1

Slide 20-1

Chem 1101

examinable

A/Prof Sébastien PerrierRoom: 351

Phone: 9351-3366

Email: s.perrier@chem.usyd.edu.au

Prof Scott KableRoom: 311

Phone: 9351-2756

Email: s.kable@chem.usyd.edu.au

A/Prof Adam BridgemanRoom: 222

Phone: 9351-2731

Email: a.bridgeman@chem.usyd.edu.au Slide 20-2

Where are we going…?

� Gases� Properties, molecular picture

� Thermochemistry� Fuels, environmental effects of fuel

combustion

� Equilibrium� Gas phase, acid-base, solubility

� Electrochemistry� Galvanic and concentration cells

� Intermolecular Forces� Polymers, biomolecules

5 main topics…

� Nitrogen chemistry

� e.g. smog, explosives

� Sources of chemicals

� e.g. ammonia, polymers

� Batteries, Electrolysis

� e.g. corrosion, fuel cells

3 interleaving applications…

for interest

Slide 20-3

Lecture resources

� Access from WebCT

or

� http://www.kcpc.usyd.edu.au/CHEM1101.html

� Login: chem1101� Password: carbon12

Slide 20-4

Gases

Blackman, Bottle, Schmid, Mocerino & Wille, Chapter 6

Blackman Figure 16.8

Slide 20-5

Boyle’s Law

The volume of a gas is inversely proportional to its pressure

Figure 6.5 Blackman Slide 20-6

Charles’ Law

The volume of a gas is directly proportional to its Kelvin temperature

Figure 6.6 Blackman

2

Slide 20-7

Avogadro’s Law

Equal volumes of gas at the same pressure and temperature contain equal numbers of molecules

Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto

Slide 20-8

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:� equal volumes of gas at the same pressure and temperature contain equal numbers of molecules

V=k3 n

Empirical Gas Laws

Slide 20-9

P V = n R T

Ideal Gas Equation

pressure

volumenumbermoles

absolutetemperature

gasconstant

Universal gas constant R = 8.314 J mol–1 K–1

(R has different values with different units – see later)

P

nTkV ×=

Slide 20-10

Ideal Gas Equation

Slide 20-11

Common Units of Pressure

1.01325 barbar

only used by primitives…

14.7 lb in–2pounds/sq inch (psi)

760 torrmm mercury = torr

1 atmatmosphere

SI unit1.01325×105 Pa; 101.325 kPa; 1013.25 hPa

pascal (Pa); kilopascal (kPa) ; hectopascal (hPa)

atmospheric pressure

unit

Slide 20-12

R = 8.314 J mol–1 K–1

= 8.314 Pa m3 mol–1 K–1

= 8.314 kPa L mol–1 K–1

R in Different Units

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.

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)

3

Slide 20-13

1 atmosphere = 101.3 kPa

⇒ R = 8.314/101.3 atm L mol–1 K–1

= 0.08206 atm L mol–1 K–1

R in Different Units cont’d

Slide 20-14

Often made with sodium azide:6NaN3(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) + 9N2(g)

Example: Car Air Bags

Example: How many grams of NaN3 are required to

produce 75 L of nitrogen at 25 °C and 1.31 atm?

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

P V = n R T n = P V / (R T)

Slide 20-15

� 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

� The density of the liquid is 100-1000 times greater than the gas state.

� The density of the solid is usually ∼10% greaterthan the liquid.

Gas Densities

n = P V / (R T)

Slide 20-16

Find the density of oxygen gas at 25 °C and 0.85 atm

Density (ρ) = m/V = M n / VM = Molar mass

So ρ = M P / RT (from PV = nRT)

L atm mol–1 K–1

g mol–1

(check units balance!)

Example

g/L 1.1298082.0

85.00.32 O ofDensity 2 =

×

×=

atm

273+25 K

Slide 20-17

The Molecular Picture(or Kinetic Theory of Gases)

The Molecular Picture

or

Kinetic Theory of Gases

Slide 20-18

Boyle’s Law

Figure 6.11 Blackman

4

Slide 20-19

Molecular Picture of Boyle’s Law

V increases

⇒⇒⇒⇒ P decreasesSlide 20-20

Charles’ Law

Figure 6.12 Blackman

Slide 20-21

Molecule Picture of Charles’ Law

T increases

⇒⇒⇒⇒ P increasesSlide 20-22

Temperature Dependence

Blackman 6.9

Slide 20-23

Kinetic energy Ek = 1/2 mv2

∝ T

= constant at fixed T

Pressure P ∝ Ek = 1/2 mv2

∝ Τ

= same for all gases

Speeds of Different Molecules

Blackman Figure 6.8 Slide 20-24

Gas Behaviour at STP

He N2 O2

Standard Molar Volume

(STP = Standard Temperature and Pressure)

5

Slide 20-25

Dalton’s Law of Partial Pressure

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."

So far we have considered pure gases.

What happens when we combine them?

...++= batot PPP

where a, b, … are different gases

Slide 20-26

Dalton’s Law of Partial Pressures

Consider 2 vessels with different gases…

What is the total pressure, P, and the partial pressures of A and B?

Depress the piston and force gas A into the other cylinder.

Assume equal

cylinder volumes

Slide 20-27

Dalton’s Law of Partial Pressures

The number of moles of each gas in the cylinder is trivial…

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 = ΧΧΧΧA Ptot

Slide 20-28

� Boyle’s Law: PV = constant for constant temperature

� Charles’ Law: V/T = constant for constant pressure

� 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!

Implications of PV = nRT

Slide 20-29

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 20-30

6

Slide 20-31

PV = nRT from Kinetic Theory

Pressure arises from the force experienced on

vessel wall by gas molecules bouncing off it

L Square of side L

– momentum of gas molecule = mass × velocity = mv– change in momentum from bouncing off wall: ∆∆∆∆p = 2mv

Beyond examinable:

Slide 20-32

no. collisions/second with wall = ½ N v L2

22

1 L N 2m

incrementtime

momentuminchange

dt

mdforce

vv

v

×=

==

N m area

forcepressure 2

v==

half molecules

moving towards wallno. molecules

/ unit volume

area

of

wall

Avogadro constantno. moles

N = no. molecules/unit volume = n NA

/ V

time increment ∆t = 1/(no. collisions/second)

PV = nRT from Kinetic Theory

Slide 20-33

define temperature T ∝ average energy = ½ mv2 by

T = NA m v2 / R

(defines absolute temperature scale in terms of molecular energy)

Learn

equation,

not derivation

P V = n NA m v2

P V = n RT

PV = nRT from Kinetic Theory