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STATES OF MATTER
Intermolecular Forces of Attraction
Kinetic Molecular Theory
•All matter is composed of atoms that are in constant motion
Kinetic Theory Facts
• All phases of matter express the degree that they reflect the kinetic theory through their kinetic energy
• kinetic energy is measured by temperature
• phase changes involve changes in temperature due to the existence threshold temperature of each phase (i.e. ice naturally is found at cold not hot temperatures)
Solids Liquids Gases
Definitesize
Definite size No Definitesize
Definiteshape
No Definiteshape
No Definiteshape
LowKineticEnergy
More KineticEnergy thanSolids, but lessthan Gases
High KineticEnergy
• While gases have a great deal of random motion, solids and liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them
• these forces are the van der Waals forces
Definitions
• Bonds are intramolecular forces of attraction
• Forces of attraction between molecules are called intermolecular forces of attraction
• intermolecular forces of attraction are commonly called van der Waals forces
The Condensed Phases
• Solids and Liquids
• Physical properties of the condensed phases reflect the degree of intermolecular forces (i.e. boiling point)
nod ipo le-d ipo le fo rces
yesH ydrogen bond ing
YesIs H bond ing to N , O , or F?
noLondon d ispersion fo rces
(induced d ipo les)
N Ovan der W aals
Are po lar m olecues invo lved?
yesIon-d ipo le forces
noIon ic bond ing
YesAre po lar m olecu les and ions present?
In teracting substancesAre ions present?
Dipole-dipole forces
• Exist between neutral polar molecules
• work best the closer the molecules are to each other
• the greater the polarity of the molecules, the greater the force of attraction
H bonding
• Special case of dipole-dipole interaction specifically between H of one polar molecule with N, O or F and an unshared electron pair of another nearby small electronegative ion (usually N, O, or F on another molecule)
• VERY STRONG
London dispersion forces
• Induced dipoles
• not really dipoles on the AVERAGE, but instantaneously dipole conditions can exist thus allowing for pseudopolar regions to occur
• No matter how strong the van der Waal force of attraction is, it is still not stronger than attractions involving ions
Ion-dipole forces
• Attraction between ions and the partial charge on the end of a polar molecule
• ex. NaCl in water solution
15
FORCESINTRAMOLECULAR:
INTERMOLECULAR:
FORCES WITHIN A MOLECULE
FORCES BETWEEN MOLECULES
O = O STRONG INTRAMOLECULMR FORCESVERY WEMK INTERMOLECULAR FORCES
ELECTROSTATIC
MUCH WEAKER THAN EITHERIONIC OR COVALENT BONDS
DICTATE WHETHER MOLEULAR SUBTANCE IS GAS, LIQUID OR SOLIDAT ROOM CONDITIONS
GAS: NEGLIGIBLE LIQUID: WEAK TO MODERATE
SOLID: MODERATE TO STRONG
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INTERMOLECULAR FORCES:
HELP DEFINE STANDARD STATE OFA SUBSTANCE
HELP DEFINE COMMON PHYSICALPROPERTIESOF A SUBSTANCE
BOILING/MELTING POINTSVISCOSITYSURFACE TENSIONCAPILLARY ACTIONARE MUCH WEAKER THAN BONDS(INTRA MOLECULAR FORCES)
IONIC BONDS: ~300 - 1000’S kJ/MOL
COVALENT BONDS: ~150 - ~800 kJ/MOL
INTERMOLCULAR FORCES: ~1 - 40 kJ/MOL
STRENGTH DIMINISHES WITH INCREASING DISTANCE
17
IONIC COMPOUNDS: + -
MOLECULARCOMPOUNDS: -+ MOLECULAR
DIPOLE
DISPERSION FORCES +- +- +-
“INDUCED” TEMPORARY DIPOLE
PRESENT IN ALL MOLECULAR SUBSTANCES!INCREASES WITH MOLECULAR MASS
F2
Cl2
I2
-188 oC58.8184
18
DIPOLE-DIPOLE FORCES
POLAR MOLECULE
NATURAL ASSYMETRIC CHAGE DISTRIBUTIONPERMANENT DIPOLE
OCCURS IF CENTERS OF CHARGES DO NOT COINCIDE
O
H H+ +
-
O C O
Cl
Cl C
Cl
ClF
THE MORE POLAR THE MOLECULE, THE STRONGER THEINTERMOLECULAR FORCE
CH3 - CH2 - CH3 CH3CN44 = MASS = 41
0.1 = DIPOLE MOMENT = 3.9231 = BOILING POINT (K) = 355
19
HYDROGEN BOND
SPECIAL CASE OF DIPOLE FORCES
H ATOM ATTACHED TO F, O, N
LARGE ELECTRONEGATIVITY DIFFERENCE
SMALL SIZE ALLOWS H TO GET CLOSE
H2OH2SH2SeH2Te
183481130
oC- 60.7-41.5-2
~ - 73100THIS PROPERTY AFFECTSLIFE AND MANY OTHER
PROPERTIES
20
DIFFUSION: MOLECULES MOVING THRU MOLECULES
VISCOSITY: RESISTANCE TO FLOW
SURFACETENSION:
21
SOLID
GAS (VAPOR)
LIQUID
SY
ST
EM
EN
ER
GY
SY
ST
EM
EN
ER
GYS
UB
LIM
AT
ION D
EP
OS
ITIO
NMELTING(FUSION)
VAPORIZATION
CONDENSATION
FREEZING
22
TE
MP
ER
AT
UR
E
TIME
EN
TH
AL
PY
SOLID
LIQUID +SOLID
MELTING POINTFREEZING POINT
Hf = HEAT REQUIRED TO MELT SUBSTANCEAT ITS MELTING POINT
HEAT REQUIRED TO BREAK DOWNINTERMOLECULAR FORCES
> 0 J/g OR J/mol
LIQUID
LIQUID +VAPOR
VAPOR
23
GASES LIQUIDS SOLIDSASSUMESSHAPE &
VOLUME OFCONTAINER
DEFINITEVOLUME;
ASSUMES SHAPEOF CONTAINER
DEFINITE SHAPEAND VOLUME
COMPRESSIBLE VIRTUALLYINCOMPRESSIBLE
VIRTUALLYINCOMPRESSIBLE
DIFFUSESRAPIDLY IN
ANOTHER GAS
DIFFUSESSLOWLY INANOTHER
LIQUID
DIFFUSES VERYSLOWLY IN
ANOTHER SOLID
FLOWS READILY FLOWS READILY DOES NOT FLOW
24
F/A = PRESSURE
14.7 lb/in2 = 1 ATM = 760 mm Hgtorr
GAS
Patm
h
P = Patm - h
bar
25
WHAT IS THE PRESSURE OF A GAS (ATM) IF IT SUPPORTS A COLUMN OF MERCURY TO A HEIGHT OF 535 mm?
535 mm Hg x 1 atm760 mm Hg
= 0.704 atm
IF THE ATMOSPHERIC PRESSURE DROPS TO 0.645 atm, HOW HIGH IS THE COLUMN OF MERCURY SUPPORTED?
0.645 atm x 760 mm Hg1 atm = 490 OR 4.9 x 102 mm Hg
IF THE BAROMETRIC PRESSURE IS 755 mm Hg AND A GAS CREATES A h OF 275 mm IN A MANOMETER, WHAT IS THE PRESSURE OF THE GAS?
P = Patm - h = 755 - 275 = 480 mm Hg = 480 torr
480 torr x1 atm
760 torr = 0.632 atm
26
P x V = k (n, T)
OR P 1/V
VO
LU
ME
PRESSURE
BOYLE’S LAW
27
VO
LU
ME
TEMPERATURE
CONSTANT PRESSURE!!!
V = k(n, P) x T
OR V T
CHARLES’ LAW
28
VO
LU
ME
TEMPERATURE0 oC
n = 1.0
n = 0.5
n = 2
n = 4
-273.15 oC
ABSOLUTE OR KELVIN SCALE
K = oC + 273
USE IN ALL CALCULATIONS!!!
29
AVOGADRO: EQUAL VOLUMES OF GASES AT SAME T & P CONTAIN EQUAL NUMBER OF MOLECULES!
V = k(n, T) x n
P x V = k (n, T)
V = k(n, P) x T
PV = nkTPV = nRT
R = 0.0821 L.atm.mol-1.K-1
IDEAL GASLAW
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THE PRESSURE IN AN AEROSOL CAN AT 25 oC IS 1.5 atm.A FIRE CAN REACH 1200 oC. WHAT IS THE PRESSURE OF THE CAN AT THAT TEMPERATURE?
P V = n R TINITFINAL
1.5X
Vi
Vi
nn
0.08210.0821
25 + 2731200 + 273
R= Pi / TiPf / Tf = Pi / Ti
Pf =1.5 atm x 1473 K
298 K7.4 atm
WHAT IS THE V OF 1.0 MOL GAS AT 1.0 atm AND 0 oC?
V = nRT
P1 mol x 0.0821 L.atm.mol-1.K-1 x 273 K
1.0 atm
= 22.4 L
STP
31
IDEAL VS. REAL GAS
MOLECULES FAR APARTNO COLLISIONS
NO INTERACTIONS
FAR APARTSOME COLLISIONS
LOW INTERACTIONS
OCCUPY NO SPACE & HAVE NO VOLUME
MATTER; MUST OCCUPY SPACE & HAVE VOLUME
PV/RT = 1 PV/RT > 1
SINCE GASES ARE REAL:
CAN NEVER ACHIEVE ABSOLUTE 0
APPROACH IDEAL GAS AT HIGH VOLUMES
LOW P, HIGH T
32
KINETIC MOLECULAR THEORY
1. THE VOLUME OF GAS MOLECULES IS NEGLIBLE COMPARED TO THE VOLUME OF THE CONTAINER
2. PARTICLES UNDERGO CONSTANT RANDOM MOTION AND DO NOT INTERACT WITH ONE ANOTHER
3. AVERAGE KINETIC ENERGY OF THE PARTICLES IS PROPORTIONAL TO ABSOLUTE TEMPERATURE
TEMPERATURE IS THE MEASURE OF THE AVG.KINETIC ENERGY OF THE PARTICLES IN THE SYSTEM
E ~ RT
R = 8.314 J.mol-1.K-1
33
LIQUID VAPOR EQUILIBRIA
DYNAMIC EQUILIBRIUM!
VAPOR PRESSURE
ALL NON-GASEOUS MATERIALS EXERT A
VAPOR PRESSURE.
FOR SOLIDS: VERY LOWASSUMED TO BE 0
EVAPORATION
34
VAPORPRESSURE
(mm Hg)
T, oC
760
NORMAL BOILING POINT
VS BOILING POINT
HVAP
AMOUNT OF HEAT REQUIRED TO
VAPORIZE SOME AMOUNT OF LIQUID
VOLATILE NON-VOLATILE
35
SOLID
LIQUID
VAPOR (GAS)P,
atm
T, oC
1.0 MELTING POINT
BOILING POINT
TRIPLE POINT
36
CAPILLARY ACTION
ADHESIVE VS COHESIVE FORCES
INTERMOLECULAR
ADHESIVE COHESIVE> <
37
BOILING
POINT
INTERMOLECULAR FORCES
DISPERSION: INCREASING MASSDIPOLE-DIPOLE: INCREASING POLARITYHYDROGEN BOND: INCREASING “NUMBER”
OR ANY OF THE PHYSICALPROPERTIES
X DISPERSION ONLY
X DISPERSION + DIPOLE
XDISPERSION, DIPOLE
& H-BONDING
XIONIC
38
WHICH OF THE FOLLOWING HAVE THE HIGHER BP?
A) C6H14 C10H22 C2H6
B) C6H14 C6H13OH C6H12(OH)2
C) CCl4 CCl3F CF4
D) HCl HF F2
WHAT FORCES ARE PRESENT IN:
A) NaBr
B) NF3
C) CH2OHCH2CH2OH
D) Ar
IONIC
DISPERSION + DIPOLE-DIPOLE
DISPERSION + DIPOLE + H-BONDDISPERSION
39
UNIT CELLS
CRYSTAL LATTICE:THE REPEATING PATTERN IN A THREE DIMENSIONAL ARRAY
A A A A
A A A A A
A A A A
A A A A A
A A A AA A A A A
NOTE: THIS ATOMIS SHARED BY MORE
THAN ONE UNIT CELLNOT 5 ATOMS PER CELLIS 1 + 1/4(4) = 2 FULL ATOMS
CONSIDER FACES, EDGES, CORNERS AND THOSE TOTALLY WITHIN CELL
40
CRYSTALLINE SOLIDS
AMORPHOUS SOLIDS
WELL DEFINED POSTIONS FOR EMCH ATOM
ILL DEFINED POSTIONS
ORDERED REPETITION OF PATTERN
ORDER EXTENDS OVER SHORT RANGE
LONG RANGE ORDER!
LOCAL ORDER!
41
UNIT CELL STOICHIOMETRY
ATOM LOCATED ENTIRELY WITHIN CELL CONTRIBUTES 1 FULL ATOM TO CELL STOICHIOMETRY
6 FACES 2 CELLS
FACE ATOM CONTRIBUTES 1/2 x 6 = 3 ATOMS TO UNIT CELL
12 EDGES 4 CELLS
EDGE ATOM CONTRIBUTES 1/4 x 12 = 3 ATOMS TO UNIT CELL
EIGHT CORNERS8 CELLS
CORNER ATOM CONTRIBUTES 1/8 x 8 = 1 ATOM TO UNIT CELL
WHAT ARE THESE UNIT CELLS?
42
SIMPLE CUBIC (SC)
ab
c
a = b = c
90o
SC = 1 ATOM/UNIT CELL
CONTAINED WITHIN CELL
BODY CENTERED CUBIC (bcc)
BCC = 2 ATOMS/UNIT CELL
FACE CENTERED CUBIC (fcc)
FCC = 4 ATOMS/UNIT CELL
43
COORDINATION NUMBER (CN) OR GEOMETRY
4 PARTICLES CONNECTED TO CENTRAL ATOMCN = 4 = TETRAHEDRON
CN = 6 CN = 8 CN = 12
DENSITY IS DETERMINED BY HOW CLOSE (EFFICIENCY)PARTICLES ARE PACKED INTO A UNIT CELL
PACKING EFFICIENCY: % OF UNIT CELL OCCUPIED BY ATOMS, IONS OR MOLECULES
52% 68% 74%
DENSITY IS A MEASURE OF HOW CONCENTRATED IS THE MASS OF A PURE SUBSTANCE ....OR HOW TIGHTLY PACKED
d =MASS OF SUBSTANCE
VOLUME OF SUBSTANCEATOMS/UNIT CELL x MASS ATOM
SIDE3
44
FOUR TYPES OF CRYSTALLINE SOLIDS:METALLICIONICCOVALENTMOLECULAR
METALLIC: COVALENT TYPE BOND IN METALS
DELOCALIZED “SEA” OF ELECTRONS
2 ATOMIC ORBITALS2 MOLECULAR ORBITALS
4 8 16 6 x 1023
ENERGYBAND
45
EN
ER
GY
EMPTYORBITALS
FILLEDORBITALS
FERMI LEVEL
METAL
BANDGAP
NON-CONDUCTOR
SEMI-CONDUCTOR
46
IONICCRYSTALS
EXTENDED SOLIDSHIGH MELTING & BOILING POINTSIONIC COMPOUNDS
COVALENT SOLIDS
LARGE NETWORKS HIGH MELTING, HARD SOLIDSDIAMOND, MOST SEMICONDUCTORS, SiO2
MOLECULARSOLIDS
INDIVIDUAL MOLECULES IN A LATTICELOW MELTING, SOFTICE, SUGAR, IODINE, SOLID HYDROGEN
GASESGASES
Importance of GasesImportance of Gases
• Airbags fill with NAirbags fill with N22 gas in an gas in an
accident. accident. • Gas is generated by the Gas is generated by the
decomposition of sodium azide, decomposition of sodium azide, NaNNaN33..
• 2 NaN2 NaN33 ---> 2 Na + 3 N ---> 2 Na + 3 N22
THREE STATES OF MATTERTHREE STATES OF MATTERTHREE STATES OF MATTERTHREE STATES OF MATTER
General Properties of General Properties of GasesGases
• There is a lot of “free” space There is a lot of “free” space in a gas.in a gas.
• Gases can be expanded Gases can be expanded infinitely.infinitely.
• Gases fill containers Gases fill containers uniformly and completely.uniformly and completely.
• Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.
Properties of GasesProperties of GasesGas properties can be modeled using Gas properties can be modeled using
math. Model depends on—math. Model depends on—• V = volume of the gas (L)V = volume of the gas (L)• T = temperature (K)T = temperature (K)
– ALL temperatures in the entire ALL temperatures in the entire chapter MUST be in Kelvin!!! chapter MUST be in Kelvin!!! No Exceptions!No Exceptions!
• n = amount (moles)n = amount (moles)• P = pressureP = pressure
(atmospheres) (atmospheres)
PressurePressurePressure of air is Pressure of air is
measured with a measured with a BAROMETER BAROMETER (developed by (developed by Torricelli in 1643)Torricelli in 1643)
Hg rises in tube until force of Hg Hg rises in tube until force of Hg (down) balances the force of (down) balances the force of atmosphere (pushing up). (Just like atmosphere (pushing up). (Just like a straw in a soft drink)a straw in a soft drink)
P of Hg pushing down related to P of Hg pushing down related to
• Hg densityHg density
• column heightcolumn height
PressurePressureColumn height measures Column height measures
Pressure of atmospherePressure of atmosphere
• 1 standard atmosphere (atm) 1 standard atmosphere (atm) * *
= 760 mm Hg (or torr) *= 760 mm Hg (or torr) *
= 14.7 pounds/in= 14.7 pounds/in2 2 (psi)(psi)
= 101.3 kPa (SI unit is = 101.3 kPa (SI unit is PASCAL) PASCAL)
= about 34 feet of water!= about 34 feet of water!
* Memorize these!* Memorize these!
Pressure Conversions
A. What is 475 mm Hg expressed in atm?
1 atm
760 mm Hg
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
760 mm Hg
14.7 psi = 1.52 x 103 mm Hg
= 0.625 atm475 mm Hg x
29.4 psi x
Pressure Conversions
A. What is 2 atm expressed in torr?
B. The pressure of a tire is measured as 32.0 psi.
What is this pressure in kPa?
Boyle’s LawBoyle’s LawP P αα 1/V 1/VThis means Pressure and This means Pressure and
Volume are Volume are INVERSELY INVERSELY PROPORTIONAL if PROPORTIONAL if moles and temperature moles and temperature are constant (do not are constant (do not change). For example, P change). For example, P goes up as V goes down.goes up as V goes down.
PP11VV11 = P = P22 V V22
Robert Boyle Robert Boyle (1627-1691). (1627-1691). Son of Earl of Son of Earl of Cork, Ireland.Cork, Ireland.
Boyle’s Law and Kinetic Boyle’s Law and Kinetic Molecular TheoryMolecular Theory
Boyle’s Law and Kinetic Boyle’s Law and Kinetic Molecular TheoryMolecular Theory
P proportional to 1/VP proportional to 1/V
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s LawA bicycle pump is a A bicycle pump is a
good example of good example of Boyle’s law. Boyle’s law.
As the volume of the As the volume of the air trapped in the air trapped in the pump is reduced, its pump is reduced, its pressure goes up, pressure goes up, and air is forced and air is forced into the tire.into the tire.
Charles’s Charles’s LawLaw
If n and P are constant, If n and P are constant, then V then V αα T T
V and T are directly V and T are directly proportional.proportional.
VV11 V V22
==
TT11 T T22
• If one temperature If one temperature
goes up, the volume goes up, the volume goes up!goes up!
Jacques Charles (1746-Jacques Charles (1746-1823). Isolated boron and 1823). Isolated boron and studied gases. Balloonist.studied gases. Balloonist.
CharlesCharles’’s original balloons original balloon
Modern long-distance balloonModern long-distance balloon
Charles’s LawCharles’s Law
Gay-Lussac’s LawGay-Lussac’s Law
If n and V are constant, If n and V are constant, then P then P αα T T
P and T are directly P and T are directly proportional.proportional.
PP11 P P22
==
TT11 T T22
• If one temperature If one temperature
goes up, the pressure goes up, the pressure goes up!goes up!
Joseph Louis Gay-Joseph Louis Gay-Lussac (1778-1850)Lussac (1778-1850)
Gas Pressure, Temperature, Gas Pressure, Temperature, and Kinetic Molecular Theoryand Kinetic Molecular TheoryGas Pressure, Temperature, Gas Pressure, Temperature, and Kinetic Molecular Theoryand Kinetic Molecular Theory
P proportional to TP proportional to T
Combined Gas Law
• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V1 P2 V2
= T1 T2
No, it’s not related to R2D2
Combined Gas Law
If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!
= P1 V1
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = 0.800 atm V1 = 180 mL T1 = 302 K
P2 = 3.20 atm V2= 90 mL T2 = ??
CalculationP1 = 0.800 atm V1 = 180 mL T1 = 302 KP2 = 3.20 atm V2= 90 mL T2 = ??
P1 V1 P2 V2
= P1 V1 T2 = P2 V2 T1
T1 T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
T2 = 604 K - 273 = 331 °C
= 604 K
Learning Check
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?
One More Practice Problem
A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
And now, we pause for this commercial message from
STPOK, so it’s really not THIS kind of STP…
STP in chemistry stands for Standard Temperature and Pressure
Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0° C (273 K)
STP allows us to compare amounts of gases between different pressures and temperatures
STP allows us to compare amounts of gases between different pressures and temperatures
Try This One
A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
Avogadro’s Avogadro’s HypothesisHypothesis
Equal volumes of gases at the same T and Equal volumes of gases at the same T and P have the same number of molecules.P have the same number of molecules.
V = n (RT/P) = knV = n (RT/P) = kn
V and n are directly related.V and n are directly related.
twice as many twice as many moleculesmolecules
Avogadro’s Hypothesis and Avogadro’s Hypothesis and Kinetic Molecular TheoryKinetic Molecular Theory
Avogadro’s Hypothesis and Avogadro’s Hypothesis and Kinetic Molecular TheoryKinetic Molecular Theory
P proportional to nP proportional to n
The gases in this The gases in this experiment are all experiment are all measured at the measured at the same T and V.same T and V.
IDEAL GAS LAWIDEAL GAS LAW
Brings together gas Brings together gas properties.properties.
Can be derived from Can be derived from experiment and theory.experiment and theory.
BE SURE YOU KNOW BE SURE YOU KNOW THIS EQUATION!THIS EQUATION!
P V = n R TP V = n R T
Using PV = nRTUsing PV = nRTP = PressureP = Pressure
V = VolumeV = Volume
T = TemperatureT = Temperature
N = number of molesN = number of moles
R is a constant, called the Ideal Gas ConstantR is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for all the Instead of learning a different value for R for all the possible unit combinations, we can just memorize possible unit combinations, we can just memorize one value and convert the units to match R.one value and convert the units to match R.
R = 0.0821R = 0.0821
L • atm Mol • K
Using PV = nRTUsing PV = nRTHow much NHow much N22 is required to fill a small room with a is required to fill a small room with a
volume of 960 cubic feet (27,000 L) to 745 mm volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 Hg at 25 ooC?C?
SolutionSolution
1. Get all data into proper units1. Get all data into proper units
V = 27,000 LV = 27,000 L
T = 25 T = 25 ooC + 273 = 298 KC + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg) P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm = 0.98 atm
And we always know R, 0.0821 L atm / mol KAnd we always know R, 0.0821 L atm / mol K
77
How much NHow much N22 is required to fill a small room with a volume of 960 cubic is required to fill a small room with a volume of 960 cubic
feet (27,000 L) to P = 745 mm Hg at 25 feet (27,000 L) to P = 745 mm Hg at 25 ooC?C?
SolutionSolution
2. Now plug in those values and solve for the 2. Now plug in those values and solve for the unknown.unknown.
PV = PV = nnRTRT
n = 1.1 x 10n = 1.1 x 1033 mol (or about 30 kg of gas) mol (or about 30 kg of gas)
RT RTRT RT
Learning Check
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?
Learning Check
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
Deviations from Deviations from Ideal Gas LawIdeal Gas Law
• Real molecules have volume.The ideal gas consumes the entire
amount of available volume. It does not account for the volume of the molecules themselves.
• There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.
– Otherwise a gas could not condense to become a liquid.
Gases in the AirThe % of gases in air Partial pressure (STP)
78.08% N2 593.4 mm Hg
20.95% O2 159.2 mm Hg
0.94% Ar 7.1 mm Hg
0.03% CO2 0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg 2 2 2
Total Pressure 760 mm Hg
Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures
What is the total pressure in the flask?What is the total pressure in the flask?
PPtotaltotal in gas mixture = P in gas mixture = PAA + P + PBB + ... + ...Therefore, Therefore,
PPtotaltotal = P = PHH22OO + P + POO22 = 0.48 atm = 0.48 atm
Dalton’s Law: total P is sum of PARTIAL Dalton’s Law: total P is sum of PARTIAL pressures. pressures.
2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)
0.32 atm 0.32 atm 0.16 atm 0.16 atm
Dalton’s Dalton’s LawLaw
John DaltonJohn Dalton1766-18441766-1844
Health NoteWhen a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank
air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood,
a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for
deep descents.
Collecting a gas “over water”
• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
Table of Vapor Pressures for Water
Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?
768 torr – 17.5 torr = 750.5 torr
GAS DENSITYGAS DENSITYGAS DENSITYGAS DENSITY
HighHigh densitydensity
Low Low densitydensity
22.4 L of ANY gas AT STP = 1 mole
Gases and StoichiometryGases and Stoichiometry
2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)
Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of in a flask with a volume of
2.50 L. What is the volume of O2.50 L. What is the volume of O22 at STP? at STP?
Bombardier beetle Bombardier beetle uses decomposition of uses decomposition of hydrogen peroxide to hydrogen peroxide to defend itself.defend itself.
Gases and Gases and StoichiometryStoichiometry
2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)
Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of 2.50 L. What is the in a flask with a volume of 2.50 L. What is the
volume of Ovolume of O22 at STP? at STP?
SolutionSolution
1.1 g1.1 g HH22OO22 1 mol H 1 mol H22OO22 1 mol O 1 mol O22 22.4 L O 22.4 L O22
34 g H34 g H22OO22 2 mol H 2 mol H22OO22 1 mol O 1 mol O22
= 0.36 L O2 at STP
Gas Stoichiometry: Practice!
A. What is the volume at STP of 4.00 g of CH4?
B. How many grams of He are present in 8.0 L of gas at STP?
What if it’s NOT at STP?
• 1. Do the problem like it was at STP. (V1)
• 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)
Try this one!
How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
• diffusion is the gradual diffusion is the gradual mixing of molecules of mixing of molecules of different gases.different gases.
• effusion is the effusion is the movement of molecules movement of molecules through a small hole through a small hole into an empty container.into an empty container.
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
Graham’s law governs Graham’s law governs effusion and diffusion effusion and diffusion of gas molecules.of gas molecules.
Thomas Graham, 1805-1869. Professor Thomas Graham, 1805-1869. Professor in Glasgow and London.in Glasgow and London.
Rate of effusion is Rate of effusion is inversely proportional to inversely proportional to its molar mass.its molar mass.
Rate of effusion is Rate of effusion is inversely proportional to inversely proportional to its molar mass.its molar mass.
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
Molecules effuse thru holes in a rubber Molecules effuse thru holes in a rubber balloon, for example, at a rate (= balloon, for example, at a rate (= moles/time) that ismoles/time) that is
• proportional to Tproportional to T• inversely proportional to M.inversely proportional to M.
Therefore, He effuses more rapidly Therefore, He effuses more rapidly than Othan O22 at same T. at same T.
HeHe
Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion
Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
