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19.07.101 © Prof. Zvi C. Koren
6נושא
גזים
19.07.102 © Prof. Zvi C. Koren
Boyle
Charles
Gay-Lussac
AvogadroTorricelli
Kelvin
DaltonGraham
Maxwell Boltzmann
19.07.103 © Prof. Zvi C. Koren
Example:
An automobile air bag is rapidly inflated by the explosion of
sodium azide (NaN3), which releases nitrogen gas when detonated:
2 NaN3(s) 2Na(s) + 3N2(g)
• How much sodium azide is required to
inflate an air bag with a volume of 50.0 L?
• Does the amount of sodium azide
required depend on the temperature of the
gas produced?
• Does altitude affect an air bag's
performance?
Gas Laws:A Practical Application - Air Bags
You will be able to answer these questions at the end of this chapter.
19.07.104 © Prof. Zvi C. Koren
Air-Bag Chemistry
Products
Reactants
(Substances in red
are initially in
the air-bag)
Gas-Generator Reaction
extremely reactive
inert
Na(s)
N2(g)NaN3(s)
Initial Reaction Triggered by
Sensor.
reactive
reactive
inert
K2O(s)
Na2O(s)
N2
Na
KNO3(s)Second Reaction.
inert
Alkaline
silicate
(glass)
K2O
Na2O
SiO2(s)
Final Reaction.
Write balanced chemical reactions for each of these steps.
The signal from the deceleration sensor ignites the gas-generator mixture by an electrical
impulse, creating the high-temperature condition necessary for NaN3 to decompose.
Air-Bag Chemistry
19.07.105 © Prof. Zvi C. Koren
The Mercury Barometer: Measuring Atmospheric Pressure
Evangelista Torricelli (1608 – 1647, Italy):
“We live submerged at the bottom of an ocean of air.”
Aristotle (4th century BCE): “Nature abhors a vacuum.”
Properties of Gases:
Gas Pressures, P
19.07.106 © Prof. Zvi C. Koren
Patm
The Mercury Barometer: Measuring Atmospheric Pressure
Evangelista Torricelli
vacuum
PHg
meniscus
1 standard atmosphere is supported by a
mercury column that is 760 mm Hg in height:
The atmospheric pressure
(or barometric) pressure
can be described in terms of “mm Hg”
1 torr 1 mm Hg
1 atm 760 mm Hg
760 mm Hg 600 mm Hg
19.07.107 © Prof. Zvi C. Koren
Pressure Units
dgh V
mgh
h
h
A
mg
A
w
A
f P
P = pressure
f = force
A = area
w = weight
m = mass
g = acceleration due to gravity = 9.80665 m/s2
h = height of column
d = density = m/V
P = d·g·h
For the mercury barometer (at 0 oC):
dHg = 13.5951 g/cm3 = 0.0135951 kg/cm3 x (102 cm / 1 m)3 = 1.35951 x 104 kg/m3
h = 760 mm Hg = 76 cm Hg = 0.76 m Hg
P = d · g · h
1 atm = (1.35951x104 kg/m3)·(9.80665 m/s2)·(0.76 m) = 1.01325x105 kg/m·s2
= 1.01325 x 105 Pa
= 101.325 kPa
= 1.01325 bar
= 14.6960 lb/in2 (“PSI”, Pounds per Square Inch)
760 (exactly) mm Hg
SI (and other) Units
Energy: 1 J = 1 kg·m2/s2
Force: 1 N = 1 kg·m/s2 (“kms”)
Pressure: 1 Pa = 1 N/m2 = 1 kg/m·s2
1 bar = 105 Pa
1 torr = 1 mm Hg
J = Joule, N = Newton, Pa = Pascal
19.07.108 © Prof. Zvi C. Koren
The Mercury Manometer: Measuring Gas Pressure
OPENManometer
Pgas = Patm + PHg
Pgas PH
g
CLOSEDManometer
Pgas = PHg
vacuum
Pgas
Patm
PH
g
gas gas
19.07.109 © Prof. Zvi C. Koren
Chemists use four basic measurements when
working with gases:
1. The quantity of the gas, n (in moles)
2. The temperature of the gas, T (in kelvins)
3. The volume of gas, V (in liters)
4. The pressure of the gas, P (in atmospheres)
• Changes in gas properties have been mathematically modeled for many years.
• These mathematical models are collectively called gas laws.
• They define the relationships between pairs of gas properties
and experiments confirm these gas laws.
Gas Laws
19.07.1010 © Prof. Zvi C. Koren
0
2
4
6
8
10
12
14
16
18
20
0 5 10 15 20 25 30 35
P
V
P • V = k or P = k •V–1 , [n,T]
P V-1, [n,T]
Boyle’s Law
hyperbolaP1V1 = P2V2 , [n,T]
gas
piston
Robert Boyle (1627 – 1691)
England
19.07.1011 © Prof. Zvi C. Koren
V
T
V = k T, [n,P]
V T, [n,P]
Charles’s and Gay-Lussac’s Law
k T
V
2
2
1
1
T
V
T
V
-273 oC
, [n,P]
, [n,P]
“Absolute
Zero”
extrapolation
Absolute temperature scale is in kelvins: T(K) = t(0C) + 273.15
Jacques Alexandre CésarCharles
(1746-1823)
Joseph LouisGay-Lussac(1778 – 1850)
William Thomson“Lord Kelvin”
(1824 -1907), Irish
The title “Baron Kelvin” was given in honor of his achievements, and named
after the River Kelvin, which flowed past his university in Glasgow, Scotland.
19.07.1012 © Prof. Zvi C. Koren
V n , [P,T]
Avogadro’s Law
V
n
V = k n , [P,T] k n
V
2
2
1
1
n
V
n
V , [P,T]
, [P,T]
Lorenzo Romano Amedeo Carlo Avogadro di Quareqa e di Carreto (1776 – 1854)
19.07.1013 © Prof. Zvi C. Koren
Combining all three gas laws:
P V
n
= k, [n,T]Boyle:
T+ Charles:
+ Avogadro:
P V= k, [n]
TP V
= k
At STP (Standard Temperature and Pressure), 0oC (= 273.15 K) & 1 atm:
1 mole of an ideal gas occupies 22.414 L of volume.
KmolatmLKmol
Latm
/ 0.082057
) 15.273() 1(
) 414.22() 1(
Tn
VP R
PV = nRT
R
= 8.31 J/mol·K
= 1.99 cal/mol·K
“Gas Constant”
Ideal Gas Law
= [energy/mol·K]
19.07.10
mole
Mole Day is celebrated on Oct. 23rd from 6:02 am to 6:02 pm
© Prof. Zvi C. Koren14 Happy Mole Day to You!!!
19.07.1015 © Prof. Zvi C. Koren
mole
Summary: Moles in Chemistry
# of items: Avogadro’s #
Mass: m/MW
Solution: MV
Gas: PV = nRT
19.07.1016 © Prof. Zvi C. Koren
Problem:
Potassium superoxide is used to purify air in a spacecraft.
What mass of KO2 is required to recycle 500.0 mL of CO2 at 25.00 oC and 750. mmHg?
4KO2(s) + 2CO2(g) 2K2CO3(s) + 3O2(g)
Solution:
Stoichiometric Flow-Chart:
RT
PV n
2CO 750. mmHg
Hgmm
atm
760
1 = 0.98684 atmP =
V = L500.0 mL
mL
L
1000
1 = 0.5000
25.00 + 273.15 = 298.15 K
2COn = 0.020168
0.020168 mol CO2
2
2
CO 2
KO 4
mol
mol
2
2
KO 1
KO 71.0971
mol
g= 2.87 g KO2
R = 0.082057 L·atm/mol·K
T =
? MWmass KO2 moles KO2 rxn PV=nRT
moles CO2 (V, T, P) of CO2
Gas Laws and Chemical Reactions: Stoichiometry
19.07.1017 © Prof. Zvi C. Koren
Ptotal = ΣPi = P1 + P2 + P3 + …
= n1RT/V + n2RT/V + n3RT/V + …
= RT/V
1
1
1
1
1
2
2
2 2
2
2
2 3
3
3
(n1 + n2 + n3 + …)
V
RTn P total
total
Pi = partial pressure of gas i
= niRT/V
RT/Vn
RT/Vn
P
P
total
i
total
i n
n
total
i Xi mole fraction of gas i ΣXi = 1
Pi = Xi•Ptotal
= ntotal RT/V
Gas Mixtures and Dalton’s Law of Partial Pressures
John Dalton (1766-1844), England
19.07.1018 © Prof. Zvi C. Koren
PV = nRTFrom:
RT
PV n
MW
m
MWRT
P
V
m
MWRT
P d d MW, [P,T]
For example:
)/ 00.32() )(273 (0.0821
1 STP @ d
2O molgKKatm/molL
atm
= 1.43 g/L
Gas Density
19.07.1019 © Prof. Zvi C. Koren
Determining the Molecular Weight of a Volatile Liquid
vent Measurements:
Mass of vapor = 1.2189 g
(measurements of empty and filled flask)
Volume of flask = 510.0 mL
(measured with water)
Temperature of vapor = 100.0 oC
Atmospheric pressure = 745 torr
MW = ?
Dumas Method
19.07.1020 © Prof. Zvi C. Koren
19.07.1021 © Prof. Zvi C. Koren
Ideal Gas Properties:
• No intermolecular forces
• Molecular volume is negligible
Tenets of KMT:
1. Gases consist of molecules whose separation is much larger than the volume of the
molecules themselves.
2. The gas molecules are in continuous random and rapid motion.
3. The average kinetic energy of the gas molecules is determined by the absolute gas
temperature.
Notes:
All gas molecules at the same T, regardless of mass, have the
same average kinetic energy.
Temperature is a bulk (macroscopic) property, and not one
based on an individual (microscopic) molecule)
4. Gas molecules collide with each other and with the walls of the container, but they
do so without loss of energy. These collisions are perfectly elastic.
Kinetic Molecular Theory (KMT) of Gases
19.07.1022 © Prof. Zvi C. Koren
KEmole3(½ RT) = = ½Mv2
1/21/2
M
3RT )(
2v
____
vrms = root-mean-square velocity (m/s)
R = (J/mol·K)
MW = (kg/mol)
Average Kinetic Energy for one mole KEmole T
Kinetic Energy for one molecule KEmolecule = ½mv2, m mass of one molecule
Kinetic Energy for one mole KEmole = ½Mv2, M MW, Molecular weight
Average Kinetic Energies, KE
Average Kinetic Energy for one mole KEmole = ½Mv2,
But also:
= 3(½ RT)
proof is not for this course
Combine these two
Graham’s Law
of Diffusion
for 2 Gases
at same T:
v1/v2 = (M2/M1)½
Thomas Graham(1805 – 1869), Scotland
19.07.1023 © Prof. Zvi C. Koren
Applications of KMT
1. Diffusion of NH3(g) vs. HCl(g)
19.07.1024 © Prof. Zvi C. Koren
2. Separation of Uranium Isotopes
19.07.1025 © Prof. Zvi C. Koren
Ludwig Eduard Boltzmann
(1844 – 1906), Austria James Clerk Maxwell
(1831 – 1879), Scotland
Maxwell-Boltzmann Distribution of Molecular Speeds
= f (T, MW)
Why do molecules, all at the same T,
have such a wide span of velocities?
19.07.1026 © Prof. Zvi C. Koren
Maxwell-Boltzmann Distribution of Molecular Velocities
MW Effects:
> >
(continued)
19.07.1027 © Prof. Zvi C. Koren
Temperature Effects 1:
(continued)
19.07.1028 © Prof. Zvi C. Koren
Temperature Effects 2:
19.07.1029 © Prof. Zvi C. Koren
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