Download - the gaseous state of matter
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The gasesThe Gaseous State of Matter
The air in a hot air balloon expands When it is heated. Some of the airescapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.
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The atmosphere
Copyright 2011 John Wiley & Sons, Inc 12-2
The atmosphere protects the planet and provides chemicals necessary for life.
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The Atmosphere
Copyright 2011 John Wiley & Sons, Inc 12-3
Each of the gases, N2, O2, CO2 and H2O among others in the atmosphere, serve a purpose
O2 supports metabolism
N2 dilutes the O2 so explosive combustion does not take place and is also part of proteins
CO2 and H2O trap heat
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General Properties
Gases • Have an indefinite volume
Expand to fill a container• Have an indefinite shape
Take the shape of a container• Have low densities
• Have high kinetic energies
Copyright 2011 John Wiley & Sons, Inc 12-4
2
air
H O
d 1.2 g / L at 25 C
d 1.0 g / mL
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Kinetic Molecular Theory (KMT)
Assumptions of the KMT and ideal gases include:
1. Gases consist of tiny particles
2. The distance between particles is large compared with the size of the particles.
3. Gas particles have no attraction for each other
4. Gas particles move in straight lines in all directions, colliding frequently with each other and with the walls of the container.
Copyright 2011 John Wiley & Sons, Inc 12-5
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Kinetic Molecular Theory
Assumptions of the KMT (continued):
5. Collisions are perfectly elastic (no energy is lost in the collision).
6. The average kinetic energy for particles is the same for all gases at the same temperature.
7. The average kinetic energy is directly proportional to the Kelvin temperature.
Copyright 2011 John Wiley & Sons, Inc 12-6
21KE = where is mass and is velocity
2mv m v
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Diffusion
Copyright 2011 John Wiley & Sons, Inc 12-7
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Effusion
Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure.
Graham’s law of effusion:
Copyright 2011 John Wiley & Sons, Inc 12-8
rate of effusion of gas A density B molar mass B= =
rate of effusion of gas B density A molar mass A
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Graham’s law of effusion
Rates of effusion of gases are inversely proportional to the square roots of their molar masses
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Measurement of Pressure
ForcePressure =
Area
Copyright 2011 John Wiley & Sons, Inc 12-10
Pressure depends on the• Number of gas molecules• Temperature of the gas• Volume the gas occupies
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Atmospheric Pressure
Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface.
Copyright 2011 John Wiley & Sons, Inc 12-11
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Barometer
Copyright 2011 John Wiley & Sons, Inc 12-12
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Pressure Conversions
Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm
Copyright 2011 John Wiley & Sons, Inc 12-13
1 atm675 mm Hg = 0.888 atm
760 mm Hg
Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr.
760 torr675 mm Hg = 675 torr
760 mm Hg
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Dependence of Pressure on Number of Molecules
Copyright 2011 John Wiley & Sons, Inc 12-14
P is proportional to n (number of molecules) at Tc (constant T) and Vc (constant V).The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.
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Dependence of Pressure on Temperature
Copyright 2011 John Wiley & Sons, Inc 12-15
P is proportional to T at nc
(constant number of moles) and Vc.
The increased pressure is due to• more frequent collisions• higher energy collisions
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Boyle’s Law
Copyright 2011 John Wiley & Sons, Inc 12-16
1 1 2 2
1At and : α or c cT n V PV PV
P
What happens to V if you double P?• V decreases by half!
What happens to P if you double V?• P decreases by half!
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Boyle’s Law
A sample of argon gas occupies 500.0 mL at 920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature.
Copyright 2011 John Wiley & Sons, Inc 12-17
1 12
2
PV
PV
1 1 2 2 PV PV
2
920. torr 500. mL = = 491 torr
937 mLP
Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL
Calculate
Set-Up
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Boyle’s Law
Another approach to the same problem:
Since volume increased from 500. mL to 937 ml, the pressure of 920. torr must decrease.
Multiply the pressure by a volume ratio that decreases the pressure:
Copyright 2011 John Wiley & Sons, Inc 12-18
2
500. mL
937 mL = 920. torr = 491 torrP
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Charles’ Law
Copyright 2011 John Wiley & Sons, Inc 12-19
1 2
1 2
At and :
α or
c cP n
V VV T
T T
• The volume of an ideal gas at absolute zero (-273°C) is zero.
• Real gases condense at their boiling point so it is not possible to have a gas with zero volume.
• The gas laws are based on Kelvin temperature.
• All gas law problems must be worked in Kelvin!
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Charles’ Law
A 2.0 L He balloon at 25°C is taken outside on a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant?
Copyright 2011 John Wiley & Sons, Inc 12-20
1 2
1 2
V V
T T
1 22
1
rearranged gives V T
VT
2
(2.0 L)(258 K) = = 1.7 L
298 KV
Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K
Calculate
Set-Up 1 2
1 2
V V
T T
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Charles’ Law
Copyright 2011 John Wiley & Sons, Inc 12-21
Another approach to the same problem:
Since T decreased from 25°C to -15°C, the volume of the 2.0L balloon must decrease.
Multiply the volume by a Kelvin temperature ratio that decreases the volume:
2
258K
298 = 2.0L = 1.7L
KP
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Gay-Lussac’s Law
Copyright 2011 John Wiley & Sons, Inc 12-22
1 2
1 2
At and : α or c c
P PV n P T
T T
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Combined Gas Laws
Copyright 2011 John Wiley & Sons, Inc 12-23
Used for calculating the results of changes in gas conditions.
1 1 2 2
1 2
PV PV
T T
• Boyle’s Law where Tc
• Charles’ Law where Pc
1 2
1 2
V V
T T
1 1 2 2 PV PV
• Gay Lussacs’ Law where Vc
1 2
1 2
P P
T T
P1 and P2 , V1 and V2 can be any units as long as they are the same. T1 and T2 must be in Kelvin.
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Combined Gas Law
Copyright 2011 John Wiley & Sons, Inc 12-24
If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr?
STP: Standard Temperature 273K or 0°CStandard Pressure 1 atm or 760 torr
1 1 22
1 2
PV T
VT P
Knowns V1 = 500. mL T1 =273K P1= 760 torrT2 = 85°C = 358K P2= 560 torr
Set-Up
2
(760 torr)(500. mL)(358K) = 890. ml
(273K)(560 torr)V Calculate
1 1 2 2
1 2
PV PV
T T
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Combined Gas Law
A sample of oxygen gas occupies 500.0 mL at 722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg.
Copyright 2011 John Wiley & Sons, Inc 12-25
1 2 22
1 1
T PVT
PV
1 1 2 2
1 2
PV PV
T T
2
491 torr 2530 ml 248K= =853K 580 C
722 torr 500.0 mlT
Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr V2 = 2.53 L = 2530 mL P2= 560 torr
Set-Up
Calculate
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Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.
PTotal = PA + PB + PC + ….
Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air.
Copyright 2011 John Wiley & Sons, Inc 12-26
2 2 2 2.Air N O Ar CO H OP P P P P P
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Collecting Gas Over Water
• Gases collected over water contain both the gas and water vapor.
• The vapor pressure of water is
constant at a given temperature• Pressure in the bottle is
equalized so that the Pinside = Patm
Copyright 2009 John Wiley & Sons, Inc 12-27
2 atm gas H OP P P
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Avogadro’s Law
Equal volumes of different gases at the same T and P contain the same number of molecules.
Copyright 2011 John Wiley & Sons, Inc 12-28
1 volume1 molecule
1 mol
1 volume1 molecule
1 mol
2 volumes2 molecules
2 mol
The ratio isthe same:
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Mole-Mass-Volume Relationships
Molar Volume: One mole of any gas occupies 22.4 L at STP.
Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP.
Copyright 2011 John Wiley & Sons, Inc 12-29
Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm
Set-Up
Calculate
22.4 L1 mol = 22.4 L so the conversion factor is
1mol
3.94 g
1.52 L
22.4 L
1 mol
= 58.1g/mol
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Density of Gases
Calculate the density of nitrogen gas at STP.
Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.
Copyright 2011 John Wiley & Sons, Inc 12-30
mass gd = =
volume L
STP
1 mold = molar mass
22.4 L
STP
28.02 g 1 mold = = 1.25g/L
1 mol 22.4 L
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Ideal Gas Law
Calculate the volume of 1 mole of any gas at STP.
Copyright 2011 John Wiley & Sons, Inc 12-31
PV nRT R
L atm where = 0.0821
mol K
Knowns n = 1 mole T = 273K P = 1 atm
Set-Up
Calculate
nRT
VP
L atm(1 mol)(0.0821 )(273 K)
mol K = (1 atm)
V
Molar volume!
= 22.4 L
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Ideal Gas Law
How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg?
Copyright 2011 John Wiley & Sons, Inc 12-32
Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm
Set-Up
Calculate
PV
nRT
(0.980 atm)(1.3 L) = =0.052 mol
L atm(0.0821 )(297 K)
mol K
n
PV nRT R
L atm where = 0.0821
mol K
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Ideal Gas Law
Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr.
Copyright 2011 John Wiley & Sons, Inc 12-33
Knowns m =4.12 g V = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm
Set-Up
Calculate
g g = so =
M Mn PV RT
L atm(4.12 g)(0.0821 )(296 K)
mol KM = =107 g/mol(0.988 atm)(0.943 L)
R T
P V
g M =
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Gas Stoichiometry
Copyright 2011 John Wiley & Sons, Inc 12-34
• Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L.
• Use the Ideal Gas Law if the conditions are not at STP.
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Gas Stoichiometry
Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm.
P4(s) + 6H2(g) 4PH3(g)
Copyright 2011 John Wiley & Sons, Inc 12-35
2 4.0 L H2 1 mol H
22.4L
4
2
1 mol P
6 mol H
4= 0.030 mol P
Knowns V = 4.0 L T = 273 K P = 1 atm
Solution Map L H2 mol H2 mol P4
Calculate
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Gas Stoichiometry
What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C2H6?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
Copyright 2011 John Wiley & Sons, Inc 12-36
Solution Map m C2H6 mol C2H6 mol O2 volume O2
Knowns m = 3.2 g C2H6 T = 25°C = 298K P = 1 atm
2 63.2g C H 2 6
2 6
1 mol C H
30.08g C H
2
2 6
7 mol O
2 mol C H
2= 0.37mol OCalculate
L atm(0.37 mol)(0.0821 )(298 K)
mol K = = 9.1 L(1 atm)
V
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Volume-Volume Calculations
Copyright 2011 John Wiley & Sons, Inc 12-37
Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm.
N2(g) + 3H2(g) 2NH3(g)
2 9.0 L H2
2
1 L N
3 L H
2= 3.0 L N
Knowns V = 9.0 L T = 450K P = 5.00 atm
Solution Map Assume T and P for both gases are the same.Use volume ratio instead of mole ratio!L H2 L N2
Calculate
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Real Gases
Most real gases behave like ideal gases under ordinary temperature and pressure conditions.
Conditions where real gases don’t behave ideally:• At high P because the distance between particles is too
small and the molecules are too crowded together. • At low T because gas molecules begin to attract each
other.
High P and low T are used to condense gases.
Copyright 2011 John Wiley & Sons, Inc 12-38
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The law of effusion and diffusion
The postulate of the kinetic theory states that the temperature of a system is proportional to the average kinetic energy of its particles and nothing else.
In other words, two gases at the same temperature, such as and must have the same average kinetic energy, hence
multiplying by 2 gives:
Copyright 2011 John Wiley & Sons, Inc 12-39
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Law of Effusion and diffusion
Then rearranging gives:
Taking out the square root both sides we get
Copyright 2011 John Wiley & Sons, Inc 12-40