Emancipation Proclamation Issued (1862) HOMEWORK – DUE THURSDAY 9/24/15 (will accept Tuesday)

HW-BW 5.1 (Bookwork) CH 5 #’s 1, 14, 16, 20-24 all, 27, 32-37 all, 95, 103 HW-WS 8 (Worksheet) (from course website)

HOMEWORK – DUE TUESDAY 9/29/15 HW-BW 5.2 (Bookwork) CH 5 #’s 43, 46, 48, 50, 53, 54, 56, 58, 59, 62, 70,

73, 75, 96, 116, 118 HW-WS 9 (Worksheet) (from course website)

Lab Wednesday/Thursday – EXP 6 (2 day lab) -- PRE LAB!!! Nomenclature Quiz!!

SO32-

(aq) + MnO4−

(aq) → SO42-

(aq) + Mn2+(aq)

S2O32-

(aq) + Cl2(g) → SO42-

(aq) + Cl-(aq)

5 SO32-

(aq) + 2 MnO4−

(aq) + 6 H+(aq) → 5 SO4

2-(aq) + 2 Mn2+

(aq) + 3 H2O(l)

S2O32-

(aq) + 5 H2O(l) + 4 Cl2(g) → 2 SO42-

(aq) + 10 H+(aq) + 8 Cl-

(aq)

Cl2(g) → ClO-(aq) + Cl-

(aq)

Balance in basic conditions

H2O2(aq) + ClO2(aq) → ClO2-(aq) + O2(g)

H2O2(aq) + 2 OH-(aq) + 2 ClO2(aq) → 2 ClO2

-(aq) + O2(g) + 2 H2O(l)

Cl2(g) + 2 OH-(aq) → ClO-

(aq) + Cl-(aq) + H2O

GasIndefinite (variable) shape

Indefinite (variable) volume

Highly compressible

HUGE amounts of space

Highly disordered!!

Lots of KINETIC energy

Low relative density

PressurePressure is:

“Amount of force per unit area”

Force = mass x acceleration

F = M x A F = M x A

how much how fast

PressureUnits of standard pressure:

atmosphere: 1 atm =

millimeters of mercury: 760 mmHg

torr: 760 torr

kilopascal: 101.325 kPa

Pressure

Pressure depends on two things:

1) Energy of collisions

HEAT!!!

LOWER pressure HIGHER pressure

LOWER temperature HIGHER temperature

PressurePressure depends on two things:

2) Frequency of collisions

Effected by how fast the particles are moving

1) Energy of collisions

HEAT!!!

LOWER pressure HIGHER pressure

LOWER temperature HIGHER temperature

“Laws”

TEMPERATURE MUST BE IN KELVIN!!!!

Gay-Lussac’s Law – Temperature and Pressure

Temperature and pressure are directly proportional

= kP

T

1 2

1 2

P P

T T

assumes volume and # of particles are constant

P must remain constant

T

PressurePressure depends on two things:

2) Frequency of collisions

Effected by how fast the particles are moving

1) Energy of collisions

HEAT!!!

Effected by how many particles there are

LOWER pressure HIGHER pressure

FEWER particles MORE particles

“Laws”Avogadro’s Law – Particles and Volume

Particles and volume are directly proportional

= kV

n

1 2

1 2

V V

n n

assumes temperature and pressure are constant

V must remain constant

n

PressurePressure depends on two things:

2) Frequency of collisions

Effected by how fast the particles are moving

1) Energy of collisions

HEAT!!!

Effected by how many particles there are

Effected by how far the particles have to travel

LOWER pressure HIGHER pressure

LARGER volume SMALLER volume

“Laws”

P x V must remain constant

assumes temperature and # of particles are constant

Boyle’s Law – Pressure and Volume

Volume and pressure are inversely related

P1V1 = P2V2

PV = k

HIGHER temperature LOWER temperature

LARGER volume SMALLER volume

“Laws”

TEMPERATURE MUST BE IN KELVIN!!!!

Charles’s Law – Temperature and Volume

Temperature and volume are directly proportional

= kV

T

1 2

1 2

V V

T T

assumes pressure and # of particles are constant

V must remain constant

T

“Laws”Combined Gas Law – Pressure, Temperature and Volume

Combination of Boyle’s, Charles’s, and Gay-Lussac’s

assumes # of particles is constant

P1V1 = P2V2

1 2

1 2

V V

T T 1 2

1 2

P P

T T

1 1 2 2

1 2

P V P V

T T

“Laws”Combined Gas Law – Pressure, Temperature and Volume

Combination of Boyle’s, Charles’s, and Gay-Lussac’s

assumes # of particles is constant

1 1 2 2

1 2

P V P V

T T

“ “Laws” ”“Super Combined Gas Law” – Pressure, Temperature, Particles, and Volume

Combination of Boyle’s, Charles’s, Avogadro’s, and Gay-Lussac’s

1 1 2 2

1 1 2 2

P V P V

T n T n

“ “Laws” ”

1 1 2 2

1 1 2 2

P V P V

T n T n

P V

T n

P T

P n

V T

V n

You need to know how these variables are related

23315.0 11.7

298

Ka

t

Km atm

A steel container filled with nitrous oxide at 15.0 atm is cooled from 25.0 ºC to -40.0 ºC. What is the relationship between temperature and pressure? What is the final pressure.

Working with Gas Laws

T so P must

1 25.0 273 298T K 2 40.0 273 233T KTEMPERATURE MUST BE KELVIN!

233

298

298

233or

15.0 atm233

15.0298

aK

tm K

12

11

2 2PTT

P

V

V

1 1 2 2

1 2

P V P V

T T 1

211

2 2PTT

P

V

V

Working with Gas Laws

1 512P torr

2 1.417V L

1 25.5 T oC

2 1.875P atm

1 750.0V mL

2 ?T oC

11

512 0.67368760

P atm

torr atmtorr

2 3

11.417 1417

1 10V

mL L mL

L

1 25.5 273.15 298.65T oC K

12

1

2

12

V 1417

V 7

T 298.6P 1.875

P 0.6736T

58

5

0.0

atm

atm

mL K

mL2

21

12

1

P 1.875

P 0.67

V 141

368

7

V 750.

T 298.65T 1570.4

03

atm

mL

mat Lm

K K

2 1570.43 273.15T K2 1570.43 273.15 1297.28T oC K 32 1570.43 273.15 1297.28 1.30 10T o oC C K

A sample of an unknown gas has a volume of 750.0 mL and a pressure of 512 torr when its temperature is 25.5 oC. What will the temperature (oC) of the gas be if the volume is expanded to 1.417 L and the pressure is increased to 1.875 atm?

A sample of an unknown gas has a volume of 750.0 mL and a pressure of 512 torr when its temperature is 25.5 oC. What will the temperature (oC) of the gas be if the volume is expanded to 1.417 L and the pressure is increased to 1.875 atm?

Working with Gas Laws

1 512P torr

1 750.0V mL

1 25.5 T oC

2 1.875P atm

2 1.417V L

2 ?T oC1 25.5 273.15 298.65T oC K

2T 21.875

T 2912

.5

8 65 atm

K

torr21.875 760

51T 298.6

25

1

atm torr

torr K

atm21.875 1.417

7

760

51 50.T 298.65

2 1 0

atm torr

torr K

atm

L

mL 321.417 1

750.0 1 10

1.875 760T 298.6

1 15

5 2

atm torr

torr

L mL

atm mL K

L321.875 760 1.417 1

750.0 1 10T 298.65 1570.4

5 13

12

atm torr

torr a

L m

tm

L

mL L K K2T 298.65 K

2 1570.43 273.15T K2 1570.43 273.15 1297.28T oC K 32 1570.43 273.15 1297.28 1.30 10T o oC C K

P

V