Download - Chapter 11: Properties of Gases Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop
Chapter 11: Properties of Gases
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Properties of Common Gases
Despite wide differences in chemical properties, ALL gases more or less obey the same set of physical properties.
Four Physical Properties of Gases Inter-related
1. Pressure (P)2. Volume (V)3. Temperature (T)4. Amount = moles (n)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Pressure, Its Measurement and Units
Pressure is force per unit area Earth exerts gravitational force on
everything with mass near it Weight
Measure of gravitational force that earth exerts on objects with mass
What we call weight is gravitational force acting on object
areaforce
Pressure
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Force vs. Pressure Consider a woman wearing flat shoes
vs. high "spike" heels Weight of woman is same = 120 lbs. Pressure on floor differs greatly
Shoe Area Pressure
Flat 10" x 3" = 30 in2
Spike 0.4" x 0.4" = 0.16 in2 Psi
in.lbs
P 750 160 120
2
Why airline stewardesses cannot wear spike heels!
Psiinlbs
P 4 30 120
2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ways to Measure Pressure Atmospheric Pressure
Resulting force per unit area When earth's gravity acts on molecules in
air Pressure due to air molecules colliding with
object Barometer
Instrument used to measure atmospheric pressure
Toricelli Barometer
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Toricelli Barometer
Simplest barometer
Tube = 80 cm in length
Sealed at one end
Filled w/ Hg In dish filled w/
Hg
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Toricelli Barometer Air Pressure
Pushes down on Hg Forces Hg up tube
Weight of Hg in tube Pushes down on Hg
in dish When 2 forces
balance Hg level stabilizes Read atmospheric
pressure
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Toricelli Barometer A.P. high
Pushes down on Hg in dish
level in tube A.P. low
P on Hg in dish < P from column
level in tubeResult: Measure height of Hg
in tube Atmospheric pressure
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Standard Atmospheric Pressure Typical range of P for most places where
people live 730 – 760 mm Hg
Top of Mt. Everest A.P. = 250 mm Hg
Standard atmosphere (atm) Average pressure at sea level Pressure needed to support column of
mercury 760 mm high measures at 0oC
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Units of Pressure
Pascal = Pa SI unit for Pressure Very small 1atm 101,325 Pa = 101 kPa 1atm 1.013 Bar = 1013 mBar 1 atm too big for most lab work
atmtorr7601
1 1 atm 760 mm Hg
At sea level 1 torr = 1 mm Hg
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Manometers
Used to measure P's inside closed reaction vessels Pressure changes caused by gases
produced or used up during chemical reaction
Open-end Manometer U tube partly filled with liquid (usually
Hg) One arm open to atm One arm exposed to trapped gas in
vessel
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Open Ended ManometerPgas = Patm Pgas > Patm
Gas pushes Hg up tube
Pgas < Patm
Atm pushes Hg down tube
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Using Open Ended Manometersa. A student collected a gas in
an apparatus connected to an open-end manometer. The mercury in the column open to the air was 120 mm higher and the atmospheric pressure was measured to be 752 torr. What was the pressure of the gas in the apparatus?
This is a case of Pgas > Patm
Pgas = 752 torr + 120 torr
= 872 torr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Using Open Ended Manometersb. In another experiment, it
was found that the mercury level in the arm of the manometer attached to the container of gas was 200 mm higher than in the arm open to the air. What was the pressure of the gas?
This is a case of Pgas < Patm
Pgas = 752 torr – 200 torr
= 552 torr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Closed-end Manometer Very convenient for measuring Pgas < 1 atm Arm farthest from vessel (gas) sealed
Tube can be shorter Tube emptied of air under vacuum and Hg
allowed in Then open system to atm and some Hg drains out No atm. P exists above Hg in sealed arm
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Closed-end Manometer
When gas sample has P < 1 atm Hg in arm , as not
enough P to hold up Hg.
Patm = 0
Pgas = PHg So directly read off
P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!Gas pressure is measured using a close-ended mercury manometer. The height of fluid in the manometer is 23.7 in Hg. What is this pressure in atm?A. 23.7 atmB. 0.792 atmC. 602 atmD. 1.61 atm
mm760
atm1cmmm10
incm54.2
inHg7.23 0.792 atm
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Using Liquids Other Than Mercury in Manometers and
BarometersProblem: Hg has d = 13.6 g/mL So dense that little difference in P
close to 1 atm Not very accurate when P ~1 atm Solution: Go to less dense liquid Then differences in liquid levels get more precision in P
measurement
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Comparison of Hg and H2O 1 mm column of Hg
and 13.6 mm column of water exert same pressure
Mercury is 13.6 times more dense than water
Both columns have same weight and diameter, so they exert same pressure d = 1.00 g/mLd = 13.6 g/mL
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Using Liquids Other Than Mercury in Manometers and
Barometers For example: use H2O (d = 1.00 g/mL) P = 1 mm Hg Now 13.6 mm change with H2O Simple relationship exists between two
systems. or
Use this relationship to convert pressure change in mm H2O to pressure change in mm Hg
AABB dhdh b
AAB d
dhh
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex 2. Converting mm Acetone to mm Hg
Acetone has a density of 0.791 g/mL. Acetone is used in an open-ended manometer to measure a gas pressure slightly greater than atmospheric pressure, which is 756 mm Hg at the time of the measurement. The liquid level is 20.4 mm higher in the open arm than in the arm nearest the gas sample. What is the gas pressure in torr?
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 2 Solution
First convert mm acetone to mm Hg
Hgmm.mL/g.
mL/g.mm.hHg 191
613 7910 420
Then add PHg to Patm to get Ptotal
Pgas = Patm + PHg
= 756.0 torr + 1.19 torr Pgas = 757.2 torr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Boyle’s Law Studied relationship
between P and V Work done at
constant T as well as constant number of moles (n)
T1 = T2
As V , P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Charles’s Law Charles worked on
relationship of how V changes with T
Kept P and n constant
Showed V as T
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Gay-Lussac’s Law Worked on relationship between pressure and
temperature Volume (V) and number of moles (n) are
constant P as T This is why we don’t heat canned foods on a
campfire without opening them! Showed that gas pressure
is directly proportional to absolute temperature
T (K)
P
Low T, Low P
High T, High PTP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Combined Gas Law
Ratio
Constant for fixed amount of gas (n)
for fixed amount (moles)
OR can equate 2 sets of conditions to give
combined gas law
TPV
CTPV
2
22
1
11
TVP
TVP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Combined Gas Law
All T's must be in K Value of P and V can be any units as
long as they are the same on both sides
Only equation you really need to remember
Gives all relationships needed for fixed amount of gas under two sets of conditions
2
22
1
11
TVP
TVP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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How Other Laws Fit into Combined Gas Law
2
22
1
11
TVP
TVP
Boyle’s Law T1 = T2 P1V1 = P2V2
Charles’ Law
P1 = P2
Gay-Lussac’s Law
V1 = V2
2
2
1
1
TP
TP
2
2
1
1
TV
TV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Combined Gas Law
K358.Vtorr 560
K 273.15
mL 500.torr 760 15
2
Used for calculating effects of changing conditions T in Kelvin P and V any units, as long as units cancel
Ex. If a sample of air occupies 500. mL at STP*, what is the volume at 85 °C and 560 torr?
*Standard Temperature (273.15K) and Pressure
(1 atm)
890 mL
2
22
1
11
TVP
TVP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex.3. Using Combined Gas Law What will be the final pressure of a sample
of nitrogen gas with a volume of 950 m3 at 745 torr and 25.0 °C if it is heated to 60.0 °C and given a final volume of 1150 m3?
First, number of moles is constant even though actual number is not given
You are given V, P and T for initial state of system as well as T and V for final state of system and must find Pfinal
This is a clear case for combined gas law
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 3 List what you know and what you don’t know Convert all Temperatures to Kelvin Then solve for unknown—here P2
P1 = 745 torr P2 = ?
V1 = 950 m3 V2 = 1150 m3
T1 = 25.0 °C + 273.15
= 298.15 K
T2 = 60.0 °C + 273.15 = 333.15 K
3
3
21
2112 115015298
15333950745mK.
K.mtorrVTTVP
P
P2 = 688 torr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4. Combined Gas Law Anesthetic gas is normally given to a
patient when the room temperature is 20.0 °C and the patient's body temperature is 37.0°C. What would this temperature change do to 1.60 L of gas if the pressure and mass stay the same? What do we know? P and n are constant So Combined Gas Law simplifies to
2
2
1
1
TV
TV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4
V1 = 1.60 L V2 = ?
T1 = 20.0 °C + 273.15 = 293.15 K
T2 = 37.0 °C + 273.15 = 310.15 K
List what you know and what you don’t know Convert all Temperatures to Kelvin Then solve for unknown—here V2
K.K.L.
TTV
V15293
15310601
1
212
V2 = 1.69 L
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!Which units must be used in all gas law
calculations?A. K B. AtmC. LD. no specific units as long as they cancel
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Relationships between Gas Volumes In reactions in which products and
reactants are gases: If T and P are constant Simple relationship among volumes
hydrogen + chlorine hydrogen chloride
1 vol 1 vol 2 vol hydrogen + oxygen water (gas) 2 vol 1 vol 2 vol ratios of simple, whole numbers
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Avogadro’s Principle When measured at same T and P, equal
V's of gas contain equal number of moles Volume of a gas is directly proportional
to its number of moles, n V n (at constant P and T)
H2 (g) + Cl2 (g) 2 HCl (g)
Coefficients
1 1 2
Volumes 1 1 2
Molecules 1 1 2 (Avogadro's Principle)
Moles 1 1 2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Standard Molar Volume
Volume of 1 mole gas must be identical for all gases under same P and T
Standard Conditions of Temperature and Pressure — STP STP = 1 atm and 273.15 K
(0.0°C) Under these conditions 1 mole gas occupies V = 22.4 L 22.4 L standard molar volume
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning Check:
Calculate the volume of ammonia formed by the reaction of 25L of hydrogen with excess nitrogen.
N2 (g) + 3H2 (g) 2NH3 (g)
32
32 NH L17 H L3
NH L2
1
H L25
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning Check:N2 (g) + 3H2 (g) 2NH3 (g)
If 125 L H2 react with 50L N2, what volume of NH3 can be expected?
H2 is limiting reagent 83.3 L
32
32 NH L 83.3H L 3
NH L 21
H L 125
32
32 NH L 100N L 1
NH L 21
N L 50
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning Check:How many liters of N2 (g) at 1.00 atm and 25.0 °C are produced by the decomposition of 150. g of NaN3? 2NaN3 (s) 2Na (s) + 3N2 (g)
23
233 N mol 461.3 NaN mol 2
N mol 3
g 65.0099
NaN mol 1
1
NaN 150.g
L 53.77STPat mol 1L 22.4
1
N mol 461.3 2
L62.84K15.372
K15.982L53.177V
2
1
212
2
2
1
1
T
TVV ;
T
V
T
V
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!How many liters of SO3 will be produced
when 25 L of sulfur dioxide reacts with 75 L of oxygen ? All gases are at STP.
A. 25 LB. 50 LC. 100 LD. 150 LE. 75 L
41
32 3
2
32 3
2
2 L SO25 L SO x 25 L SO
2 L SO
2 L SO75 L O x 150 L SO
1 L O
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ideal Gas Law
With Combined Gas Law we saw that
With Avogadro’s results we see that this is modified to
Where R = a new constant = Universal Gas constant
RnTPV
CTPV
nRTPV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ideal Gas Law PV = nRT Equation of state of a gas: If we know 3 of these variables, then
we can calculate 4th Can define state of the gas by defining
3 of these values Ideal Gas Hypothetical gas that obeys Ideal Gas
Law relationship over all ranges of T, V, n and P
As T and P, real gases ideal gases
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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What is the value of R?
Plug in values of T, V, n and P for 1 mole of gas at STP (1 atm and 0.0°C) T = 0.0°C = 273.15 K P = 1 atm V = 22.4 L n = 1 mol
K.molL.atm
nTPV
R1527314221
R = 0.082057 L·atm·mol1·K1
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: PV = nRTHow many liters of N2(g) at 1.00 atm and 25.0 °C are produced by the decomposition of 150. g of NaN3? 2NaN3(s) 2Na(s) + 3N2(g)
V = ?V = nRT/P
3
2332
NaN mol 2N mol 3
g 65.0099 NaN mol 1
1 NaN 150.g
N moln
atm1
K15.298082057.0N mol461.3V Kmol
atmL2
V=84.62L
P = 1 atm T = 25C + 273.15 = 298.15 K
n = 3.461 mol N2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 5. Ideal Gas Law Problem
What volume in milliliters does a sample of nitrogen with a mass of 0.245 g occupy at 21°C and 750 torr?
What do I know? Mass and identity (so molecular mass MM) of
substance – can find moles Temperature Pressure
What do I need to find? Volume in mL
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 5 Solution
V = ? (mL)mass = 0.245 g MM = 214.0 = 28.0
g/mol Convert T from °C to KT = 21°C + 273.15 K = 294 K Convert P from torr to atm
Convert mass to moles
atm.torr
atmtorrP 9870
7601
750
mol.mol/g.g.
MMm
n 310758028
2450
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 5 Solution
PnRT
V
atm.KKmolatmL.moles.
V9870
294082057010758 113
= 214 mL LmL
L.V1
10002140
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Solid CaCO2 decomposes to solid CaO and
CO2 when heated. What is the pressure, in atm, of CO2 in a 50.0 L container at 35 oC when 75.0 g of calcium carbonate decomposes?
A. 0.043 atmB. 0.010 atmC. 0.38 atmD. 0.08 atmE. 38 atm
49
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
50
3 22
3
1 mol CaCO 1 mol CO L atm75.0 g CaO x x x 0.0821 x 308 K
100.1 g 1 mol CaCO K mol50.0 L
0.38 atm
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Determining Molecular Mass of GasIf you know P, T, V and mass of gas
Use Ideal Gas Law to determine moles (n) of gas
Then use mass and moles to get MMIf you know T, P, and Density (d) of a
gas Use density to calculate volume and
mass of gas Use Ideal Gas Law to determine moles
(n) of gas Then use mass and moles to get MM
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 6.
The label on a cylinder of an inert gas became illegible, so a student allowed some of the gas to flow into a 300 mL gas bulb until the pressure was 685 torr. The sample now weighed 1.45 g; its temperature was 27.0°C. What is the molecular mass of this gas? Which of the Group 0 gases (inert gases) was it?
What do I know? V, mass, T and P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 6 Solution
Mass = 1.45 g Convert T from °C to K. T = 27.0°C + 273.15 K = 300.2 K Convert P from torr to atm
Use V, P, and T to calculate n
atm.torr
atmtorrP 9010
7601
685
K.)Kmol/Latm(.L.atm.
RTPV
n23000820570
30009010 0.01098 mole
LmL
LmLV 3000
10001
300 .
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 6 Solution (cont)
Now use the mass of the sample and the moles of the gas (n) to calculate the molecular mass (MM)
Gas = Xe (At. Mass = 131.29 g/mol)
mol.g.
nmass
MM010980
451132 g/mol
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7.
A gaseous compound of phosphorus and fluorine with an empirical formula of PF2 was found to have a density of 5.60 g/L at 23.0 °C and 750 torr. Calculate its molecular mass and its molecular formula.
Know Density Temperature Pressure
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7. Solution
d = 5.60 g/L 1 L weighs 5.60 g So assume you have 1 L of gas V = 1.000 L Mass = 5.60 g Convert T from °C to K T = 23.0°C + 273.15 K = 296.2 K Convert P from torr to atm
atm.torr
atmtorrP 98680
7601
750
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7. Solution (cont)
Use n and mass to calculate MM
K.)Kmol/Latm(.L.atm.
RTPV
n22960820570
000198680
0.04058 mole
mol.g.
nmass
MM040580
605138 g/mol
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7. Solution (cont)
Now to find molecular formula given empirical formula and MM
First find mass of empirical formula unit
1 P = 1 31g/mol = 31g/mol 2 F = 2 19 g/mol = 38 g/mol Mass of PF2 = 69 g/mol
269138
mol/gmol/g
massempiricalmassmolecular
MM
the correct molecular formula is P2F4
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Which Gas Law to Use? Which Gas Law to Use in Calculations?
If you know Ideal Gas Law, you can get all the rest
Gas Law
Problems
Use CombinedGas LawUse Ideal Gas
Law
nRTPV 22
22
11
11
TnVP
TnVP
Amount of gasgiven or asked
for in moles or g
Amount of gas remains constant or not mentioned
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!7.52 g of a gas with an empirical formula of
NO2 occupies 2.0 L at a pressure of 1.0 atm and 25 oC. Determine the molar mass and molecular formula of the compound.
A. 45.0 g/mol, NO2
B. 90.0 g/mol, N2O4
C. 7.72 g/mol, NOD. 0.0109 g/mol, N2O
E. Not enough data to determine molar mass
60
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
61
2 2
2 4
L atm7.52 g x 0.0821 x 298 K
K molMW = 90.0 g/mol 1.0 atm x 2.0 L
1 mol NO 2 mol NO g90 x =
mol 45.0 g mol
Molecular formula is N O
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Stoichiometry of Reactions Between Gases
Can use stoichiometric coefficients in equations to relate volumes of gases Provided T and P are constant Volume moles V n
Ex. 8. Methane burns according to the following equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
1 vol 2 vol 1 vol 2 vol
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 8
The combustion of 4.50 L of CH4 consumes how many liters of O2? (Both volumes measured at STP.)
P and T are all constant so just look at ratio of stoichiometric coefficients
= 9.00 L O2
4
22 CH1
O2504O of Volume
LL
L.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 9. In one lab, the gas collecting apparatus
used a gas bulb with a volume of 250 mL. How many grams of Na2CO3 (s) would be needed to prepare enough CO2 (g) to fill this bulb when the pressure is at 738 torr and the temperature is 23 °C? The equation is:
Na2CO3(s) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 9. Solution
What do I know? T, P, V and MM of Na2CO3
What do I need to find? grams Na2CO3
How do I find this? Use Ideal Gas Law to calculate moles
CO2
Convert moles CO2 to moles Na2CO3
Convert moles Na2CO3 to grams Na2CO3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 9 Solution (cont)
1. Use Ideal Gas Law to calculate moles CO2
a. First convert mL to L
b. Convert torr to atm
c. Convert °C to KT = 23.0°C + 273.15 K = 296.2 K
LmL
LmLV 250.0
10001
250
atm.torr
atmtorrP 9710
7601
738
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 9 Solution (cont)
1. Use Ideal Gas Law to calculate moles CO2
2. Convert moles CO2 to moles Na2CO3
K.)Kmol/Latm(.L.atm.
RTPV
n22960820570
25009710
= 9.989 x 103 mole CO2
2
322
3
11
109899COmol
CONamolCOmol.
= 9.989 x 103 mol Na2CO3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 9 Solution (cont)
3. Convert moles Na2CO3 to grams Na2CO3
= 1.06 g Na2CO3
32
3232
3
1 106
109899CONamolCONag
CONamol.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!2Na(s) + 2H2O(l ) → 2NaOH(aq) + H2(g )
How many grams of sodium are required to produce 20.0 L of hydrogen gas at 25.0 C, and 750 torr ?
A. 18.6 gB. 57.0 gC. 61.3 gD. 9.62 gE. 37.1 g
69
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution Moles of H2 produced:
Grams of sodium required:
70
2
1 atm750 torr x x 20.0 L
760 torrn = = 0.807 mol HL atm
0.0821 x 298 KK mol
22
2 mol Na 23.0 g g Na = 0.807 mol H x x = 37.1 g
mol H mol Na
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Dalton's Law of Partial Pressure For mixture of non-reacting gases in
container Total pressure exerted is sum of the
individual partial pressures that each gas would exert alone
Ptotal = Pa + Pb + Pc + ···
Where Pa, Pb, and Pc = partial pressures
Partial pressure Pressure that particular gas would exert if it
were alone in container
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Dalton’s Law of Partial Pressures Assuming each gas behaves ideally
Partial pressure of each gas can be calculated from Ideal Gas Law
So Total Pressure is VRTn
P aa
VRTn
P bb
VRTn
P cc
VRTn
VRTn
VRTn
PPPP
cba
cbatotal
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Dalton’s Law of Partial Pressures
Rearranging
Or
Where ntotal = na + nb + nc + ··· ntotal = sum of # moles of various gases
in mixture
VRT
)nnn(P cbatotal
VRT
nP totaltotal
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Dalton’s Law of Partial PressuresMeans for Mixture of Ideal Gases
Total number of moles of particles is important Not composition or identity of involved particles
Pressure exerted by ideal gas not affected by identity of gas particles
Reveals 2 important facts about ideal gases1. Volume of individual gas particles must be
important2. Forces among particles must not be important
If they were important, P would be dependent on identity of gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
75
Ex. 10
Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25 °C and 1.0 atm and 12 L O2 at 25 °C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25 °C.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 10 Solution
Have 2 sets of conditions Before and after being put into the tank
He O2
Pi = 1.0 atm
Pf = PHe Pi = 1.0 atm
Pf = PO2
Vi = 46 L Vf = 5.0 L
Vi = 12 L Vf = 5.0 L
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 10 Solution (cont) First calculate pressure of each gas in
5 L tank (Pf) using combined gas law
Then use these partial pressures to calculate total pressure
atm.L
LatmVVP
Pf
iiHe 29
5461
atm.L
LatmVVP
Pf
iiO 42
5121
2
atm.atm.atm.PPP OHetotal 61142292
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!250 mL of methane, CH4, at 35 oC and
0.55 atm and 750 mL of propane, C3H8, at 35 oC and 1.5 atm, were introduced into a 10.0 L container. What is the final pressure, in torr, of the mixture?
A. 95.6 torrB. 6.20 x 104 torrC. 3.4 x 103 torrD. 760 torrE. 59.8 torr
78
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
79
4
3 8
0.55 atm x 0.250 LP(CH ) = 0.0138 atm
10.0 L1.5 atm x 0.750 L
P(C H ) = 0.112 atm10.0 L
760 torrP = 0.0138 + 0.112 atm x = 95.6 torr
atmT
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Mole Fractions and Mole Percents
Mole Fraction Ratio of number moles of given
component in mixture to total number moles in mixture
Mole Percent (mol%)
total
A
ZCBA
AA n
nnnnn
nX
100% Mole AX
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Mole Fractions of Gases from Partial Pressures
If V and T are constant then, = constant
For mixture of gases in one container
RTV
Pn AA
RTV
RTV
PRTV
PRTV
PRTV
P
RTV
PX
ZCBA
A
A
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Mole Fractions of Gases from Partial Pressures
cancels, leaving RTV
ZCBA
AA PPPP
PX
or
total
A
total
AA n
nPP
X
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Ex. 11 The partial pressure of oxygen was
observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present
Use
total
AA P
PX
2100743156
2 .torrtorr
XO
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Partial Pressures and Mole Fractions
Partial pressure of particular component of gaseous mixture
Equals mole fraction of that component times total pressure
totalAA PXP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
85
Ex. 12 The mole fraction of nitrogen in the air
is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr.
totalNN PXP 22
torrtorr.PN 593760780802
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!250 mL of methane, CH4, at 35 oC and 0.55
atm and 750 mL of propane, C3H8, at 35 oC and 1.5 atm were introduced into a 10.0 L container. What is the mole fraction of methane in the mixture?
A. 0.50B. 0.11C. 0.89D. 0.25E. 0.33
86
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
87
4
4
3 8
CH
0.55 atm x 0.250 LP(CH ) = 0.0138 atm
10.0 L1.5 atm x 0.750 L
P(C H ) = 0.112 atm10.0 L
0.0138 atm = = 0.110
(0.0138 + 0.112) atmX
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Collecting Gases over Water Application of Dalton’s Law of Partial Pressures Gases that don’t react with water can be
trapped over water Whenever gas is collected by displacement of
water, mixture of gases results Gas in bottle is mixture of water vapor and gas
being collected
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Collecting Gases over Water Water vapor is present because molecules of
water escape from surface of liquid and collect in space above liquid
Molecules of water return to liquid When rate of escape = rate of return
Number of water molecules in vapor state remains constant
Gas saturated with water vapor = “Wet” gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Vapor Pressure
Pressure exerted by vapor present in space above any liquid Constant at constant T
When wet gas collected over water, we usually want to know how much “dry” gas this corresponds to
Ptotal = Pgas + Pwater Rearranging Pgas = Ptotal – Pwater
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 13A sample of oxygen is collected over water at 20.0 °C and a pressure of 738 torr. Its volume is 310 mL. The vapor pressure of water at 20°C is 17.54 torr.
a. What is the partial pressure of O2? b. What would the volume be when
dry at STP? a. PO2
= Ptotal – Pwater
= 738 torr – 17.5 torr = 720 torr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 13 Solutionb. Use the combined gas law to calculate PO2 at STP
P1 = 720 torr P2 = 760 torr
V1 = 310 mL V2 = ?
T1 = 20.0 + 273.12 = 293 K
T2 = 0.0 + 273 K = 273 K
2
22
1
11
TVP
TVP
V2 = 274 mL torrKKmLtorr
V760293
2733107202
21
2112 PT
TVPV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!An unknown gas was collected by water
displacement. The following data was recorded: T = 27.0 oC; P = 750 torr; V = 37.5 mL; Gas mass = 0.0873 g; Pvap(H2O) = 26.98 torr
Determine the molecular weight of the gas.A. 5.42 g/molB. 30.2 g/molC. 60.3 g/molD. 58.1 g/molE. 5.81 g/mol
93
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
94
L atm0.0873 g x 0.0821 x 300 KgRT K mol
PV (750 - 26.98)torr x 0.0375 L
60.3 g/mol
MW
MW
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Diffusion Complete
spreading out and intermingling of molecules of one gas into and among those of another gas Ex. Perfume
in room
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Effusion Movement of gas
molecules Through
extremely small opening into vacuum
Vacuum No other gases
present in other half
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Thomas Graham
Studied relationship between effusion rates and molecular masses for series of gases
Wanted to minimize collisions Slow molecules down Make molecules bump aside or move to
rear
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Graham's Law of Effusion Rates of effusion of gases are inversely
proportional to square roots of their densities, d, when compared at identical pressures and temperatures
d
1Rate Effusion (constant P and T)
kd Rate Effusion
k is virtually identical for all gases
(constant P and T)
kdd BA (B) Rate Effusion(A) Rate Effusion
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Graham's Law of Effusion
Rearranging
Finally, dA MM (constant V and n)
Result: Rate of effusion is inversely proportional to molecular mass of gas
A
B
A
B
dd
d
d
BA
)( Rate Effusion)( Rate Effusion
A
B
A
B
MM
dd
BA
)( Rate Effusion)( Rate Effusion
kMM Rate Effusion (constant P and T)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Graham's Law of Effusion
Heavier gases effuse more slowly Lighter gases effuse more rapidly
Ex.14. Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6) - a gas used in the enrichment process to produce fuel for nuclear reactors.
kMM Rate Effusion
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex.14. Solution
First must compute MM's MM (H2) = 2.016 g/mol
MM (UF6) = 352.02 g/mol
Thus the very light H2 molecules effuse ~13 times as fast as the massive UF6 molecules.
2130162
02352)(UF Rate Effusion)(H Rate Effusion
2
6
6
2 ..
.M
M
H
UF
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!If it takes methane 3.0 minutes to diffuse
10.0 m, how long will it take sulfur dioxide to travel the same distance ?
A. 1.5 minB. 12.0 minC. 1.3 minD. 0.75 minE. 6.0 min
102
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
103
1 1
2 2
2
2
Remember, velocity and time are inversely related.
3.0 min 16.04 g/mol64.06 g/mol
6.0 min
t MWt MW
t
t
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex.15.
For the series of gases He, Ne, Ar, H2, and O2 what is the order of increasing rate of effusion?
Lightest are fastest So H2 > He > Ne > O2 >Ar
substance
He Ne Ar H2 O2
MM 4 20 40 2 32
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Kinetic Theory and Gas Laws So far, considered gases from
experimental point of view At P < 1 atm, most gases approach ideal
Ideal gas law predicts behavior Does NOT explain it
Recall scientific method Law is generalization of many observations Laws allow us to predict behavior Don't explain WHY
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Kinetic Theory and the Gas Law To answer WHY it happens—must
construct Theory or Model Models consist of speculations about what
individual atoms or molecules might be doing to cause observed behavior of macroscopic system (large number of atoms/molecules)
For model to be successful: Must explain observed behavior in question Predict correctly results of future
experiments
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Kinetic Theory and the Gas Law Model can never be proved absolutely
true Approximation by its very nature Bound to fail at some point
One example of model is kinetic theory of gases Attempts to explain properties of ideal
gases. Speculates on behavior of individual gas
particles
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Postulates of Kinetic Theory of Gases
1. Particles are so small compared with distances between them, so volume of individual particles can be assumed to be negligible.
Vgas ~ 0
2. Particles are in constant motion Collisions of particles with walls of
container are cause of pressure exerted by gas
number collisions Pgas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Postulates of Kinetic Theory of Gases
3. Particles are assumed to exert no force on each other
Assumed neither to attract nor to repel each other
4. Average kinetic energy of collection of gas particles is assumed to be directly proportional to Kelvin Temperature
KEavg TK
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Kinetic Theory of Gases Kinetic theory of matter and heat transfer
(ch 7) Heat PV KEave But for constant # moles of ideal gas PV = nRT
where nR is proportionality constant This means T KEave
Specifically
As increase T, KEave, number collisions with walls, thereby
increasing P
RT23
KEave
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Real Gases
Don’t conform to these assumptions Have finite volumes Do exert forces on each other However, KTG does explain Ideal Gas
behavior True test of model is how well its
predictions fit experimental observations
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Postulates of Kinetic Theory of Gases
Picture ideal gas consisting of particles having no volume and no attractions for each other
Assumes that gas produces pressure on its container by collisions with walls
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
113
Kinetic Theory Explains Gas Laws P and V (Boyle's Law)
For given sample of ideal gas at given T (n and T constant) If V , P
By KT Gases V, means gas particles hit wall more often P V
1)nRT(P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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P and T (Gay-Lussac's Law) For given sample
of ideal gas at constant V (n and V constant)
P is directly proportional to T
T V
nRP
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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P and T (Gay-Lussac's Law)
KT Gases accounts for this
As T KEave Speeds of molecules
Gas particles hit wall
more often as V same
So P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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T and V (Charles' Law) For given sample
of ideal gas at constant P (n and P constant)
V is directly proportional to T
T P
nRV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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T and V (Charles' Law)
KT Gases account for this
As T KEave Speeds of
molecules Gas particles hit
wall more often as P same
So V
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
118
V and n (Avogadro's Principle) For ideal gas at constant T and P
V is directly proportional to n
KT Gases account for this As n (number gas particles), at same T Since P constant Must V
n P
RTV
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
119
Dalton's Theory of Partial Pressures
Expected from Kinetic Theory of Gases All gas particles are independent of each other Volume of individual particles is unimportant Identities of gases do not matter
Conversely, can think of Dalton's Law of Partial Pressures as evidence for KTG Gas particles move in straight lines, neither
attracting nor repelling each other Particles act independently
Only way for Dalton's Law to be valid
gases individualPPtotal
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Law of Effusion (Graham's Law)
Key conditions: Comparing 2 gases at same P and T Conditions where gases don't hinder each
other Hence, particles of 2 gases have same KEave
Let = average of velocity squared of molecules of gases
Then
A
B
MM
)B( Rate Effusion)A( Rate Effusion
21 KEKE 2v
2222
12112
1 vmvm
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
121
Law of Effusion (Graham's Law)
Rearranging
Taking square root of both sides Since
NowRate of Effusion So Effusion Rate = k
2
1
22
21
mm
v
v
2
1
2
1
mm
v
v
11 Mm
2
1
2
1
2
1
MM
mm
v
v
v
v
1
2
2 Gas of Rate Effusion1 Gas of Rate Effusion
MM
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
122
Absolute Zero
If KEave = 0, then T must = 0.
Only way for KEave = 0, is if v = 0 since m 0.
When gas molecules stop moving, then gas as cold as it can get = Absolute Zero
)v(mKET ave2
21
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
123
Real Gases: Deviations from Ideal Gas Law
Combined Gas Law
Ideal Gas Law
Real Gases deviate Why?
constantT
PV
RnTPV
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Real Gases Deviate from Ideal Gas Law
1. Gas molecules have finite V's
Take up space Less space of
kinetic motions Vmotions < Vcontainer particles hit
walls of container more often
P over ideal
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Real Gases
2. Particles DO attract each other Even weak attractions means they hit
walls of container less often P over ideal gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Effect of Attractive Forces on Real Gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
127
van der Waal's equation for Real Gases
a and b are van der Waal's constants
Obtained by measuring P, V, and T for real gases over wide range of conditions
nRTnbVV
anP
2
2
corrected P corrected V
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
128
van der Waal's equation for Real Gases
a — Pressure correction Indicates some attractions between
molecules Large a
Means strong attractive forces between molecules
Small a Means weak attractive forces between
molecules
nRTnbVV
anP
2
2
corrected P
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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van der Waal's equation for Real Gases
b — Volume correction Deals with sizes of molecules Large b
Means large molecules Small b
Means small molecules Gases that are most easily liquefied have
largest van der Waal's constants
nRTnbVV
anP
2
2
corrected V