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