Chem 1101
Where are we going…?
5 main topics…
A/Prof Sébastien Perrier
Room: 351
Gases
Phone: 9351-3366
Properties, molecular picture
Fuels, environmental effects of fuel
combustion
examinable
Prof Scott Kable
Room: 311
A/Prof Adam Bridgeman
Electrochemistry
e.g. ammonia, polymers
Galvanic and concentration cells
Room: 222
Phone: 9351-2731
Email: [email protected] 20-1
Lecture resources
Batteries, Electrolysis
Intermolecular Forces
e.g. corrosion, fuel cells
Polymers, biomolecules
Slide 20-2
Gases
Access from WebCT
Blackman, Bottle, Schmid,
Mocerino & Wille, Chapter 6
or
e.g. smog, explosives
Sources of chemicals
Email: [email protected]
Nitrogen chemistry
Equilibrium
Gas phase, acid-base, solubility
Phone: 9351-2756
for
interest
Thermochemistry
Email: [email protected]
3 interleaving
applications…
http://www.kcpc.usyd.edu.au/CHEM1101.html
Login: chem1101
Password: carbon12
Slide 20-3
Charles’ Law
Boyle’s Law
The volume of a gas is inversely proportional to its pressure
Figure 6.5 Blackman
Slide 20-4
Blackman Figure 16.8
Slide 20-5
The volume of a gas is directly proportional to its Kelvin temperature
Figure 6.6 Blackman
Slide 20-6
1
Avogadro’s Law
Empirical Gas Laws
Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto
Equal volumes of gas at the same pressure and
temperature contain equal numbers of molecules
Boyle’s Law:
the volume of a gas is inversely proportional to its
pressure
V=k1/P
Charles’ Law:
the volume of a gas is directly proportional to its
Kelvin temperature
V=k2T
Avogadro’s Law:
Slide 20-7
Ideal Gas Equation
V =k×
volume
V=k3 n
Slide 20-8
Ideal Gas Equation
nT
P
P V = n R T
pressure
equal volumes of gas at the same pressure and
temperature contain equal numbers of molecules
number
moles
absolute
temperature
gas
constant
Universal gas constant R = 8.314 J mol–1 K–1
(R has different values with different units – see later)
Slide 20-9
Common Units of Pressure
atmospheric
pressure
pascal (Pa);
1.01325×105 Pa;
kilopascal (kPa) ; 101.325 kPa;
hectopascal (hPa) 1013.25 hPa
atmosphere
1 atm
SI unit
1.01325 bar
Energy = J = kg m2 s–2 (eg E = ½ mv2)
Pressure = Pa = force/area = kg m-1 s–2 (force =
kg m s–2)
⇒ Pa m3 has units kg m2 s–2, same as J.
= 8.314 Pa m3 mol–1 K–1
760 torr
14.7 lb in–2
R in Different Units
R = 8.314 J mol–1 K–1
unit
mm mercury =
torr
pounds/sq inch
(psi)
bar
Slide 20-10
only used by
primitives…
Slide 20-11
1 kPa = 103 Pa, 1 L = 10–3 m3, ∴1 kPa L = 1 Pa m3
i.e. we can use P in kPa, V in L (= dm3)
= 8.314 kPa L mol–1 K–1
Slide 20-12
2
R in Different Units cont’d
1 atmosphere = 101.3 kPa
Example: Car Air Bags
Often made with sodium azide:
⇒ R = 8.314/101.3 atm L mol–1 K–1
= 0.08206 atm L mol–1 K–1
6NaN3(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) + 9N2(g)
Example: How many grams of NaN3 are required to
produce 75 L of nitrogen at 25 °C and 1.31 atm?
P V = n R T
n = P V / (R T)
example
n = 1.31 atm x 75 L / (0.082 x 298 K) = 4.0 mol N2
Stoichiometry => 6/9 x 4 = 2.67 mol NaN3
Mass NaN3 = 2.67 x 65.0 = 174 g
Slide 20-13
Gas Densities
Slide 20-14
Example
Gases have low densities because the molecules
are far apart.
For example:
N2 (g) at
20 °C, ρ = 1.25 g/L
N2 (l) at –196 °C, ρ = 808 g/L
N2 (s) at –210 °C, ρ = 1030 g/L
Find the density of oxygen gas at 25 °C and 0.85 atm
Density (ρ) = m/V = M n / V
M = Molar mass
So ρ = M P / RT (from PV = nRT)
n = P V / (R T)
The density of the liquid is 100-1000 times
greater than the gas state.
g mol–1
atm
32.0 × 0.85
Density of O 2 =
= 1.1 g/L
0.082 × 298
L atm mol–1 K–1 273+25 K
The density of the solid is usually ∼10% greater
than the liquid.
Slide 20-15
The Molecular Picture
(check units balance!)
Slide 20-16
Boyle’s Law
(or Kinetic Theory of Gases)
The Molecular Picture
or
Kinetic Theory of Gases
Figure 6.11 Blackman
Slide 20-17
Slide 20-18
3
Molecular Picture of Boyle’s Law
V increases
Charles’ Law
Figure 6.12 Blackman
⇒ P decreases
Slide 20-19
Molecule Picture of Charles’ Law
Slide 20-20
Temperature Dependence
T increases
⇒ P increases
Slide 20-21
Speeds of Different Molecules
Blackman 6.9
Slide 20-22
Gas Behaviour at STP
(STP = Standard Temperature and Pressure)
Kinetic energy Ek = 1/2 mv2
∝T
= constant at fixed T
Pressure P
Blackman Figure 6.8
∝ Ek = 1/2 mv2
∝Τ
= same for all gases
Slide 20-23
Standard Molar Volume
He
N2
O2
Slide 20-24
4
Dalton’s Law of Partial Pressure
Dalton’s Law of Partial Pressures
Consider 2 vessels with different gases…
Depress the piston and force gas A into the other cylinder.
So far we have considered pure gases.
What happens when we combine them?
Dalton’s Law: “The total
pressure of a mixture of
gases equals the sum of the
pressures that each would
exert if it were present
alone."
Assume
equal
cylinder
volumes
Ptot = Pa + Pb + ...
where a, b, … are different gases
What is the total pressure, P, and the partial pressures of A and B?
Slide 20-25
Slide 20-26
Dalton’s Law of Partial Pressures
The number of moles of each gas in the cylinder is trivial…
Implications of PV = nRT
Boyle’s Law: PV = constant for constant temperature
Charles’ Law: V/T = constant for constant pressure
ntot = nA + nB = 0.3 + 0.6 = 0.9 mol
The mole fraction of A is ΧA = 0.3/0.9 = 0.33
The mole fraction of B is ΧB = 0.6/0.9 = 0.67
Note the relationship between mole
fraction and partial pressure:
pA = ΧA Ptot
Slide 20-27
Lowest temperature: -273 ºC (ie V = 0 at this
temperature)
Gay-Lussac: (combining volumes)
Avogadro’s hypothesis: (equal volumes of gas have the
same number of molecules)
Avogadro’s constant: NA = # molecules/mol = 6.02 ×
1023 mol–1
Dalton’s law of partial pressures (see later)
Diverse observations from single unifying theory:
one of the goals of science!
Slide 20-28
Example questions
CONCEPTS
Classical behaviour of gases at different pressure,
volume, temperatures and moles
Boyle’s Law, Charles’ Law, Avagadro Law
CALCULATIONS
PV=nRT
Gas densities
Slide 20-29
Slide 20-30
5
PV = nRT from Kinetic Theory
Beyond examinable:
no. collisions/second with wall
Pressure arises from the force experienced on
vessel wall by gas molecules bouncing off it
half molecules
moving towards wall
= ½ N v L2
area
of
wall
no. molecules
/ unit volume
time increment ∆t = 1/(no. collisions/second)
force =
Square of side L
L
PV = nRT from Kinetic Theory
dmv changein momentum
=
timeincrement
dt
= 2mv × 12 N v L2
– momentum of gas molecule = mass × velocity = mv
– change in momentum from bouncing off wall: ∆p = 2mv
Slide 20-31
pressure=
force
= m v2 N
area
no. moles
Avogadro constant
N = no. molecules/unit volume = n NA / V
Slide 20-32
PV = nRT from Kinetic Theory
P V = n NA m v2
define temperature T ∝ average energy = ½ mv2 by
T = NA m v2 / R
(defines absolute temperature scale in terms of molecular energy)
P V = n RT
Learn
equation,
not derivation
Slide 20-33
6