for a fixed amount at constant pressure
• V / T nR / P constant Charles Law
• for a fixed volume and temperature
• P / n R T / V constant Avogadros Law

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Evaluation of the Ideal Gas Constant, R
PV nT
Ideal gas Equation PV nRT R
at Standard Temperature and Pressure, the molar
volume 22.4 L P 1.00 atm (by
Definition!) T 0 oC 273.15 K (by
Definition!) n 1.00 moles (by Definition!)
(1.00 atm) ( 22.414 L) ( 1.00 mole) ( 273.15 K)
L atm mol K
R
0.08206
L atm mol K
or to three significant figures R 0.0821
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Values of R in Different Units
Atm x L Mol x K
R 0.08206 R 62.36 R 8.314 R
8.314
Torr x L Mol x K
kPa x dm3 Mol x K
J Mol x K
Most calculations in this text use values of R
to 3 significant figures. J is the
abbreviation for joule, the SI unit of energy.
The joule is a derived unit composed of the base
units kg x m2/s2
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Gas Law Solving for Pressure
Problem Calculate the pressure in a container
whose Volume is 87.5 L and it is filled with
5.038kg of Xenon at a temperature of 18.8
oC. Plan Convert all information into the units
required, and substitute into the Ideal Gas
equation ( PVnRT ). Solution
5038 g Xe 131.3 g Xe / mol
nXe 38.37014471
mol Xe
T 18.8 oC 273.15 K 291.95 K
PV nRT P nRT
V
(38.37 mol )(0.0821 L atm)(291.95 K)
P
_______ atm atm
87.5 L
(mol K)
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Ideal Gas Calculation - Nitrogen

• Calculate the pressure in a container holding 375
  g of Nitrogen gas. The volume of the container is
  0.150 m3 and the temperature is 36.0 oC.
• n 375 g N2/ 28.0 g N2 / mol 13.4 mol N2
• V 0.150 m3 x 1000 L / m3 150 L
• T 36.0 oC 273.15 309.2 K
• PVnRT P nRT/V
• P
• P atm

( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 K)
150 L
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Mass of Air in a Hot Air Balloon - Part I

• Calculate the mass of air in a spherical hot air
  balloon that has a volume of 14,100 cubic feet
  when the temperature of the gas is 86 oF and the
  pressure is 748 mm Hg?
• P 748 mm Hg x 1atm / 760 mm Hg 0.984 atm
• V 1.41 x 104 ft3x (12 in/1 ft)3x(2.54 cm/1 in)3
  x
• x (1ml/1 cm3) x ( 1L / 1000 cm3) 3.99 x
  105 L
• T oC (86-32)5/9 30 oC
• 30 oC 273 303 K

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Mass of Air in a Hot Air Balloon - Part II

• PV nRT n PV / RT
• n
  1.58 x 104 mol
• mass 1.58 x 104 mol air x 29 g air/mol air
• 4.58 x 105 g Air
• 458 Kg Air

( 0.984 atm) ( 3.99 x 105 L)
( 0.08206 L atm/mol K) ( 303 K )
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Inflated dual air bag
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Sodium Azide Decomposition-I

• Sodium Azide (NaN3) is used in some air bags in
  automobiles. Calculate the volume of Nitrogen gas
  generated at 21 oC and 823 mm Hg by the
  decomposition of 60.0 g of NaN3 .
• 2 NaN3 (s) 2
  Na (s) 3 N2 (g)
• mol NaN3 60.0 g NaN3 / 65.02 g NaN3 / mol
• 0.9228 mol NaN3
• mol N2 0.9228 mol NaN3 x 3 mol N2/2 mol NaN3
• ___________mol N2

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Sodium Azide Calc - II

• PV nRT V nRT/P
• V
• V __________ liters

( 1.38 mol) (0.08206 L atm / mol K) (294 K)
( 823 mm Hg / 760 mmHg / atm )
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Ammonia Density Problem

• Calculate the Density of ammonia gas (NH3) in
  grams per liter at 752 mm Hg and 55 oC.
• Density mass per unit volume
  g / L
• P 752 mm Hg x (1 atm/ 760 mm Hg) 0.989 atm
• T 55 oC 273 328 K
• n mass / Molar mass
  g / M
• d
• d __________ g / L

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Like Example 5.6 (P 149)
Problem Calculate the volume of carbon dioxide
produced by the thermal decomposition of
36.8 g of calcium carbonate at 715mm Hg,
and 370 C according to the reaction below
CaCO3(s)
CaO(s) CO2 (g) Solution
1 mol CaCO3 100.1g CaCO3
36.8g x 0.368 mol
CaCO3
Since each mole of CaCO3 produces one 1 mole of
CO2 , 0.368 moles of CO2 at STP, we now have to
convert to the conditions stated
nRT P
(0.368 mol)(0.08206L atm/mol K)(310.15K)
(715mmHg/760mm Hg/atm)
V
V __________ L
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Calculation of Molar Mass

• n
• n

Mass
Molar Mass
P x V
Mass
R x T
Molar Mass
Mass x R x T
Molar Mass
MM
P x V
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Dumas Method of Molar Mass
Problem A volatile liquid is placed in a flask
whose volume is 590.0 ml and allowed to boil
until all of the liquid is gone, and only vapor
fills the flask at a temperature of 100.0 oC and
736 mm Hg pressure. If the mass of the flask
before and after the experiment was 148.375g and
149.457 g, what is the molar mass of the
liquid? Plan Use the gas law to calculate the
molar mass of the liquid. Solution
1 atm 760 mm Hg
Pressure 736 mm Hg x
0.9684 atm
mass 149.457g - 148.375g 1.082 g
(1.082 g)(0.0821 Latm/mol K)(373.2 K)
Molar Mass
58.02 g/mol
( 0.9684 atm)(0.590 L)
note the compound is acetone C3H6O MM 58
g/mol !
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Calculation of Molecular Weight of a Gas
Natural Gas - Methane
Problem A sample of natural gas is collected at
25.0 oC in a 250.0 ml flask. If the sample had a
mass of 0.118 g at a pressure of 550.0 Torr, what
is the molecular weight of the gas? Plan Use the
Ideal gas law to calculate n, then calculate the
molar mass. Solution
1mm Hg 1 Torr
1.00 atm 760 mm Hg
P 550.0 Torr x x
0.724 atm
1.00 L 1000 ml
V 250.0 ml x 0.250 L
P V R T
n
T 25.0 oC 273.15 K 298.2 K
n
0.007393 mol
(0.0821 L atm/mol K) (298.2 K)
MM 0.118 g / 0.007393 mol ____________ g/mol
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Gas Mixtures

• Gas behavior depends on the number, not the
  identity, of gas molecules.
• Ideal gas equation applies to each gas
  individually and the mixture as a whole.
• All molecules in a sample of an ideal gas behave
  exactly the same way.

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Figure 5.9 Partial pressure of each gas in a
mixture of gases depends on the number of moles
of that gas.
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Daltons Law of Partial Pressures - I

• Definiton In a mixture of gases, each gas
  contributes to the total pressure the amount it
  would exert if the gas were present in the
  container by itself.
• To obtain a total pressure, add all of the
  partial pressures P total p1p2p3...pi

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Daltons Law of Partial Pressure - II

• Pressure exerted by an ideal gas mixture is
  determined by the total number of moles
• P(ntotal RT)/V
• ntotal sum of the amounts of each gas pressure
• the partial pressure is the pressure of gas if it
  was present by itself.
• P (n1 RT)/V (n2 RT)/V (n3RT)/V ...
• the total pressure is the sum of the partial
  pressures.

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Daltons Law of Partial Pressures - Prob 1

• A 2.00 L flask contains 3.00 g of CO2 and 0.10 g
  of Helium at a temperature of 17.0 oC.
• What are the Partial Pressures of each gas, and
  the total Pressure?
• T 17 oC 273 290 K
• nCO2 3.00 g CO2/ 44.01 g CO2 / mol CO2
• 0.0682 mol CO2
• PCO2 nCO2RT/V
• PCO2
• PCO2 _____________ atm

( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K)
(2.00 L)
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Daltons Law Problem - 1 cont.

• nHe 0.10 g He / 4.003 g He / mol He
• 0.025 mol He
• PHe nHeRT/V
• PHe
• PHe 0.30 atm
• PTotal PCO2 PHe 0.812 atm 0.30 atm
• PTotal 1.11 atm

(0.025 mol) ( 0.08206 L atm / mol K) ( 290 K )
( 2.00 L )
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Daltons Law Problem 2 using mole fractions

• A mixture of gases contains 4.46 mol Ne, 0.74 mol
  Ar and 2.15 mol Xe. What are the partial
  pressures of the gases if the total pressure is
  2.00 atm ?
• Total moles 4.46 0.74 2.15 7.35 mol
• XNe 4.46 mol Ne / 7.35 mol 0.607
• PNe XNe PTotal 0.607 ( 2.00 atm) 1.21 atm
  for Ne
• XAr 0.74 mol Ar / 7.35 mol 0.10
• PAr XAr PTotal 0.10 (2.00 atm) 0.20 atm for
  Ar
• XXe 2.15 mol Xe / 7.35 mol 0.293
• PXe XXe PTotal 0.293 (2.00 atm) 0.586 atm
  for Xe

82
Relative Humidity

• Rel Hum
  x 100
• Example the partial pressure of water at 15oC
  is 6.54 mm Hg, what is the Relative Humidity?
• Rel Hum (6.54 mm Hg/ 12.788 mm Hg )x100
• ___________

Pressure of Water in Air
Maximum Vapor Pressure of Water
83
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84
Figure 5.10 Production of oxygen by thermal
decomposition of KCIO3.
85
Molecular sieve framework of titanium (blue),
silicon (green), and oxygen (red) atoms contract
on heatingat room temperature (left) d 4.27 Å
at 250C (right) d 3.94 Å.
86
Collection of Hydrogen gas over Water - Vapor
pressure - I

• 2 HCl(aq) Zn(s) ZnCl2
  (aq) H2 (g)
• Calculate the mass of Hydrogen gas collected over
  water if 156 ml of gas is collected at 20oC and
  769 mm Hg.
• PTotal PH2 PH2O PH2 PTotal -
  PH2O
• PH2 769 mm Hg - 17.5 mm Hg
• 752 mm Hg
• T 20oC 273 293 K
• P 752 mm Hg /760 mm Hg /1 atm 0.989 atm
• V ___________ L

87
COLLECTION OVER WATER CONT.-II

• PV nRT n PV / RT
• n
• n 0.00641 mol
• mass 0.00641 mol x 2.01 g H2 / mol H2
• mass ______________ g Hydrogen

(0.989 atm) (0.156 L)
(0.0821 L atm/mol K) (293 K)
88
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
Molarity
PV nRT
moles / liter
Solutions
Gases
89
Gas Law Stoichiometry - I - NH3 HCl
Problem A slide separating two containers is
removed, and the gases are allowed to mix and
react. The first container with a volume of 2.79
L contains Ammonia gas at a pressure of 0.776 atm
and a temperature of 18.7 oC. The second with a
volume of 1.16 L contains HCl gas at a pressure
of 0.932 atm and a temperature of 18.7 oC. What
mass of solid ammonium chloride will be formed,
what will be remaining in the container, and
what is the pressure? Plan This is a limiting
reactant problem, so we must calculate the
moles of each reactant using the gas law to
determine the limiting reagent. Then we can
calculate the mass of product, and determine what
is left in the combined volume of the container,
and the conditions. Solution
Equation NH3 (g) HCl(g)
NH4Cl(s)
TNH3 18.7 oC 273.15 291.9 K
90
Gas Law Stoichiometry - II - NH3 HCl
PV
n
RT
(0.776 atm) (2.79 L)
nNH3
0.0903 mol NH3
(0.0821 L atm/mol K) (291.9 K)
limiting
(0.932 atm) (1.16 L)
nHCl
0.0451 mol HCl
(0.0821 L atm/mol K) (291.9 K)
reactant
Therefore the product will be 0.0451 mol NH4Cl or
2.28 g NH4Cl
Ammonia remaining 0.0903 mol - 0.0451 mol
0.0452 mol NH3
V 1.16 L 2.79 L 3.95 L
(0.0452 mol) (0.0821 L atm/mol K) (291.9 K)
nRT
PNH3
____
atm
V
(3.95 L)
91
Postulates of Kinetic Molecular Theory
I The particles are so small compared with
the distances between them that the volume
of the individual particles can be assumed
to be negligible (zero). II The particles are
in constant motion. The collisions of the
particles with the walls of the container
are the cause of the pressure exerted by
the gas. III The particles are assumed to exert
no forces on each other they are assumed
to neither attract nor repel each other. IV
The average kinetic energy of a collection of gas
particles is assumed to be directly
proportional to the Kelvin temperature of
the gas.
92
Figure 5.11 An ideal gas particle in a cube
whose sides are of length L (in meters).
93
Figure 5.12 (a) The Cartesian coordinate axes.
94
Figure 5.12 (b) The velocity u of any gas
particle can be broken down into three mutually
perpendicular components, ux, uy, and uz.
95
Figure 5.12 (c) In the xy plane,ux2 uy2
uyx2 by the Pythagorean theorem.
96
Figure 5.13 (a) Only the x component of the gas
particles velocity affects the frequency of
impacts on the shaded walls. (b) For an elastic
collision, there is an exact reversal of the x
component of the velocity and of the total
velocity.
97
Figure 5.14 Path of one particle in a gas.
98
A balloon filled with air at room temperature.
99
The balloon is dipped into liquid nitrogen at 77
K.
100
The balloon collapses as the molecules inside
slow down because of the decreased temperature.
101
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102
Velocity and Energy

• Kinetic Energy 1/2mu2
• Average Kinetic Energy (KEavg)
• add up all the individual molecular energies and
  divide by the total number of molecules!
• The result depends on the temp. of the gas
• KEavg3RT/2NA
• Ttemp. in Kelvin , NA Avogadros number,
  Rnew quantity (gas constant)
• kEtotal (No. of molecules)(KEavg)
  (NA)(3RT/2NA) 3/2RT
• So, 1 mol of any gas has a total molecular
  Kinetic Energy KE of 3/2RT!!!

103
Figure 5.15 A plot of the relative number of O2
molecules that have a given velocity at STP.
104
Figure 5.16 A plot of the relative number of
N2 molecules that have a given velocity at three
temperatures.
105
Figure 5.17 A velocity distribution for
nitrogen gas at 273 K.
106
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107
Molecular Mass and Molecular Speeds - I
Problem Calculate the molecular speeds of the
molecules of Hydrogen, Methane, and carbon
dioxide at 300K! Plan Use the equations for the
average kinetic energy of a molecule, and the
relationship between kinetic energy and velocity
to calculate the average speeds of the three
molecules. Solution for Hydrogen, H2
2.016 g/mol
3 R 2 NA
8.314 J/mol K
EK x T 1.5 x
x 300K
6.022 x 1023 molecules/mol
EK 6.213 x 10 - 21 J/molecule 1/2 mu2
2.016 x 10 -3 kg/mole 6.022 x 1023
molecules/mole
m
3.348 x 10 - 27 kg/molecule
6.213 x 10 - 21 J/molecule 1.674 x 10 - 27
kg/molecule(u2)
u _______________ m/s ___________ m/s
108
Molecular Mass and Molecular Speeds - II
for methane CH4 16.04 g/mole
3 R 2 NA
8.314 J/mol K 6.022 x 1023
molecules/mole
EK 6.213 x 10 - 21 J/molecule 1/2 mu2
16.04 x 10 - 3 kg/mole 6.022 x 1023
moleules/mole
m
2.664 x 10 - 26 kg/molecule
6.213 x 10 - 21 J/molecule 1.332 x 10 - 26
kg/molecule (u2)
u ___________________ m/s __________ m/s
109
Molecular Mass and Molecular Speeds - III
for Carbon dioxide CO2 44.01 g/mole
EK 6.213 x 10 - 21 J/molecule 1/2mu2
44.01 x 10 - 3 kg/mole 6.022 x 1023
molecules/mole
6.213 x 10 - 21 J/molecule 3.654 x 10 - 26
kg/molecule (u2)
u ___________________ m/s _________ m/s
110
Molecular Mass and Molecular Speeds - IV
Molecule H2
CH4 CO2
Molecular Mass (g/mol)
2.016 16.04
44.01
Kinetic Energy (J/molecule)
6.213 x 10 - 21 6.213 x 10 - 21 6.213
x 10 - 21
Velocity (m/s)
1,926 683.0
412.4
111
Important Point !

• At a given temperature, all gases have the same
  molecular Kinetic energy distributions.
• or
• The same average molecular Kinetic Energy!

112
Diffusion vs. Effusion

• Diffusion - One gas mixing into another gas, or
  gases, of which the molecules are colliding with
  each other, and exchanging energy between
  molecules.
• Effusion - A gas escaping from a container into a
  vacuum. There are no other (or few ) for
  collisions.

113
Figure 5.18 The effusion of a gas into an
evacuated chamber.
114
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115
Relative Diffusion of H2 versus O2 and N2 gases

• Average Molecular weight of air
• 20 O2 32.0 g/mol x 0.20 6.40
• 80 N2 28.0 g/mol x 0.80 22.40
• 
  28.80
• 28.80 g/mol
• or approximately 29 g/mol

116
Graham's Law calc.

• RateHydrogen RateAir x (MMAir / MMHydrogen)1/2
• RateHydrogen RateAir x ( 29 / 2 )1/2
• RateHydrogen RateAir x 3.95
• Or RateHydrogen RateAir x 4 !!!!!!

117
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118
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119
NH3 (g) HCl(g) NH4Cl (s)

• HCl 36.46 g/mol NH3 17.03 g/mol
• RateNH3 RateHCl x ( 36.46 / 17.03 )1/ 2
• RateNH3 RateHCl x 1.463

120
Figure 5.19 (a) demonstration of the relative
diffusion rates of NH3 and HCl molecules through
air.
121
Figure 5.19 (b) When HCl(g) and NH3 (g) meet
in the tube, a white ring of NH4Cl(s) forms.
122
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123
Figure 5.20 Uranium-enrichment converters from
the Paducah gaseous diffusion plant in Kentucky.
124
Gaseous Diffusion Separation of Uranium 235 / 238

• 235UF6 vs 238UF6
• Separation Factor S
• after Two runs S 1.0043
• after approximately 2000 runs
• 235UF6 is gt 99 Purity !!!!!
• Y - 12 Plant at Oak Ridge National Lab

(238.05 (6 x 19))0.5
(235.04 (6 x 19))0.5
125
Example 5.9 (P167)
Calculate the impact rate on a 1.00-cm2 section
of a vessel containing oxygen gas at a pressure
of 1.00 atm and 270C.
Solution
Calculate ZA ZA A A 1.00 x
10-4 m2

1.00 atm
N P V RT
( )
L atm K mol

0.08206
(300. K)
N V
Mol L
molecules mol
1000 L m3
4.06 x 10-2 x 6.022 x 1023
x
2.44 x 1025 molecules/m3
1 kg 1000g
M 32.0 g/mol x 3.2 x 10-2 kg/mol
J K mol
8.3145
(300. K)
ZA (1.00 x 10-4 m2) (2.44 x 1025 m-3) x
2(3.14)( 3.20 x 10-2 kg/mol)
ZA 2.72 x 1023 collisions/s
126
Figure 5.21 The cylinder swept out by a gas
particle of diameter d.
127
Like Example 5.10(P 169)
Problem Calculate the collision frequency for a
Nitrogen molecule in a sample of pure nitrogen
gas at 280C and 2.0 atm. Assume that the
diameter of an N2 molecule is 290 pm. Solution
2.0 atm
P RT
N V

8.097 x 10-2 mol/L
(301. K)
(8.097 x 10-2 mol/L)( 6.022 x 1023
molecules/mol)(1000L/m3)
4.876 x 1025 molecules/m3
d 290 pm 2.90 x 10-10 m
Also, for N2, M 2.80 x 10-2 kg/mol
(8.3145 J K-1mol-1)(301 K)
Z 4p(4.876 x 1025 m-3)(1.45 x 10-10 m)2 x
2p(2.80 x 10-2 kg/mol)
Z collisions/sec
128
Like Example 5.11 (P171)
Problem Calculate the mean free path in a
sample of nitrogen gas at 280 C and 2.00
atm. Solution Using data from the previous
example, we have
1
l
2 (N/V)(pd2)
1
l
5.49 x 10-8
2 ( 4.876 x 1025 )(p)(2.90 x 10-10 )2
Note that a nitrogen molecule only travels 0.5
nano meters before it collides with another
nitrogen atom.
129
Figure 5.22 Plots of PV/nRT versus P for
several gases (200 K).
130
Figure 5.23 Plots of PV/nRT versus P for
nitrogen gas at three temperatures.
131
Figure 5.24 (a) gas at low concentration -
relatively few interactions between particles (b)
gas at high concentration - many more
interactions between particles.
132
Figure 5.25 Illustration of pairwise
interactions among gas particles.
133
Table 5.3 (P172) Values of Van der
Waals Constants for
some Common Gases
( )
atm L2 mol2
( )
L mol
Gas a
b
He 0.034
0.0237 Ne
0.211
0.0171 Ar
1.35
0.0322 Kr
2.32
0.0398 Xe
4.19
0.0511 H2
0.244
0.0266 N2
1.39
0.0391 O2
1.36
0.0318 Cl2
6.49
0.0562 CO2
3.59
0.0427 CH4
2.25
0.0428 NH3
4.17
0.0371 H2O
5.46 0.0305
134
Figure 5.26 The volume occupied by the gas
particles themselves is less important at (a)
large container volumes (low pressure) than at
(b) small container volumes (high pressure).
135
Van der Waals Calculation of a Real gas
Problem A tank of 20.0 liters contains Chlorine
gas at a temperature of 20.000C at a pressure of
2.000 atm. if the tank is pressurized to a new
volume of 1.000 L and a temperature of 150.000C.
What is the new pressure using the ideal gas
equation, and the Van der Waals equation? Plan
Do the calculations! Solution
PV (2.000 atm)(20.0L) RT
(0.08206 Latm/molK)(293.15 K)
n
1.663 mol
nRT (1.663 mol) (0.08206 Latm/molK) (423.15
K) V (1.000 L)
P

atm
nRT n2a (1.663 mol) (0.08206
Latm/molK)(423.15 K) (V-nb) V2
(1.00 L) - (1.663 mol)(0.0562)
P -

-
(1.663 mol)2(6.49) (1.00 L)2
63.699 - 17.948 atm
136
Table 5.4(P173) Atmospheric Composition
near Sea Level
(dry air)
Component
Mole Fraction
N2
0.78084 O2
0.20946 Ar

0.00934 CO2
0.000345 Ne

0.00001818 He
0.00000524 CH4

0.00000168 Kr
0.00000114 H2

0.0000005 NO
0.0000005 Xe

0.000000087
The atmosphere contains various amounts of
water vapor, depending on conditions.
137
Figure 5.27 The variation of temperature and
pressure with altitude.
138
Severe Atmospheric Environmental Problems we
must deal with in the next period of time
Urban Pollution Photochemical Smog Acid
Rain Greenhouse Effect Stratospheric Ozone
Destruction
139
Figure 5.28 Concentration (in molecules per
million molecules of air) of some smog
components versus time of day.
140
Figure 5.29 Our various modes of transportation
produce large amounts of nitrogen oxides, which
facilitate the formation of photochemical smog.
141
This photo was taken in 1990. Recent renovation
has since replaced the deteriorating marble.
142
Figure 5.31 Diagram of the process for
scrubbing sulfur dioxide from stack gases in
power plants.
143
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