Chapter 13 - States of Matter
Kinetic Theory and the Nature of Glass
All matter is composed of tiny particles in constant motion .
For Gases: (PME)
1. Made up of particles.
2. Particles moving rapidly in constant, random motion.
3. All collisions are perfectly elastic. (No energy lost.)
Oxygen trivias: (Gas Trivias)
1. Diameter of O
2 molecule:
0.339 nm = .000000000339 meters!
(very small)
2. Average distance traveled between collisions: 106 nm
(Mean free path)
Travels over 300x its own diameter before it hits something else.
(mostly empty space)

Oxygen trivias: (Gas Trivias)
1. Diameter of O
2 molecule:
.339 nm = .000000000339 meters! (very small)
2. Average distance traveled between collisions: 106 nm
(Mean free path)
Travels over 300x its own diameter before it hits something else.
(mostly empty space)
3. Average velocity of 0
2
443m/s (~1000mph!)
(very fast)
4. Frequency of collisions 4.5 billon collisions per second
(A lot of collisions)

Gas Particles
Gas pressure is caused by billions of gas particles colliding with an object at the same time.
- When no particles are present  empty space  vacuum
Atmospheric pressure - gas particles in the air colliding with objects.
Standard atmospheric pressure (at 0 ° C): 760 mm of Hg
Or
(Pascal is the SI unit of pressure.) 1 atmosphere
Or
101.3 kPa
Barometer – device used to measure atmospheric pressure.

Ex 1) Convert 1400 mm Hg to kPa & atm.
1400mm Hg
=
1400mm Hg
=

Ex 1) Convert 1400 mm Hg to kPa & atm.
1400mm Hg kPa mm Hg
= kPa
1400mm Hg
= atm

Heat something temp goes up energy goes into speeding up the particles (increases KE)
Temperature - measure of average Kinetic Energy
Thermometer
What would happen if cooled thermometer until there was no molecular motion?
Absolute zero – temp at which there’s no molecular motion.
0 Kelvin ( - 273 o C) (not achievable)
Outer space 2 – 4 K, Lab conditions 0.0000001 K
At 200 K, gas particles have twice the KE of these at 100K.
Ch13 HW#1 1 - 5

Ch 13 HW#1 1-5
1.Convert 190 mm Hg to a) kPa b)atm
2. Mt. Everest pressure is 33.7 kPa. Greater or less than .25atm?
3. Explain relationship between absolute temp & KE of particles.

Ch 13 HW#1 1-5
1.Convert 190 mm Hg to a) kPa b)atm a)190 mm Hg 101.3 kPa = 25.1 kPa
760 mm Hg b)190 mm Hg 1 atm = .25 ATM
760 mm Hg
2. Mt. Everest pressure is 33.7 kPa. Greater or less than .25atm?
3. Explain relationship between absolute temp & KE of particles.

Ch 13 HW#1 1-5
1.Convert 190 mm Hg to a) kPa b)atm a)190 mm Hg 101.3 kPa = 25.1 kPa
760 mm Hg b)190 mm Hg 1 atm = .25 ATM
760 mm Hg
2. Mt. Everest pressure is 33.7 kPa. Greater or less than .25atm?
33.7 kPa 1 atm = .33 ATM
101.3 kPa
3. Explain relationship between absolute temp & KE of particles.

Ch 13 HW#1 1-5
1.Convert 190 mm Hg to a) kPa b)atm a)190 mm Hg 101.3 kPa = 25.1 kPa
760 mm Hg b)190 mm Hg 1 atm = .25 ATM
760 mm Hg
2. Mt. Everest pressure is 33.7 kPa. Greater or less than .25atm?
33.7 kPa 1 atm = .33 ATM
101.3 kPa
3. Explain relationship between absolute temp & KE of particles.
Directly related as temp goes up, so does KE.
(From 100K to 200K (temp doubled), the KE doubles)

4. How does average KE of H
2
0 affected as you pour hot into equal amount of cold H
2
0?
5. By what factor does ave KE go up as gas increases from 300K to 900K?

4. How does average KE of H
2
0 affected as you pour hot into equal amount of cold H
2
0?
Hot cools as cold warms as they reach middle temp.
Average stays constant.
5. By what factor does ave KE go up as gas increases from 300K to 900K?

4. How does average KE of H
2
0 affected as you pour hot into equal amount of cold H
2
0?
Hot cools as cold warms as they reach middle temp.
Average stays constant.
5. By what factor does ave KE go up as gas increases from 300K to 900K?
3X

Ch13.2 - The Nature of Liquids
+
Particles in liquids are free to move around.
- They have enough KE that the intermolecular forces are not strong enough to form bonds and make a structure.
Don’t have enough KE to be completely free of each other.

Evaporation
- occurs at the surface of the liquid
- liquid to gas
- faster at higher temps because more KE.
- each evaporating particle has to gain enough energy to be able to break the intermolecular forces holding it.

Evaporation
- occurs at the surface of the liquid
- liquid to gas
- faster at higher temps because more KE.
- each evaporating particle has to gain enough energy to be able to break the intermolecular forces holding it.
- the particles that leave first during evaporation are the first ones – high KE (high temp)
The ones left are slower – low KE(low temp) so evaporation causes the liquid to become cooler.
- cooling process

Vapor pressure
As more particles evaporate, creates a pressure.
- each liquid has a unique pressure for each temp.
- slower particles can condense back to liquid.
Dynamic equilibrium H
2
0
(l)
H
2
0
(g)
(both condensation and evaporation at same time.)

Boiling Point of Liquid
Boiling Point special point at which the vapor pressure of the liquid is equal to the external pressure.
- Normal boiling point - temp at which a liquid boils at standard pressure, 101.3kPa.
(Water is 100 °C at 101.3kPa.)
100
80
- the temp of a liquid never rises above its boiling point temp. chloroform water
VP 60
(kPa)40
20 ethanol ethanoic acid
20 40 60 80 100 120
Temp ( °C)
FIRE
Ch13 HW#2 6-9

Ch13 HW#2 6 – 9
6. Explain vapor pressure.
7. What happens when v.p. reaches ATM pressure?
8. What is the boiling point of water?
9. a)B.P. of water at 60kPa: ____ b)B.P. of water at 80kPa: ____ c)What pressure water boil at 60 °C:___ d)For each liquid, what pressure is required to boil at 60 °C:

Ch13 HW#2 6 – 9
6. Explain vapor pressure.
As the substance evaporates, it creates a pressure above it.
7. What happens when v.p. reaches ATM pressure?
8. What is the boiling point of water?
9. a)B.P. of water at 60kPa: ____ b)B.P. of water at 80kPa: ____ c)What pressure water boil at 60 °C:___ d)For each liquid, what pressure is required to boil at 60 °C:

Ch13 HW#2 6 – 9
6. Explain vapor pressure.
As the substance evaporates, it creates a pressure above it.
7. What happens when v.p. reaches ATM pressure?
The substance starts to boil.
8. What is the boiling point of water?
9. a)B.P. of water at 60kPa: ____ b)B.P. of water at 80kPa: ____ c)What pressure water boil at 60 °C:___ d)For each liquid, what pressure is required to boil at 60 °C:

Ch13 HW#2 6 – 9
6. Explain vapor pressure.
As the substance evaporates, it creates a pressure above it.
7. What happens when v.p. reaches ATM pressure?
The substance starts to boil.
8. What is the boiling point of water?
It depends on the pressure! At standard atm pressure, its 100 °C.
100
80 chloroform water
VP 60
(kPa)40 ethanol ethanoic acid
20
20 40 60 80 100 120
Temp ( °C)
9. a)B.P. of water at 60kPa: ____ b)B.P. of water at 80kPa: ____ c)What pressure water boil at 60 °C:___ d)For each liquid, what pressure is required to boil at 60 °C: chloroform : ____ water :____ ethanol :____ ethanoic acid:____

Ch13 HW#2 6 – 9
6. Explain vapor pressure.
As the substance evaporates, it creates a pressure above it.
7. What happens when v.p. reaches ATM pressure?
The substance starts to boil.
8. What is the boiling point of water?
It depends on the pressure! At standard atm pressure, its 100 °C.
100
80 chloroform water
VP 60
(kPa)40 ethanol ethanoic acid
20
20 40 60 80 100 120
Temp ( °C)
9. a)B.P. of water at 60kPa: (90 °C) b)B.P. of water at 80kPa: (95 °C) c)What pressure water boil at 60 °C: (20kPa) d)For each liquid, what pressure is required to boil at 60 °C: chloroform : (100kPa) ethanol : (50kPa) water : (20kPa) ethanoic acid: (10kPa)

Ch13.3 – The Nature of Solids
- particles in solids are packed together in highly organized
3-D patterns. Call them crystalline solids.
- vibrate about fixed positions
- when heat a solid, vibrate more rapidly, break structure, melts.
- ionic solids have high melting points - stronger bonds.
NaCl 801 °C
- molecular solids (covalent bonds) - weaker bonds, lower melting points.
- some solids decompose instead of melt. (like wood)

Crystals have different structures based on the sizes of the particles bonding. (table of crystals pg398.)
- crystals have repeating patterns. The simplest repeating pattern is called the unit cell.
Simple Cubic (usually only 1 type of atom)
Atoms located at corners of a cube.
Body Centered Cube (usually only 1 type of ions)
2 types of atoms are about the same size, so crystal forms shape where 1 type is at corners of cube, other is located in center.
Face Centered Cubic (usually 2 types of atoms)
1 is a lot larger than other , small ones can fill the holes between large ones on faces of cube. Because of closeness of atoms, strong bonds, high m.p.
Ex: NaCl

Allotropes – different structures of the same element.
Ex: Carbon has diamond and graphite
Not all solids have a crystalline form.
Amorphous solid – lack an ordered internal structure.
Exs: glass, butter
Sometimes referred to as super-cooled liquids.
Sublimation – solid that goes directly to the gaseous phase.
Exs: dry ice, moth balls, popsicles, solid air fresheners.

Phase Diagram - Water
22,100
Pressure
(kPa)
101.3
Solid
Liquid
Gas
0
Temperature (C °)
100 374

Phase Diagram - Water
22,100
Pressure
(kPa)
101.3
Solid
Liquid
Critical Point
Triple Point
0
Temperature (C °)
Gas
100 374
Critical point for water – At 374°C and 22,100kPa is the last point where water can still be condensed to a liquid.
Triple point – all 3 phases are in equilibrium

Phase Diagram - Water
22,100
Pressure
(kPa)
101.3
Solid
Liquid
Critical Point
Triple Point
0
Temperature (C °)
Gas
100 374
Ex: What is the physical state of H
2
O at 50 °C and 70 kPa?_____
Name a temp and pressure that H
2
O boils at: _____________
Ch13 HW#3 10 – 15

Ch13 HW#3 10 – 15
10) Name a physical property to distinguish molecular solid from ionic.
11) Substance goes directly from solid to gas: _________
12) Popsicle in freezer for long time, wrapper sticky - Why?
13) Triple point:

Ch13 HW#3 10 – 15
10) Name a physical property to distinguish molecular solid from ionic.
melting point
11) Substance goes directly from solid to gas: _________
12) Popsicle in freezer for long time, wrapper sticky - Why?
13) Triple point:

Ch13 HW#3 10 – 15
10) Name a physical property to distinguish molecular solid from ionic.
melting point
11) Substance goes directly from solid to gas: sublimation
12) Popsicle in freezer for long time, wrapper sticky - Why?
13) Triple point:

Ch13 HW#3 10 – 15
10) Name a physical property to distinguish molecular solid from ionic.
melting point
11) Substance goes directly from solid to gas: sublimation
12) Popsicle in freezer for long time, wrapper sticky - Why?
The ice in the popsicle sublimes
13) Triple point:

Ch13 HW#3 10 – 15
10) Name a physical property to distinguish molecular solid from ionic.
melting point
11) Substance goes directly from solid to gas: sublimation
12) Popsicle in freezer for long time, wrapper sticky - Why?
The ice in the popsicle sublimes
13) Triple point:
Point where all three phases in equilibrium

22,100
Phase Diagram - Water
Liquid
Pressure
(kPa)
101.3
Solid
Critical Point
Gas
Triple Point
0 100 374
Temperature (C °)
14) Physical state from graph: a) 80 °C & 100 kPa ____ c) .005
°C & 20 kPa ____ b) 20 °C & 20 kPa ____
15) Water boils at 100 °C at pressure of 101.3 kPa a) Name another set __________ b) Name a temp & pressure where water sublimes __________

22,100
Phase Diagram - Water
Liquid
Pressure
(kPa)
101.3
Solid
Critical Point
Gas
Triple Point
0 100 374
Temperature (C °)
14) Physical state from graph: a) 80 °C & 100 kPa liquid c) .005
°C & 20 kPa solid b) 20 °C & 20 kPa
15) Water boils at 100 °C at pressure of 101.3 kPa liquid/gas a) Name another set __________ b) Name a temp & pressure where water sublimes __________

22,100
Phase Diagram - Water
Liquid
Pressure
(kPa)
101.3
Solid
Critical Point
Gas
Triple Point
0 100 374
Temperature (C °)
14) Physical state from graph: a) 80 °C & 100 kPa liquid c) .005
°C & 20 kPa solid b) 20 °C & 20 kPa
15) Water boils at 100 °C at pressure of 101.3 kPa liquid/gas a) Name another set (any point on green line between triple and crit) b) Name a temp & pressure where water sublimes __________

Ch13.4 – More Phase diagrams
22,100
Phase Diagram - Water
Pressure
(kPa)
101.3
Solid
Liquid
Gas
0
Temperature (C °) 100 374
1) What state is H
2
O at: a) 50 °C and 50 kPa ____ b) 75 °C and 50 kPa ___ c) 0°C, 25 kPa ____ d) -10°C, 0.5 kPa ____
2) What is a critical point?
3) What is a triple point? b) What is the triple point of H
2
O? _______
4) What is the boiling point of water? b) What is the normal b.p. of H
2
O?
5) What is the melting point of H
2
O? b) What is the normal m.p.?
6) Are there any conditions where H
2
O sublimes?

Plasma
KE of particles becomes so large, motion so radical, that even the electrons get ripped off the atoms.
Left with a bunch of +/- charges.
- Outer space is mostly a plasma because the matter comes from stars (hot)
- Partial plasmas in fluorescent lights, neon signs, lightning bolts, flames.
- Aurora Borealis
- Cold plasmas 50,000K – 100,000K (few gases)
- Stars 10million K – 1billion K
Ch13 HW#4 WS

Ch13 HW#4 Worksheet

Chapter 14.1 – Behavior of Gases
Factors that affect the behavior of gases:
1) Number of particles - More particles = Higher gas pressure
Double the particles,
Double the pressure

Chapter 14.1 – Behavior of Gases
Factors that affect the behavior of gases:
1) Number of particles - More particles = Higher gas pressure
Double the particles,
Double the pressure
2) Size of container - reducing the size of the container increases pressure
Big V
Small P
Small V
Big P
Boyle’s Law:
V
1
.
P
1
= V
2
.
P
2

Chapter 14.1 – Behavior of Gases
Factors that affect the behavior of gases:
1) Number of particles - More particles = Higher gas pressure
Double the particles,
Double the pressure
2) Size of container - reducing the size of the container increases pressure
Big V
Small P
Small V
Big P
Boyle’s Law:
P
1
.
V
1
= P
2
.
V
2
3) Temperature - increases temp particles have more KE. To adjust, the gas must have higher pressure OR must occupy larger volume!
Double the temp = double the pressure
Fire
Must be in Kelvin!
O C+273 = K or double the volume
Charles’ Law
P
T
1
1

P
T
2
2
V
T
1
1

V
T
2
2

Ex 1) A balloon contains 30 L of helium gad at 100kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25kPa?
Ex 2) A balloon filled at 27 o C had a volume of 4.0L. It is heated to 57 o C.
What is the new volume?

Ex 1) A balloon contains 30 L of helium gad at 100kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25kPa?
P
1
= 100 kPa
P
1
V
1
= P
2
V
2
(100kPa)(30L)=(25kPa)(V
2
)
V
1
= 30L
P
2
= 25 kPa V
2
=120L
V
2
= ?
Ex 2) A balloon filled at 27 o C had a volume of 4.0L. It is heated to 57 o C.
What is the new volume?
T
1
= 27 o C 300K
V
1
= 4.0L
T
2
= 57 o C 330K
V
2
= ?
V
1
T
1
4 L
300 K


V
T
2
2
V
330
2
K V
2
= 4.4L

Ex 3) The gas left in a used aerosol can is at a pressure of 100kPa at 27 o C. If it is thrown into a fire, what will the pressure be if it reaches a temp of 927 o C?
Ch14 HW#1 1 – 6

Ex 3) The gas left in a used aerosol can is at a pressure of 100kPa at 27 o C. If it is thrown into a fire, what will the pressure be if it reaches a temp of 927 o C?
P
1
= 100kPa
T
1
= 27 o C 300K
P
2
= ?
T
2
= 927 o C 1200K
300
P
T
1
K
1
100 kPa


P
T
2
2
P
2
1200 K
P
2
= 400 kPa
Ch14 HW#1 1 – 6

CH 14 HW#1 1-6
1. Gas occupies 2.5L when at 100 kPa. If pressure changes to 40 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 40kPa
2. Gas occupies 2.5L when at 100 kPa. If pressure changes to 250 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 250kPa

CH 14 HW#1 1-6
1. Gas occupies 2.5L when at 100 kPa. If pressure changes to 40 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 40kPa
V
1

P
1
( 2 .
5 L )( 100 kPa )

V
2

P
2

V
2
( 40 kPa )
V
2
= 6.5L (OUT)
2. Gas occupies 2.5L when at 100 kPa. If pressure changes to 250 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 250kPa

CH 14 HW#1 1-6
1. Gas occupies 2.5L when at 100 kPa. If pressure changes to 40 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 40kPa
V
1

P
1
( 2 .
5 L )( 100 kPa )

V
2

P
2

V
2
( 40 kPa )
V
2
= 6.5L (OUT)
2. Gas occupies 2.5L when at 100 kPa. If pressure changes to 250 kPa, volume? Piston in or out?
V
1
= 2.5L
V
2
= ?
P
1
= 100kPa P
2
= 250kPa
V
1

P
1
( 2 .
5 L )( 100 kPa )

V
2

P
2

V
2
( 250 kPa )
V
2
= 1L (IN)

3. Gas occupies 6.8L when at 327 °C.
Temp drops to 27 °C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
=300K
4. Gas occupies 6.8L when at 327 °C. Temp rises to 427°C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
= 427 °C = 700K

3. Gas occupies 6.8L when at 327 °C.
Temp drops to 27 °C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
=300K
V
1
T
1

V
2
T
2
6 .
8 L
600 K

V
2
300 K
V
2
= 3.4L (IN)
4. Gas occupies 6.8L when at 327 °C. Temp rises to 427°C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
= 427 °C = 700K

3. Gas occupies 6.8L when at 327 °C.
Temp drops to 27 °C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
=300K
V
1
T
1

V
2
T
2
6 .
8 L
600 K

V
2
300 K
V
2
= 3.4L (IN)
4. Gas occupies 6.8L when at 327 °C. Temp rises to 427°C, volume? in / out?
V
1
= 6.8L
V
2
= ?
T
1
= 327 °C = 600K T
2
= 427 °C = 700K
V
1
 
T
1
V
2
T
2
6 .
8 L
600 K
V
2
700 K
V
2
= 7.9L
(OUT)

5. Gas occupies fixed volume, at pressure of 6.58kPa at 540K. Cooled to 210K, pressure?
P
1
= 6.58kPa P
2
= ?
T
1
= 540K T
2
= 210K

5. Gas occupies fixed volume, at pressure of 6.58kPa at 540K. Cooled to 210K, pressure?
P
1
= 6.58kPa P
2
= ?
T
1
= 540K T
2
= 210K
P
T
1
1

P
T
2
2
6 .
58 kPa

600 K
P
2
300 K
P
2
= 2.56kPa
6. Why do aerosol cans say “ Do Not Incinerate ”?

5. Gas occupies fixed volume, at pressure of 6.58kPa at 540K. Cooled to 210K, pressure?
P
1
= 6.58kPa P
2
= ?
T
1
= 540K T
2
= 210K
P
T
1
1

P
T
2
2
6 .
58 kPa

600 K
P
2
300 K
P
2
= 2.56kPa
6. Why do aerosol cans say “ Do Not Incinerate ”?
Fixed volume!
Temp = Pressure = explosion!
P
T
1
1

P
2
T
2

Ch14 HW#2 7 – 17
7) A sealed container with a movable piston contains a gas that occupies a volume when at a pressure. If the pressure
.
P
1
V
1
= P
2
V
2
8) A sealed container with a movable piston contains a gas that occupies a volume when at a temp. If the temp increases, does the volume increase or decrease?
V
T
1
1

V
2
T
2
9) A sealed container occupies a volume when at a pressure
& temp. If temp increases, what happens to pressure?
P
T
1
1

P
2
T
2

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?
V
1
= 10.0L
V
2
=
P
1
= 10kPa P
2
= 101.3kPa
11) A sealed container with a movable piston contains a gas that occupies 10.0L when at 100 °C. Temp changes to standard temp, volume?
V
1
= 10.0L V
T
1
= 100 °C = 373K T
2
2
= ?
= 0 °C = 273K
12) fixed volume of gas at STP. Find pressure if temp changes to 373K?
P
1
T
1
= 101.3kPa P
2
= 273K T
2
= ?
= 373K

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?
V
1
= 10.0L
V
2
= V
1
.
P
1
= V
2
.
P
2
P
1
= 10kPa P
2
= 101.3kPa (10L)(10kPa)=V
2
(101.3kPa)
V
2
= .987L
11) A sealed container with a movable piston contains a gas that occupies 10.0L when at 100 °C. Temp changes to standard temp, volume?
V
1
= 10.0L V
T
1
= 100 °C = 373K T
2
2
= ?
= 0 °C = 273K
12) fixed volume of gas at STP. Find pressure if temp changes to 373K?
P
1
T
1
= 101.3kPa P
2
= 273K T
2
= ?
= 373K

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?
V
1
= 10.0L
V
2
= V
1
.
P
1
= V
2
.
P
2
P
1
= 10kPa P
2
= 101.3kPa (10L)(10kPa)=V
2
(101.3kPa)
V
2
= .987L
11) A sealed container with a movable piston contains a gas that occupies 10.0L when at 100 °C. Temp changes to standard temp, volume?
V
1
V
1
= 10.0L V
T
1
= 100 °C = 373K T
2
2
= ?
= 0 °C = 273K
T
1

V
2
T
2
10 .
0 L
373 K

V
2
273 K
V
2
= 7.3L
12) fixed volume of gas at STP. Find pressure if temp changes to 373K?
P
1
T
1
= 101.3kPa P
2
= 273K T
2
= ?
= 373K

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?
V
1
= 10.0L
V
2
= V
1
.
P
1
= V
2
.
P
2
P
1
= 10kPa P
2
= 101.3kPa (10L)(10kPa)=V
2
(101.3kPa)
V
2
= .987L
11) A sealed container with a movable piston contains a gas that occupies 10.0L when at 100 °C. Temp changes to standard temp, volume?
V
1
V
1
= 10.0L V
T
1
= 100 °C = 373K T
2
2
= ?
= 0 °C = 273K
T
1

V
2
T
2
10 .
0 L
373 K

V
2
273 K
V
2
= 7.3L
12) fixed volume of gas at STP. Find pressure if temp changes to 373K?
P
1
T
1
= 101.3kPa P
2
= 273K T
2
= ?
= 373K
P
T
1
1

P
T
2
2
101 .
3
273 kPa
K

P
2
373 K
P
2
= 138kPa

13) Cylinder with moveable piston at 200kPa and 500mL.
Find volume at standard pressure.
14)Cylinder with moveable piston temp of 300 °C volume of 500mL. Find volume if temp corrected to standard temp.
15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?

13) Cylinder with moveable piston at 200kPa and 500mL.
Find volume at standard pressure.
P
1
= 200kPa P
2
= 101.3kPa
V
1
= 500mL V
2
= ?
V
1
.
P
1
= V
(500mL)(200kPa) = V
2
.
P
2
2
(101.3kPa)
V
2
= ____ mL
14)Cylinder with moveable piston temp of 300 °C volume of 500mL. Find volume if temp corrected to standard temp.
15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?

13) Cylinder with moveable piston at 200kPa and 500mL.
Find volume at standard pressure.
P
1
= 200kPa P
2
= 101.3kPa
V
1
= 500mL V
2
= ?
V
1
.
P
1
= V
(500mL)(200kPa) = V
2
.
P
2
2
(101.3kPa)
V
2
= ____ mL
14)Cylinder with moveable piston temp of 300 °C volume of 500mL. Find volume if temp corrected to standard temp.
T
1
= 300 °C = 573K T
2
= 273K
V
1
= 500mL V
2
= ?
V
T
1
1

V
T
2
2
500
573 mL
K

V
2
273 K
V
2
= 238mL
15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?

13) Cylinder with moveable piston at 200kPa and 500mL.
Find volume at standard pressure.
P
1
= 200kPa P
2
= 101.3kPa
V
1
= 500mL V
2
= ?
V
1
.
P
1
= V
(500mL)(200kPa) = V
2
.
P
2
2
(101.3kPa)
V
2
= ____ mL
14)Cylinder with moveable piston temp of 300 °C volume of 500mL. Find volume if temp corrected to standard temp.
T
1
= 300 °C = 573K T
2
= 273K
V
1
= 500mL V
2
= ?
V
T
1
1

V
T
2
2
500
573 mL
K

V
2
273 K
V
2
= 238mL
15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?
V
1
=2000L V
2
= ?
V
1
.
P
1
= V
2
.
P
2
P
1
=101.3kPa P
2
= 50kPa (2000L)(101.3kPa)=V
2
(50kPa)
V
2
= 4052L

16) V
1
V
2
= 244mL T
1
= ?
T
2
= 96.2
° C = 369.2
K
= 10 ° C = 283 K
17) How much water was sucked up into the flask?

16) V
1
V
2
= 244mL T
1
= ?
T
2
= 96.2
° C = 369.2
K
= 10 ° C = 283 K
V
1
T
1

V
2
T
2
244 mL
369 .
2 K

V
2
283 K
V
2
= 187mL
17) How much water was sucked up into the flask?
244mL - 187mL = 57mL

Lab14.1 – Charles’ Law
- due tomorrow
- Ch14 HW#2 due at beginning of period

Ch14.2 – Combined Gas Law
V
1

P
1
T
1

V
2

T
2
P
2
Ex1) The volume of a balloon is 30.0L at 40 °C and 150kPa.
What volume will the balloon have at STP?

Ch14.2 – Combined Gas Law
V
1

P
1
T
1

V
2

T
2
P
2
Ex1) The volume of a balloon is 30.0L at 40 °C and 150kPa.
What volume will the balloon have at STP?
V
1
= 30.0L
V
2
= ?
P
1
= 150kPa P
2
T
1
= 40 °C or 313K T
2
= 101.3kPa
=0 °C or 273K
V
1

T
1
P
1

V
2
T

2
P
2
( 30 .
0 L )( 150 kPa )

( 313 K )
( V
2
)( 101 .
3 kPa )
( 273 K )
V
2
= 38.7L

Ex2) The volume of a gas is 10.0L at 75.6kPa and 60 °C.
Correct this volume to STP.

Ex2) The volume of a gas is 10.0L at 75.6kPa and 60 °C.
Correct this volume to STP.
V
1
= 10.0L
V
2
= ?
P
1
= 75.6kPa
P
2
T
1
= 60 °C or 333K T
2
= 101.3kPa
= 0 °C or 273K
V
1

T
1
P
1

V
2

P
2
T
2
( 10 .
0 L )( 75 .
6 kPa )

( 333 K )
( V
2
)( 101 .
3 kPa )
( 273 K )
V
2
= 6.12L

Ex3) Correct the volume: 1.49L at 18 °C and 94.7kPa to 68 °C and 82.4kPa.

Ex3) Correct the volume: 1.49L at 18 °C and 94.7kPa to 68 °C and 82.4kPa.
V
1
= 1.49L
V
2
=?
P
1
= 94.7kPa
P
2
T
1
= 18 °C or 291K T
2
=82.4kPa (V )
=68 °C or 341K (V )
V
1

T
1
P
1

V
2

T
2
P
2
( 1 .
49 L )( 94 .
7 kPa )

( 291 K )
( V
2
)( 82 .
4 kPa )
( 341 K )
V
2
= 2.0L
Ch14 HW#3 18-21

Ch14 HW#3
18) Flask with volume of 300ml at pressure of 200 kPa and 100 °C. Correct to STP.
V
1
= 300mL V
2
= ?
P
1
= 200kPa P
2
= 101.3kPa
T
1
= 100 °C = 373K T
2
= 273K
( 300 mL )( 200
V
1

T
1
P
1 kPa )
( 373 K )


(
V
V
2
T
2

2
)(
(
P
2
101
273
.
3
K kPa )
)
19) Container with initial volume of 1.0L occupied by a gas at pressure of 150kPa at 25 °C. What’s the volume if pressure = 600kPa and temp raised to 100 °C?
V
1
= 1.0L
V
2
=?
P
1
= 150kPa P
2
= 600kPa
T
1
= 298K T
2
= 373K

Ch14 HW#3
18) Flask with volume of 300ml at pressure of 200 kPa and 100 °C. Correct to STP.
V
1
= 300mL V
2
= ?
P
1
= 200kPa P
2
= 101.3kPa
T
1
= 100 °C = 373K T
2
= 273K
( 300 mL )( 200
V
1

T
1
P
1 kPa )
( 373 K )


(
V
V
2
T
2

2
)(
(
P
2
101
273
.
3
K kPa )
)
V
2
= 434mL
19) Container with initial volume of 1.0L occupied by a gas at pressure of 150kPa at 25 °C. What’s the volume if pressure = 600kPa and temp raised to 100 °C?
V
1
= 1.0L
V
2
=?
P
1
= 150kPa P
2
= 600kPa
T
1
= 298K T
2
= 373K

Ch14 HW#3
18) Flask with volume of 300ml at pressure of 200 kPa and 100 °C. Correct to STP.
V
1
= 300mL V
2
= ?
P
1
= 200kPa P
2
= 101.3kPa
T
1
= 100 °C = 373K T
2
= 273K
( 300 mL )( 200
V
1

T
1
P
1 kPa )
( 373 K )


(
V
V
2
T
2

2
)(
(
P
2
101
273
.
3
K kPa )
)
V
2
= 434mL
19) Container with initial volume of 1.0L occupied by a gas at pressure of 150kPa at 25 °C. What’s the volume if pressure = 600kPa and temp raised to 100 °C?
V
1
= 1.0L
V
2
=?
P
1
= 150kPa P
2
= 600kPa
T
1
= 298K T
2
= 373K
( 1 .
0
(
L )( 150
298 K ) kPa )

( V
2
(
)( 600
373 K kPa )
)
V
2
=.31L

20) 5.0L of gas at STP heated to 150 °C and expands to 6.0L, pressure?
V
1
= 5.0L
V
2
= 6.0L
P
1
= 101.3kPa P
2
= ?
T
1
= 273K T
2
= 423K
( 5 .
0 L )( 101 .
3 kPa )
( 273 K )

( 6 .
0 L )( P
2
)
( 423 K )
21) A container contains a gas at a given V,P,T.
How does the volume change if pressure increased 4x greater
& temp half as much?
V
1
= V V
2
= ?
P
1
= P P
2
= 4P
T
1
= T T
2
= .5T
V
T

P

V
2
(.

5
( 4
T )
P )

20) 5.0L of gas at STP heated to 150 °C and expands to 6.0L, pressure?
V
1
= 5.0L
V
2
= 6.0L
P
1
= 101.3kPa P
2
= ?
T
1
= 273K T
2
= 423K
( 5 .
0 L )( 101 .
3 kPa )
( 273 K )

( 6 .
0 L )( P
2
)
( 423 K )
P
2
= 130.8kPa
21) A container contains a gas at a given V,P,T.
How does the volume change if pressure increased 4x greater
& temp half as much?
V
1
= V V
2
= ?
P
1
= P P
2
= 4P
T
1
= T T
2
= .5T
V
T

P

V
2
(.

5
( 4
T )
P )

20) 5.0L of gas at STP heated to 150 °C and expands to 6.0L, pressure?
V
1
= 5.0L
V
2
= 6.0L
P
1
= 101.3kPa P
2
= ?
T
1
= 273K T
2
= 423K
( 5 .
0 L )( 101 .
3 kPa )
( 273 K )

( 6 .
0 L )( P
2
)
( 423 K )
P
2
= 130.8kPa
21) A container contains a gas at a given V,P,T.
How does the volume change if pressure increased 4x greater
& temp half as much?
V
1
= V V
2
= ?
P
1
= P P
2
= 4P
T
1
= T T
2
= .5T
V
T

P

V
2
(.

5
( 4
T )
P )
V
2
= 1/8 V

Ch14.4 – Ideal Gas Law
Pressure, volume, and temp affect each other.
So does the # of particles, n (moles):

Ch14.4 – Ideal Gas Law
Pressure, volume, and temp affect each other.
So does the # of particles, n (moles):
P .
V = n .
R .
T temp in Kelvin pressure volume moles Ideal Gas Law Constant
Ex1) 1 mole of any gas occupies 22.4 L at STP. Based on this, what must be the value of R?

Ch14.4 – Ideal Gas Law
Pressure, volume, and temp affect each other.
So does the # of particles, n (moles):
P .
V = n .
R .
T temp in Kelvin pressure volume moles Ideal Gas Law Constant
Ex1) 1 mole of any gas occupies 22.4 L at STP. Based on this, what must be the value of R?
n = 1 mole PV
 nRT
V = 22.4L
P = 101.3kPa
T = 0 °C = 273K
R

PV nT

( 101 .
3 kPa )( 22 .
4 L )
( 1 mol )( 273 K )
 kPa

8 .
31 mol

L
K

Ex 2) A rigid container is filled with 20.0L of nitrogen gas to a pressure of 20,000 kPa at 27 °C.
How many moles of N
2 are in the cylinder?
Ex 3) A deep underground cavern contains 2.24x10
6 L of methane gas at 1500kPa and 42 °C, how many moles?
b) How many grams is this?

Ex 2) A rigid container is filled with 20.0L of nitrogen gas to a pressure of 20,000 kPa at 27 °C.
How many moles of N
2
V = 20.0L
PV
 are in the cylinder?
nRT
P = 20,000kPa
T = 27 n = ?
°C = 300K n

PV
RT

(
(
8
20
.
,
31
000
L
 kPa mol kPa

K
)(
)(
20 .
0
300
L
K
)
)

160
Ex 3) A deep underground cavern contains 2.24x10
6 L
.
4 mol of methane gas at 1500kPa and 42 °C, how many moles?
V = 2.24x10
6 L PV
 nRT
P = 1500kPa
T = 315K n = ?
n

PV
RT

( 1500 kPa )( 2 .
24 x 10
6
L
 kPa
( 8 .
31
B) How many grams is this?
mol

K
)( 315 K
L )
)
1.3x10
6 mol CH
4
16.0 g CH
1 mol CH
4
4

1 .
3 x 10
6 mol
= 2.1x10
7 g CH
4
Ch14 HW#4 22-25

Ch14 HW#4 22 – 25
22. If you were to increase the number of particles inside a container by a hundred times, what do you think it would do to the pressure?
23. When a rigid hollow sphere containing 680 L of helium gas is heated to 600 K, the pressure of the gas increases to 1800 kPa.
How many moles of helium are in the sphere?

Ch14 HW#4 22 – 25
22. If you were to increase the number of particles inside a container by a hundred times, what do you think it would do to the pressure?
PV
 nRT

( 100 n )

RT
23. When a rigid hollow sphere containing 680 L of helium gas is heated to 600 K, the pressure of the gas increases to 1800 kPa.
How many moles of helium are in the sphere?

Ch14 HW#4 22 – 25
22. If you were to increase the number of particles inside a container by a hundred times, what do you think it would do to the pressure?
PV
 nRT

( 100 n )

RT
23. When a rigid hollow sphere containing 680 L of helium gas is heated to 600 K, the pressure of the gas increases to 1800 kPa.
How many moles of helium are in the sphere?
PV
 nRT n

PV
RT

( 1800 kPa )( 680
L
 kPa
L )
( 8 .
31 mol

K
)( 600 K )

245 .
5 mol

24. A child has a lung capacity of 2.2 L. How many moles of air do her lungs hold at a pressure of 100 kPa and a normal body temp of 37ºC? Air is a mixture of mostly O
2 and N
2, which you may assume has an average molar mass of 29 g/mol. How many grams of air are held in the lungs?
25. When a rigid hollow sphere containing 10.0 L of helium gas is cooled to -200

C, the pressure of the gas decreases. If there are
5.0 moles of gas present, what is the pressure of helium in the sphere?

24. A child has a lung capacity of 2.2 L. How many moles of air do her lungs hold at a pressure of 100 kPa and a normal body temp of 37ºC? Air is a mixture of mostly O
2 and N
2, which you may assume has an average molar mass of 29 g/mol. How many grams of air are held in the lungs?
n

PV
RT

( 100 kPa )(
L
 kPa
2 .
2
( 8 .
31 mol

K
L
)( 310
)
K )
0.085 mol air 29.0 g air
1 mol air

0 .
085 mol
= 2.48 g air
25. When a rigid hollow sphere containing 10.0 L of helium gas is cooled to -200

C, the pressure of the gas decreases. If there are
5.0 moles of gas present, what is the pressure of helium in the sphere?

24. A child has a lung capacity of 2.2 L. How many moles of air do her lungs hold at a pressure of 100 kPa and a normal body temp of 37ºC? Air is a mixture of mostly O
2 and N
2, which you may assume has an average molar mass of 29 g/mol. How many grams of air are held in the lungs?
n

PV
RT

( 100 kPa )(
L
 kPa
2 .
2
( 8 .
31 mol

K
L
)( 310
)
K )
0.085 mol air 29.0 g air
1 mol air

0 .
085 mol
= 2.48 g air
25. When a rigid hollow sphere containing 10.0 L of helium gas is cooled to -200

C, the pressure of the gas decreases. If there are
5.0 moles of gas present, what is the pressure of helium in the sphere?
L
 kPa
P
 nRT
V

( 5 .
0 mol )( 8 .
31
( 10 .
mol
0 L )

K
)( 73 K )

303 kPa

Ch14 Review Quiz – Balancing Equations Name _________
Form A
1. Magnesium oxide reacts with hydrogen fluoride to produce magnesium fluoride and water vapor.
2. Hydrochloric acid reacts with barium nitrate to produce nitric acid and barium chloride.
Form B
1. Magnesium chloride reacts with calcium fluoride to produce magnesium fluoride and calcium chloride.
2. Sulfuric acid reacts with barium nitrate to produce nitric acid and barium sulfate.

Ch14 Review Quiz Practice
1. Sodium hydroxide reacts with nitric acid to produce sodium nitrate and water vapor.

Ch14 HW#5 26 – 38
26. If you were to cut the size of a container to 10 times smaller, what do you think it would do to the pressure?
27. In the last problem, if you kept the pressure the same, what would you have to do to the temperature of the particles?
28. If you were to double the temperature of the particles inside a container, what do you think it would do to the pressure?

29. If you were to double the temperature of the particles inside a container, and wanted to keep the pressure constant, what would you have to do to the volume?
30. Explain why cooling a balloon down causes the size of the balloon to decrease?
31. A tank of nitrous oxide containing 4.5 liters measured 55 kPa.
The pressure was increased to 83 kPa. What is the volume of gas?

32. A small balloon needs 1250 ml of He in order to rise. A 500 ml sample of He at 30 °C does not rise. To what temp does this sample need to be heated in order to rise?
33. A gas at 44 °C exerts a pressure of 47 kPa. What pressure will be exerted by the same gas at 26 °C?
34. A gas occupies a volume of 300.0 cm 3 at a pressure of 80 kPa and room temp, 20 °C. Correct this volume to STP.
(Rest for HW)

Ch14 Review Quiz – Balancing Equations Name _________
Form A
1. Magnesium oxide reacts with hydrogen fluoride to produce magnesium fluoride and water vapor.
2. Hydrochloric acid reacts with barium nitrate to produce nitric acid and barium chloride.
Form B
1. Magnesium chloride reacts with calcium fluoride to produce magnesium fluoride and calcium chloride.
2. Sulfuric acid reacts with barium nitrate to produce nitric acid and barium sulfate.

Ch14 HW#5 26 – 38 cont.
35. A 2.50 L tank at STP contains pure oxygen. How many moles of oxygen are in the tank? What is the mass of oxygen in the tank?
V = 2.50L
P = 101.3kPa
T = 0 °C = 273K n = ?
m = ?
36. Lab question: A 2.50 L flask full of air is heated to 96.2
°C.
It is removed from the heat source and immediately placed in a room temp water bath, at 20 °C. Water is then sucked up into a flask.
What is the volume of air still in the flask?
V
1
= 2.50L
V
2
= ?
T
1
= 96.2
°C = 369.2K T
2
= 293K

Ch14 HW#5 26 – 38 cont.
35. A 2.50 L tank at STP contains pure oxygen. How many moles of oxygen are in the tank? What is the mass of oxygen in the tank?
V = 2.50L
P = 101.3kPa
T = 0 °C = 273K n n = ?

PV
RT

( 8
( 101 .
3 kPa )( 2 .
50 L )
.
31
L
 mol kPa

K
)( 273 K )

0 .
112 mol m = ?
0.112 mol O
2
32.0 g O
2
1 mol O
2
= 3.58 g O
36. Lab question: A 2.50 L flask full of air is heated to 96.2
°C.
2
It is removed from the heat source and immediately placed in a room temp water bath, at 20 °C. Water is then sucked up into a flask.
What is the volume of air still in the flask?
V
1
= 2.50L
V
2
= ?
T
1
= 96.2
°C = 369.2K T
2
= 293K

Ch14 HW#5 26 – 38 cont.
35. A 2.50 L tank at STP contains pure oxygen. How many moles of oxygen are in the tank? What is the mass of oxygen in the tank?
V = 2.50L
P = 101.3kPa
T = 0 °C = 273K n n = ?

PV
RT

( 8
( 101 .
3 kPa )( 2 .
50 L )
.
31
L
 mol kPa

K
)( 273 K )

0 .
112 mol m = ?
0.112 mol O
2
32.0 g O
2
1 mol O
2
= 3.58 g O
36. Lab question: A 2.50 L flask full of air is heated to 96.2
°C.
2
It is removed from the heat source and immediately placed in a room temp water bath, at 20 °C. Water is then sucked up into a flask.
What is the volume of air still in the flask?
V
1
= 2.50L
V
2
= ?
T
1
= 96.2
°C = 369.2K T
2
= 293K
(
( 2 .
50
369 .
2
L
K
V
T
)
)
1
1


V
T
2
2
( V
( 293
2
)
K )
V
2
= 1.99L

37. A sample of CO fills a 5.0 L balloon at standard temp and pressure.
What will be the volume of the balloon if temp is lowered to –25°C and the pressure increased to 200 kPa?
V
1
= 5.0L
V
2
= ?
V
1
P
1
= 101.3kPa P
2
= 200kPa
T
1
= 273K T
2
= 248K
( 5 .
0 L )( 101 .
3

T
1 kPa
P
1
)


(
V
V
2
2

T
2
)(
P
2
200
( 273 K ) ( 248
38. 6.00 g of NH
V
2
4
+
= 2.3L
are introduced into a 0.50 L flask at –200°C.
K kPa )
)
What is the pressure of the gas in kPa?
Hint: You’ll need to first convert 6.00 g of NH
4
+
Periodic Table, of course!
to moles. How?
V = 0.50L
6.00 g NH
4
+ 1 mol NH
4
+
18.0 g NH
4
+ = 0.33 mol
T = 73K
P = ?
P
 nRT
V

L
 kPa
( 0 .
33 mol )( 8 .
31 mol

K
)( 73 K )
( 0 .
5 L )

404 kPa

Chemistry – Ch13,14 Review + phase diagrams
1. What is meant by the term elastic collision ?
2. How can you raise the average kinetic energy of the water molecules in a glass of water?
3. Express 545 mm Hg in kilopascals.
4. What is significant about the temperature absolute zero?

5. A cylinder of oxygen gas is cooled from 300 K (27 °C) to 150 K (-123°C).
By what factor does the average kinetic energy of the oxygen molecules in the cylinder decrease?
6. Why does evaporation lower the temperature of a liquid?
7. Explain why increasing the temperature of a liquid increases its rate of evaporation?
8. Explain how boiling is a cooling process.

9. What is the crystal lattice of a solid?
10. A metal cylinder contains 1mol of nitrogen gas at STP.
What will happen to he pressure if another mole of gas is added to the cylinder, but the temp and volume stay the same?
11. If a gas is compressed from 4.0L to 1.0L and the temp stays constant, what happens to the pressure?

12. The gas in a container has a pressure of 300kPa at 27

C.
What will be the pressure if the temp is lowered to –173 
C?
13. Five liters of air at –50 
C is warmed to 100

C.
What is the new volume if the pressure remains constant?
14. A 5.0L sample of air at a temp of –50 
C has a pressure of 107 kPa.
What is the new pressure if the temp is raised to 100

C and the volume expands to 7.0L?

15. A 5.00L flask at 25

C contains 0.200 mol of Cl
2
.
What is the pressure of the flask?
16. What volume will 12.0g of oxygen gas occupy at 25

C and a pressure of 52.7 kPa?

22,100
Phase Diagram - Water
Pressure
(kPa)
101.3
Solid
Liquid
Gas
0
Temperature (C °) 100 374
1) Determine: a) the critical temp for water b) critical pressure c) triple point temp d) triple point pressure e) normal melting point f) normal boiling point
2) What is the physical state at: a)50 °C and 0.1kPa
b) 105 °C and 1000kPa