Transcript Chapter 5 Gases Chapter 5: Gases 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 Early Experiments The gas laws of Boyle, Charles, and Avogadro The Ideal Gas Law Gas Stoichiometry Dalton’s Law of Partial.

Chapter 5
Gases
Chapter 5: Gases
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
Early Experiments
The gas laws of Boyle, Charles, and Avogadro
The Ideal Gas Law
Gas Stoichiometry
Dalton’s Law of Partial Pressures
The Kinetic molecular Theory of Gases
Effusion and Diffusion
Collisions of Gas Particles with the Container Walls
Intermolecular Collisions
Real Gases
Chemistry in the Atmosphere
Hurricanes, such as this one off the coast of
Florida, are evidence of the powerful forces
present in the earth's atmosphere.
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.
Substances that are Gases under
Normal Conditions
Substance
•
•
•
•
•
•
•
•
•
•
•
Helium
Neon
Argon
Hydrogen
Nitrogen
Nitrogen Monoxide
Oxygen
Hydrogen Chloride
Ozone
Ammonia
Methane
Formula
He
Ne
Ar
H2
N2
NO
O2
HCL
O3
NH3
CH4
MM(g/mol)
4.0
20.2
39.9
2.0
28.0
30.0
32.0
36.5
48.0
17.0
16.0
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
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.
Pressure =
Force
Area
• Measured using a Barometer! - A device that can weigh the
atmosphere above us!
Figure 5.1:
A torricellian
barometer.
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
HeightWater
= DensityMercury
HeightMercury
DensityWater
Density of Mercury
Density of Water
• Hw = 760 mm Hg x 13.6/1.00 = 1.03 x 104 mm
• Hw = 10.3 m = __________ ft
Common Units of Pressure
Unit
Atmospheric Pressure
Pascal (Pa);
kilopascal (kPa)
1.01325 x 105 Pa
101.325 kPa
Scientific Field Used
SI unit; physics,
chemistry
Atmosphere (atm)
1 atm
Millimeters of mercury
(mmHg)
Torr
760 mmHg
Pounds per square inch
(psl or lb/in2)
14.7 lb/in2
Engineering
1.01325 bar
Meteorology, chemistry
Bar
760 torr
Chemistry
Chemistry, medicine
biology
Chemistry
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:
PCO2 (torr) = 341.6 mm Hg x 1 torr
= 341.6 torr
1 mm Hg
converting from torr to atm:
1 atm
PCO2( atm) = 341.6 torr x
= 0.4495 atm
760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x 101.325 kPa = _____________ kPa
1 atm
Figure 5.2: A simple manometer, a device for
measuring the pressure of a gas in a container.
Figure 5.3:
A J-tube similar to
the one used by
Boyle.
Boyle’s Law : P - V relationship
• Pressure is inversely proportional to Volume
k
k
•
P=
or V =
or PV=k
V
P
• Change of Conditions Problems
if n and T are constant !
• P1V1 = k
P2V2 = k’
k = k’
• Then :
P1V1 = P2V2
Figure 5.4: Plotting Boyle’s data from Table 5.1.
Figure 5.5: Plot of PV versus P for several gases.
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
3 = 1 mL
1cm
3
V1 = 15.8 cm
V2 = unknown
T and n remain constant
V (ml)
1
V1 = 15.8 cm3 x 1 mL3 x 1 L
= 0.0158 L
1 cm 1000mL
V2 = V1 x P1 = 0.0158 L x 1.23 atm = ________ L
P2
3.16 atm
1000mL = 1L
V1 (L)
x P1/P2
V2 (L)
Boyle’s 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 underwater 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
V2 = ?
P1 = ?
P 2 = 1.00 atm
Calculation Continued
2O
Pressure at depth = 6 x 103 ft x 0.3048 m x 100 cm x 1.026 g SH
1 ft
1m
1 cm3
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:
3 Hg
1.00
atm
10
mm
Area
1.00
cm
Pressure =
x
x
x
x 187,635 g =
2
760 mm Hg 1 cm
1 cm
13.6 g Hg
Pressure = _____ atm Due to the added atmospheric pressure = ___ atm!
3 x ____ atm
V1 x P1
1.00
cm
V2 = P
=
= ____ cm3 =
liters
2
1.00 atm
Boyle’s 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
Figure 5.6:
PV plot verses
P for 1 mole of
ammonia
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
T1
V1
=
T2
V2
T1 = V1 x
or:
T2
V2
Temperatures must be expressed in Kelvin to avoid negative values!
Figure 5.7: Plots of V versus T (c) for several
gases.
Figure 5.8: Plots of V versus T as in Fig. 5.7
except that here the Kelvin scale is used for
temperature.
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 = ___ K
oC
T2 = 398 K x
1.54 L
3.20 L
=______K
= K - 273.15 = ______ - 273
oC = ______oC
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
• T1 = 75o F
V2 = ?
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
Antarctic Balloon Problem - II
• V1 / T1 = V2 / T2
• V2 = 20.0 L x
V2 = V1 x ( T2 / T1 )
216.4 K
297.0 K
• V2 = __________ L
• The Balloon shrinks from 20 L to 15 L !!!!!!!
• Just by going outside !!!!!
Applying the Temperature - Pressure
Relationship (Amonton’s 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:
P1 = P2
T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC
T1
T2
T = -17.68 oC + 273.15 K = 255.47 K
1
T2 = 95.6 oC + 273.15 K = 368.8 K
P2 = 4.28 atm x 368.8 K = _______ atm
255.47 K
P2 = P1 x
T2
=?
T1
Avogadro’s Law - Amount and Volume
The Amount of Gas (Moles) is directly proportional to the volume
of the Gas.
n = kV
nV
or
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
n2 = final moles of gas
n1 =
V1
n2
V2
or:
V1 = initial volume of gas
V2 = final volume of gas
n1 = n2 x V1
V2
Avogadro’s 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 Avogadro’s 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
= 0.0183 mol SF6
146.07g SF6/mol
3
V
543.9
m
n2 = n1 x 2 = 0.0183 mol SF6 x
= 3.40 mol SF6
3
V1
2.93 m
mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = __________ g SF6
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 Avogadro’s law, and the change of conditions form.
Solution:
V1 = 28.75 L
T = constant
n1 = 1.14 moles H2
V2 = 112.46 L
P = constant
n2 = 1.14 moles + ? moles
n1
n2
V1 = V2
V2
n2 = n1 x
V1
= 1.14 moles H2 x
112.46 L
28.75 L
mass = moles x molecular mass
mass = 4.46 moles x 2.016 g / mol
added mass = 8.99g - 2.30g = 6.69g mass = ___________ g H2 gas
n2 = 4.4593 moles = 4.46 moles
Change of Conditions, with no change in
the amount of gas !
•
PxV
T
= constant
P1 x V1
•
T1
Therefore for a change
of conditions :
=
P2 x V2
T2
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 Hg=0.978 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
Change of Condition Problem I :
continued
• P1 x V1
T1
=
P2 x V2
T2
• ( 0.978 atm) ( 45.9 L)
( 298 K)
•
•
•
P2 =
=
P2 (0.00310 L)
( 428 K)
( 428 K) ( 0.978 atm) ( 45.9 L)
( 298 K) ( 0.00310 L)
= ________ atm
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
Change of Conditions Problem II:
continued
• P1 x V1
T1
• V2 =
=
P2 x V2
T2
V2 =
( 1.0 atm) ( 100 m3) (183 K)
P1V1T2
T1P2
=
(298 K) (0.0197 atm)
• V2 = 3117.23 m3 = __________ m3 or 31 times the
volume !!!
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
Change of Conditions Problem
III:continued
• V1 = 5.00 L
• P1 = 10.0 atm
• T1 = 298K
• P1V1/T1 = P2V2/T2
V2 = ?
P2 = 1.00 atm
T2 = 1173 K
V2 = V1P1T2/ P2T1
( 5.00 L) (10.00 atm) (1173 K)
• V2 =
= 197 L
( 1.00 atm) ( 298 K)
• VCO2 = (197 L C3H8) x (3 L CO2 / 1 L C3H8) =
VCO2 = _________ L CO2
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
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
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:
PV=nRT
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
Variations on the Gas Equation
• During chemical and physical processes, any of the
four variables in the ideal gas equation may be fixed.
• Thus, PV=nRT can be rearranged for the fixed
variables:
– for a fixed amount at constant temperature
• P V = nRT = constant
Boyle’s Law
– for a fixed amount at constant volume
• P / T = nR / V = constant
Amonton’s 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
Avogadro’s Law
Evaluation of the Ideal Gas Constant, R
Ideal gas Equation
PV = nRT
R = PV
nT
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!)
R=
(1.00 atm) ( 22.414 L) = 0.08206 L atm
( 1.00 mole) ( 273.15 K)
mol K
or to three significant figures R = 0.0821 L atm
mol K
Values of R in Different Units
Atm x L
= 0.08206
Mol x K
Torr x L
R = 62.36
Mol x K
3
kPa
x
dm
R = 8.314
Mol x K
J#
R = 8.314
Mol x K
R*
*
#
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
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 ( PV=nRT ).
Solution:
nXe = 5038 g Xe
= 38.37014471 mol Xe
131.3 g Xe / mol
T = 18.8 oC + 273.15 K = 291.95 K
PV = nRT
P=
P = nRT
V
(38.37 mol )(0.0821 L atm)(291.95 K)
= _______ atm =
87.5 L (mol K)
atm
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
PV=nRT
P= nRT/V
• P=
•
• P=
( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 K)
150 L
atm
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
Mass of Air in a Hot Air Balloon - Part II
• PV = nRT
n = PV / RT
( 0.984 atm) ( 3.99 x 105 L)
• n=
( 0.08206 L atm/mol K) ( 303 K )
= 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
Inflated
dual air
bag
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
Sodium Azide Calc: - II
• PV = nRT
• V =
V = nRT/P
( 1.38 mol) (0.08206 L atm / mol K) (294 K)
( 823 mm Hg / 760 mmHg / atm )
• V = __________ liters
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
P xM
• d=
=
R xT
( 0.989 atm) ( 17.03 g/mol)
( 0.08206 L atm/mol K) ( 328 K)
• d = __________ g / L
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
36.8g x
= 0.368 mol CaCO3
100.1g 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
(0.368 mol)(0.08206L atm/mol K)(310.15K)
V=
=
P
(715mmHg/760mm Hg/atm)
V = __________ L
Calculation of Molar Mass
• n=
Mass
Molar Mass
PxV
Mass
• n=
=
RxT
Molar Mass
• Molar Mass =
Mass x R x T
PxV
= MM
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
Pressure = 736 mm Hg x
= 0.9684 atm
760 mm Hg
mass = 149.457g - 148.375g = 1.082 g
Molar Mass =
(1.082 g)(0.0821 Latm/mol K)(373.2 K)
= 58.02 g/mol
( 0.9684 atm)(0.590 L)
note: the compound is acetone C3H6O = MM = 58 g/mol !
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:
P = 550.0 Torr x 1mm Hg x 1.00 atm
= 0.724 atm
1 Torr
760 mm Hg
V = 250.0 ml x 1.00 L = 0.250 L
1000 ml
n =PV
RT
T = 25.0 oC + 273.15 K = 298.2 K
n = (0.724 atm)(0.250 L)
= 0.007393 mol
(0.0821 L atm/mol K) (298.2 K)
MM = 0.118 g / 0.007393 mol = ____________ g/mol
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.
Figure 5.9: Partial pressure of each gas in a
mixture of gases depends on the number of
moles of that gas.
Dalton’s 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 = p1+p2+p3+...pi
Dalton’s 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.
Dalton’s 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
•
( 0.0682 mol CO2) ( 0.08206 L atm/mol K) ( 290 K)
• PCO2 =
(2.00 L)
• PCO2 = _____________ atm
Dalton’s Law Problem - #1 cont.
• nHe = 0.10 g He / 4.003 g He / mol He
•
= 0.025 mol He
• PHe = nHeRT/V
• PHe =
(0.025 mol) ( 0.08206 L atm / mol K) ( 290 K )
( 2.00 L )
• PHe = 0.30 atm
• PTotal = PCO2 + PHe = 0.812 atm + 0.30 atm
• PTotal = 1.11 atm
Dalton’s 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
Relative Humidity
• Rel Hum =
Pressure of Water in Air
x 100%
Maximum Vapor Pressure of Water
• 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%
= ___________ %
Figure 5.10: Production of oxygen by
thermal decomposition of KCIO3.
Molecular sieve framework of titanium (blue),
silicon (green), and oxygen (red) atoms
contract on heating–at room temperature (left)
d = 4.27 Å; at 250°C (right) d = 3.94 Å.
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
COLLECTION OVER WATER CONT.-II
• PV = nRT
• n=
n = PV / RT
(0.989 atm) (0.156 L)
(0.0821 L atm/mol K) (293 K)
• n = 0.00641 mol
• mass = 0.00641 mol x 2.01 g H2 / mol H2
• mass = ______________ g Hydrogen
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Avogadro’s
Number
6.02 x 1023
Reactants
Molarity
moles / liter
Solutions
Molecules
Moles
Molecular
g/mol
Weight
Products
PV = nRT
Gases
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)
TNH3 = 18.7 oC + 273.15 = 291.9 K
NH4Cl(s)
Gas Law Stoichiometry - II - NH3 + HCl
n = PV
RT
(0.776 atm) (2.79 L)
nNH3 =
= 0.0903 mol NH3
(0.0821 L atm/mol K) (291.9 K)
(0.932 atm) (1.16 L)
nHCl =
= 0.0451 mol HCl
(0.0821 L atm/mol K) (291.9 K)
limiting
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)
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.
Figure 5.11: An ideal gas particle in a
cube whose sides are of length L
(in meters).
Figure 5.12: (a) The Cartesian
coordinate axes.
Figure 5.12: (b) The velocity u of any gas
particle can be broken down into three
mutually perpendicular components, ux, uy,
and uz.
Figure 5.12: (c)
In the xy plane,
ux2 + uy2 = uyx2
by the
Pythagorean
theorem.
Figure 5.13: (a) Only the x component of the gas
particle’s 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.
Figure 5.14: Path of one particle in a gas.
A balloon
filled with air
at room
temperature.
The balloon is
dipped into
liquid nitrogen
at 77 K.
The balloon
collapses as the
molecules
inside slow
down because
of the
decreased
temperature.
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
–
KEavg=3RT/2NA
– T=temp. in Kelvin , NA =Avogadro’s number,
R=new 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!!!
Figure 5.15: A plot of the relative number of O2
molecules that have a given velocity at STP.
Figure 5.16:
A plot of the
relative number
of N2 molecules
that have a given
velocity at three
temperatures.
Figure 5.17:
A velocity
distribution for
nitrogen gas at
273 K.
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
8.314 J/mol K
EK = 3 R x T = 1.5 x
x 300K =
23
2 NA
6.022 x 10 molecules/mol
EK = 6.213 x 10 - 21 J/molecule = 1/2 mu2
-3 kg/mole
2.016
x
10
- 27 kg/molecule
m=
=
3.348
x
10
6.022 x 1023 molecules/mole
6.213 x 10 - 21 J/molecule = 1.674 x 10 - 27 kg/molecule(u2)
u = _______________ m/s = ___________ m/s
Molecular Mass and Molecular Speeds - II
for methane CH4 = 16.04 g/mole
8.314 J/mol K
EK = 3 R x T = 1.5 x
x 300K =
23
2 NA
6.022 x 10 molecules/mole
EK = 6.213 x 10 - 21 J/molecule = 1/2 mu2
m=
16.04 x 10 - 3 kg/mole
- 26 kg/molecule
=
2.664
x
10
23
6.022 x 10 moleules/mole
6.213 x 10 - 21 J/molecule = 1.332 x 10 - 26 kg/molecule (u2)
u = ___________________ m/s = __________ m/s
Molecular Mass and Molecular Speeds - III
for Carbon dioxide CO2 = 44.01 g/mole
8.314 J/mol K
EK = 3 R x T = 1.5 x
x 300K =
23
2 NA
6.022 x 10 molecules/mole
EK = 6.213 x 10 - 21 J/molecule = 1/2mu2
m=
44.01 x 10 - 3 kg/mole
- 26 kg/molecule
=
7.308
x
10
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
Molecular Mass and Molecular Speeds - IV
Molecule
Molecular
Mass (g/mol)
Kinetic Energy
(J/molecule)
Velocity
(m/s)
H2
CH4
2.016
16.04
6.213 x 10 - 21
1,926
6.213 x 10 - 21
683.0
CO2
44.01
6.213 x 10 - 21
412.4
Important Point !
• At a given temperature, all gases have the
same molecular Kinetic energy
distributions.
or
• The same average molecular Kinetic
Energy!
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.
Figure 5.18: The effusion of a gas into
an evacuated chamber.
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
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 !!!!!!
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
Figure 5.19: (a) demonstration of the
relative diffusion rates of NH3 and HCl
molecules through air.
Figure 5.19: (b) When HCl(g) and
NH3 (g) meet in the tube, a white ring of
NH4Cl(s) forms.
Figure 5.20: Uranium-enrichment
converters from the Paducah gaseous
diffusion plant in Kentucky.
Gaseous Diffusion Separation of Uranium 235 / 238
• 235UF6 vs 238UF6
(238.05 + (6 x 19))0.5
• Separation Factor = S =
0.5
(235.04
+
(6
x
19))
•
• after Two runs
S = 1.0043
•
after approximately 2000 runs
• 235UF6 is > 99% Purity !!!!!
•
Y - 12 Plant at Oak Ridge National Lab
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:
RT
Calculate ZA :
ZA = A N
V 2pM
1.00 atm
N = P =
L atm (300. K)
A = 1.00 x 10-4 m2
0.08206
V
RT
K mol
N = 4.06 x 10-2 Mol x 6.022 x 1023 molecules x 1000 L =
V
L
mol
m3
= 2.44 x 1025 molecules/m3
M = 32.0 g/mol x 1 kg = 3.2 x 10-2 kg/mol
1000g
J
8.3145
(300. K)
-4
2
25
-3
ZA = (1.00 x 10 m ) (2.44 x 10 m ) x
K mol
-2 kg/mol)
2(3.14)(
3.20
x
10
23
Z = 2.72 x 10 collisions/s
(
A
)
Figure 5.21: The cylinder swept out by
a gas particle of diameter d.
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
N
P
=
=
= 8.097 x 10-2 mol/L
V
RT
0.08206 L atm (301. K)
K mol
N = (8.097 x 10-2 mol/L)( 6.022 x 1023 molecules/mol)(1000L/m3)
V
(
)
N = 4.876 x 1025 molecules/m3
V
d= 290 pm = 2.90 x 10-10 m
Also, for N2, M = 2.80 x 10-2 kg/mol
-1
-1
Z = 4π(4.876 x 1025 m-3)(1.45 x 10-10 m)2 x (8.3145 J K mol )(301 K)
2π(2.80 x 10-2 kg/mol)
Z=
collisions/sec
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:
l=
l=
1
2 (N/V)(pd2)
1
2 ( 4.876 x 1025 )(p)(2.90 x 10-10 )2
= 5.49 x 10-8
Note that a nitrogen molecule only travels 0.5 nano meters before
it collides with another nitrogen atom.
Figure 5.22: Plots of PV/nRT versus P
for several gases (200 K).
Figure 5.23: Plots of PV/nRT versus P
for nitrogen gas at three temperatures.
Figure 5.24: (a) gas at low concentration relatively few interactions between particles
(b) gas at high concentration - many more
interactions between particles.
Figure 5.25:
Illustration of
pairwise
interactions
among gas
particles.
Table 5.3 (P172)
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
Values of Van der Waals
Constants for some Common Gases
2
atm
L
a
mol2
(
0.034
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46
)
b
L
( mol
)
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305
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).
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:
(2.000 atm)(20.0L)
n = PV =
= 1.663 mol
RT
(0.08206 Latm/molK)(293.15 K)
nRT
P=
V
(1.663 mol) (0.08206 Latm/molK) (423.15 K)
=
=
(1.000 L)
2a
nRT
n
P=
- 2
(V-nb) V
atm
=(1.663 mol) (0.08206 Latm/molK)(423.15 K) (1.00 L) - (1.663 mol)(0.0562)
(1.663 mol)2(6.49) = 63.699 - 17.948 =
(1.00 L)2
atm
Table 5.4(P173)
Component
N2
O2
Ar
CO2
Ne
He
CH4
Kr
H2
NO
Xe
Atmospheric Composition
near Sea Level (dry air)#
Mole Fraction
0.78084
0.20946
0.00934
0.000345
0.00001818
0.00000524
0.00000168
0.00000114
0.0000005
0.0000005
0.000000087
# The atmosphere contains various amounts of water vapor, depending on conditions.
Figure 5.27:
The variation of
temperature
and pressure
with altitude.
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
Figure 5.28: Concentration (in molecules per
million molecules of “air”) of some smog
components versus time of day.
Figure 5.29:
Our various
modes of
transportation
produce large
amounts of
nitrogen
oxides, which
facilitate the
formation of
photochemical
smog.
This photo was
taken in 1990.
Recent renovation
has since
replaced the
deteriorating
marble.
Figure 5.31: Diagram of the process
for scrubbing sulfur dioxide from stack
gases in power plants.