chapter 13 - states of matter kinetic theory and the nature of glass all matter is composed of tiny...

Chapter 13 - States of Matter Kinetic Theory and the Nature of GlassAll 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 O2 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 O2 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 02 443m/s (~1000mph!)(very fast)

4. Frequency of collisions 4.5 billon collisions per second(A lot of collisions)

Gas ParticlesGas 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 1 atmosphere

Or 101.3 kPa

(Pascal is the SI unit of pressure.)

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

1400mm Hg

kPa kPa=mm 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 nomolecular motion?

Absolute zero – temp at which there’s no molecular motion.0 Kelvin ( - 273o 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) kPab)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) kPab)atma)190 mm Hg 101.3 kPa = 25.1 kPa

760 mm Hgb)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



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



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 H20 affected as you pour hot into equal amount of cold H20?

5. By what factor does ave KE go up as gas increases from 300K to 900K?

4. How does average KE of H20 affected as you pour hot into equal amount of cold H20?Hot cools as cold warms as they reach middle temp.

Average stays constant.


4. How does average KE of H20 affected as you pour hot into equal amount of cold H20?Hot cools as cold warms as they reach middle temp.

Average stays constant.


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 enoughto 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 H20(l) H20(g)

(both condensation and evaporation at same time.)

Boiling Point of LiquidBoiling 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.)

- the temp of a liquid never rises above its boiling point temp.

100 80 chloroform water

VP 60 ethanol ethanoic acid

(kPa)40 20

20 40 60 80 100 120Temp (°C)

FIRE

Ch13 HW#2 6-9

Ch13 HW#2 6 – 96. 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:


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?




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 waterVP 60 ethanol ethanoic acid(kPa)40 20

20 40 60 80 100 120Temp (°C)


chloroform : ____ ethanol :____ water :____ ethanoic acid:____




It depends on the pressure! At standard atm pressure, its 100°C. 100 80 chloroform waterVP 60 ethanol ethanoic acid(kPa)40 20

20 40 60 80 100 120Temp (°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.

Pressure(kPa)

101.3

22,100

3741000

SolidLiquid

Gas

Temperature (C°)

Phase Diagram - Water

Pressure(kPa)

101.3

22,100 Critical Point

3741000

SolidLiquid

GasTriple Point

Temperature (C°)


Critical point for water – At 374°C and 22,100kPa is the last point wherewater can still be condensed to a liquid.

Triple point – all 3 phases are in equilibrium

Pressure(kPa)

101.3


3741000

SolidLiquid

GasTriple Point

Temperature (C°)


Ex: What is the physical state of H2O at 50°C and 70 kPa?_____ Name a temp and pressure that H2O 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


melting point

11) Substance goes directly from solid to gas: _________


13) Triple point:

Ch13 HW#3 10 – 15


melting point

11) Substance goes directly from solid to gas: sublimation


13) Triple point:

Ch13 HW#3 10 – 15


melting point



The ice in the popsicle sublimes

13) Triple point:

Ch13 HW#3 10 – 15


melting point



The ice in the popsicle sublimes

13) Triple point:

Point where all three phases in equilibrium

14) Physical state from graph:

a) 80°C & 100 kPa ____ b) 20°C & 20 kPa ____

c) .005°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 __________

Pressure(kPa)

101.3


3741000

SolidLiquid

GasTriple Point

Temperature (C°)



a) 80°C & 100 kPa liquid b) 20°C & 20 kPa liquid/gas

c) .005°C & 20 kPa solid


a) Name another set __________


Pressure(kPa)

101.3


3741000

SolidLiquid

GasTriple Point

Temperature (C°)



a) 80°C & 100 kPa liquid b) 20°C & 20 kPa liquid/gas

c) .005°C & 20 kPa solid


a) Name another set (any point on green line between triple and crit)


Pressure(kPa)

101.3


3741000

SolidLiquid

GasTriple Point

Temperature (C°)


Ch13.4 – More Phase diagrams

1) What state is H2O 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 H2O? _______

4) What is the boiling point of water? b) What is the normal b.p. of H2O?

5) What is the melting point of H2O? b) What is the normal m.p.?

6) Are there any conditions where H2O sublimes?

Pressure(kPa)

101.3

22,100

3741000

SolidLiquid

Gas

Temperature (C°)


PlasmaKE 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

Factors that affect the behavior of gases:1) Number of particles - More particles = Higher gas pressure

Chapter 14.1 – Behavior of Gases

Double the particles,Double the pressure


2) Size of container - reducing the size of the container increases pressure



Big VSmall P

Boyle’s Law: V1

.P1 = V2.P2 Small V

Big P


2) Size of container - reducing the size of the container increases pressure

3) Temperature - increases temp particles have more KE. To adjust, the gas must have higher pressure OR must occupy larger volume!



Fire

Big VSmall P

Boyle’s Law: P1

.V1 = P2.V2

Double the temp = double the pressure or double the volume

Charles’ Law Must be in Kelvin!OC+273 = K

Small VBig P

2

2

1

1

T

V

T

V

2

2

1

1

T

P

T

P

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 27oC had a volume of 4.0L. It is heated to 57oC. 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?

P1V1 = P2V2

P1= 100 kPa (100kPa)(30L)=(25kPa)(V2) V1= 30L P2= 25 kPa V2=120L V2= ?Ex 2) A balloon filled at 27oC had a volume of 4.0L. It is heated to 57oC. What is the new volume?

T1= 27oC 300KV1 = 4.0LT2 = 57oC 330KV2 = ?

V2 = 4.4LK

V

K

L

T

V

T

V

330300

4

2

2

2

1

1

Ex 3) The gas left in a used aerosol can is at a pressure of 100kPa at 27oC. If it is thrown into a fire, what will the pressure be if it reaches a temp of 927oC?

Ch14 HW#1 1 – 6

Ex 3) The gas left in a used aerosol can is at a pressure of 100kPa at 27oC. If it is thrown into a fire, what will the pressure be if it reaches a temp of 927oC?P1 = 100kPaT1 = 27oC 300KP2 = ? T2 = 927oC 1200K

P2 = 400 kPa

Ch14 HW#1 1 – 6

K

P

K

kPa

T

P

T

P

1200300

100

2

2

2

1

1

CH 14 HW#1 1-61. Gas occupies 2.5L when at 100 kPa. If pressure changes

to 40 kPa, volume? Piston in or out?

V1 = 2.5L V2 = ?P1 = 100kPa P2 = 40kPa

2. Gas occupies 2.5L when at 100 kPa. If pressure changes to 250 kPa, volume? Piston in or out?V1 = 2.5L V2 = ?P1 = 100kPa P2 = 250kPa



V1 = 2.5L V2 = ?P1 = 100kPa P2 = 40kPa

V2 = 6.5L (OUT)2. Gas occupies 2.5L when at 100 kPa. If pressure changes

to 250 kPa, volume? Piston in or out?V1 = 2.5L V2 = ?P1 = 100kPa P2 = 250kPa

)40()100)(5.2(

2

2211

kPaVkPaL

PVPV



V1 = 2.5L V2 = ?P1 = 100kPa P2 = 40kPa

V2 = 6.5L (OUT)2. Gas occupies 2.5L when at 100 kPa. If pressure changes

to 250 kPa, volume? Piston in or out?V1 = 2.5L V2 = ?P1 = 100kPa P2 = 250kPa

V2 = 1L (IN)

)40()100)(5.2(

2

2211

kPaVkPaL

PVPV

)250()100)(5.2(

2

2211

kPaVkPaL

PVPV

3. Gas occupies 6.8L when at 327°C. Temp drops to 27°C, volume? in / out?V1 = 6.8L V2 = ?T1 = 327°C = 600K T2=300K

4. Gas occupies 6.8L when at 327°C. Temp rises to 427°C, volume? in / out?

V1 = 6.8L V2 = ? T1 = 327°C = 600K T2 = 427°C = 700K


V2 = 3.4L (IN)4. Gas occupies 6.8L when at 327°C. Temp rises to 427°C,

volume? in / out?V1 = 6.8L V2 = ? T1 = 327°C = 600K T2 = 427°C = 700K

300600

8.6 2

2

2

1

1

K

V

K

L

T

V

T

V


V2 = 3.4L (IN)4. Gas occupies 6.8L when at 327°C. Temp rises to 427°C,

volume? in / out?V1 = 6.8L V2 = ? T1 = 327°C = 600K T2 = 427°C = 700K

V2 = 7.9L (OUT)

300600

8.6 2

2

2

1

1

K

V

K

L

T

V

T

V

700600

8.6 2

2

2

1

1

K

V

K

L

T

V

T

V

5. Gas occupies fixed volume, at pressure of 6.58kPa at 540K. Cooled to 210K, pressure?

P1 = 6.58kPa P2 = ?T1 = 540K T2 = 210K


P1 = 6.58kPa P2 = ?T1 = 540K T2 = 210K

P2 = 2.56kPa

300600

58.6

2

2

2

1

1

K

P

K

kPa

T

P

T

P

6. Why do aerosol cans say “Do Not Incinerate”?


P1 = 6.58kPa P2 = ?T1 = 540K T2 = 210K

P2 = 2.56kPa

300600

58.6

2

2

2

1

1

K

P

K

kPa

T

P

T

P

6. Why do aerosol cans say “Do Not Incinerate”?

Fixed volume!Temp = Pressure = explosion!

2

2

1

1

T

P

T

P

Ch14 HW#2 7 – 177) A sealed container with a movable piston contains a gas that occupies a volume when at a pressure. If the pressure increases, does the volume increase or decrease? P1V1 = P2V2

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?

9) A sealed container occupies a volume when at a pressure & temp. If temp increases, what happens to pressure?

. .

2

2

1

1

T

V

T

V

2

2

1

1

T

P

T

P

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?V1= 10.0L V2=P1= 10kPa P2= 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?V1 = 10.0L V2 = ?T1 = 100°C = 373K T2= 0°C = 273K

12) fixed volume of gas at STP. Find pressure if temp changes to 373K?P1 = 101.3kPa P2 = ?T1 = 273K T2 = 373K

10) A sealed container with a movable piston occupies 10L at 10.0kPa. If a pressure changes to standard pressure, volume? in/out?V1= 10.0L V2= V1

.P1 = V2.P2

P1= 10kPa P2= 101.3kPa (10L)(10kPa)=V2(101.3kPa) V2 = .987L




.P1 = V2.P2



V2= 7.3L


273373

0.10 2

2

2

1

1

K

V

K

L

T

V

T

V


.P1 = V2.P2



V2= 7.3L


P2 = 138kPa

273373

0.10 2

2

2

1

1

K

V

K

L

T

V

T

V

373273

3.101 2

2

2

1

1

K

P

K

kPa

T

P

T

P

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?


P1 = 200kPa P2 = 101.3kPa V1.P1 = V2

.P2

V1 = 500mL V2 = ? (500mL)(200kPa) = V2(101.3kPa)V2 = ____ 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?


P1 = 200kPa P2 = 101.3kPa V1.P1 = V2

.P2


14)Cylinder with moveable piston temp of 300°C volume of 500mL. Find volume if temp corrected to standard temp.T1 = 300°C = 573K T2 = 273KV1 = 500mL V2 = ?

V2 = 238mL15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?

273573

500 2

2

2

1

1

K

V

K

mL

T

V

T

V


P1 = 200kPa P2 = 101.3kPa V1.P1 = V2

.P2


14)Cylinder with moveable piston temp of 300°C volume of 500mL. Find volume if temp corrected to standard temp.T1 = 300°C = 573K T2 = 273KV1 = 500mL V2 = ?

V2 = 238mL15) Hot air balloon contains 2000L of hot gas at standard pressure. Find volume when at 50 kPa?V1=2000L V2 = ? V1

.P1 = V2.P2

P1=101.3kPa P2 = 50kPa (2000L)(101.3kPa)=V2(50kPa) V2 = 4052L

273573

500 2

2

2

1

1

K

V

K

mL

T

V

T

V

16) V1 = 244mL T1 = 96.2°C = 369.2K V2 = ? T2 = 10°C = 283K

17) How much water was sucked up into the flask?

16) V1 = 244mL T1 = 96.2°C = 369.2K V2 = ? T2 = 10°C = 283K

V2 = 187mL

17) How much water was sucked up into the flask? 244mL - 187mL = 57mL

2832.369

244 2

2

2

1

1

K

V

K

mL

T

V

T

V

Lab14.1 – Charles’ Law

- due tomorrow

- Ch14 HW#2 due at beginning of period

Ch14.2 – Combined Gas Law

Ex1) The volume of a balloon is 30.0L at 40°C and 150kPa. What volume will the balloon have at STP?

2

22

1

11

T

PV

T

PV

Ch14.2 – Combined Gas Law

Ex1) The volume of a balloon is 30.0L at 40°C and 150kPa. What volume will the balloon have at STP?

V1 = 30.0L V2 = ?P1 = 150kPa P2 = 101.3kPaT1 = 40°C or 313K T2=0°C or 273K

V2 = 38.7L

2

22

1

11

T

PV

T

PV

)273(

)3.101)((

)313(

)150)(0.30(

2

2

22

1

11

K

kPaV

K

kPaL

T

PV

T

PV

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.

V1 = 10.0L V2 = ?P1 = 75.6kPa P2 = 101.3kPaT1 = 60°C or 333K T2 = 0°C or 273K

V2 = 6.12L

)273(

)3.101)((

)333(

)6.75)(0.10(

2

2

22

1

11

K

kPaV

K

kPaL

T

PV

T

PV

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.V1 = 1.49L V2=?P1 = 94.7kPa P2=82.4kPa (V ) T1 = 18°C or 291K T2=68°C or 341K (V )

V2 = 2.0L

Ch14 HW#3 18-21

)341(

)4.82)((

)291(

)7.94)(49.1(

2

2

22

1

11

K

kPaV

K

kPaL

T

PV

T

PV

Ch14 HW#318) Flask with volume of 300ml at pressure of 200 kPa

and 100°C. Correct to STP.V1 = 300mL V2 = ?P1 = 200kPa P2 = 101.3kPaT1 = 100°C = 373K T2 = 273K

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?

V1 = 1.0L V2=?P1 = 150kPa P2 = 600kPa T1 = 298K T2 = 373K

)273(

)3.101)((

)373(

)200)(300( 2

K

kPaV

K

kPamL

2

22

1

11 T

PV

T

PV



V2 = 434mL19) 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?

V1 = 1.0L V2=?P1 = 150kPa P2 = 600kPa T1 = 298K T2 = 373K

)273(

)3.101)((

)373(

)200)(300( 2

K

kPaV

K

kPamL

2

22

1

11 T

PV

T

PV



V2 = 434mL19) 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?

V1 = 1.0L V2=?P1 = 150kPa P2 = 600kPa T1 = 298K T2 = 373K V2 =.31L

)273(

)3.101)((

)373(

)200)(300( 2

K

kPaV

K

kPamL

2

22

1

11 T

PV

T

PV

)373(

)600)((

)298(

)150)(0.1( 2

K

kPaV

K

kPaL

20) 5.0L of gas at STP heated to 150°C and expands to 6.0L, pressure?

V1 = 5.0L V2 = 6.0LP1 = 101.3kPa P2 = ?T1 = 273K T2 = 423K

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?V1 = V V2 = ?P1 = P P2 = 4PT1 = T T2 = .5T

)423(

))(0.6(

)273(

)3.101)(0.5( 2

K

PL

K

kPaL

)5(.

)4(2

T

PV

T

PV


V1 = 5.0L V2 = 6.0LP1 = 101.3kPa P2 = ?T1 = 273K T2 = 423K

P2 = 130.8kPa21) 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?V1 = V V2 = ?P1 = P P2 = 4PT1 = T T2 = .5T

)423(

))(0.6(

)273(

)3.101)(0.5( 2

K

PL

K

kPaL

)5(.

)4(2

T

PV

T

PV


V1 = 5.0L V2 = 6.0LP1 = 101.3kPa P2 = ?T1 = 273K T2 = 423K

P2 = 130.8kPa21) 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?V1 = V V2 = ?P1 = P P2 = 4PT1 = T T2 = .5T

V2 = 1/8 V

)423(

))(0.6(

)273(

)3.101)(0.5( 2

K

PL

K

kPaL

)5(.

)4(2

T

PV

T

PV

Ch14.4 – Ideal Gas LawPressure, volume, and temp affect each other.

So does the # of particles, n (moles):



P.V = n.R.T

pressurevolume

moles Ideal Gas Law Constant

temp in Kelvin

Ex1) 1 mole of any gas occupies 22.4 L at STP. Based on this, what must be the value of R?



P.V = n.R.T

pressurevolume

moles Ideal Gas Law Constant

temp in Kelvin

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 V = 22.4LP = 101.3kPaT = 0°C = 273K Kmol

LkPa

Kmol

LkPa

nT

PVR

nRTPV

31.8)273)(1(

)4.22)(3.101(

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 N2 are in the cylinder?

Ex 3) A deep underground cavern contains 2.24x106L 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 N2 are in the cylinder?V = 20.0LP = 20,000kPa T = 27°C = 300Kn = ?Ex 3) A deep underground cavern contains 2.24x106L of methane gas at 1500kPa and 42°C, how many moles?V = 2.24x106LP = 1500kPaT = 315Kn = ?B) How many grams is this?

1.3x106 mol CH4 16.0 g CH4 = 2.1x107 g CH4 1 mol CH4 Ch14 HW#4 22-25

molK

KmolkPaL

LkPa

RT

PVn

nRTPV

4.160)300)(31.8(

)0.20)(000,20(

molxK

KmolkPaL

LxkPa

RT

PVn

nRTPV

66

103.1)315)(31.8(

)1024.2)(1500(

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?



RTn

nRTPV

)100(



RTn

nRTPV

)100(

molK

KmolkPaL

LkPa

RT

PVn

nRTPV

5.245)600)(31.8(

)680)(1800(

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 O2 and N2, 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 -200C, the pressure of the gas decreases. If there are 5.0 moles of gas present, what is the pressure of helium in the sphere?


0.085 mol air 29.0 g air 1 mol air = 2.48 g air


molK

KmolkPaL

LkPa

RT

PVn 085.0

)310)(31.8(

)2.2)(100(


0.085 mol air 29.0 g air 1 mol air = 2.48 g air


molK

KmolkPaL

LkPa

RT

PVn 085.0

)310)(31.8(

)2.2)(100(

kPaL

KKmol

kPaLmol

V

nRTP 303

)0.10(

)73)(31.8)(0.5(

You will be taking one of these quizzes tomorrow!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 producenitric acid and barium chloride.

Form B

1. Magnesium chloride reacts with calcium fluoride to producemagnesium fluoride and calcium chloride.

2. Sulfuric acid reacts with barium nitrate to producenitric 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 cm3 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 producenitric acid and barium chloride.

Form B

1. Magnesium chloride reacts with calcium fluoride to producemagnesium fluoride and calcium chloride.

2. Sulfuric acid reacts with barium nitrate to producenitric 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 = 2.50LP = 101.3kPa T = 0°C = 273Kn = ?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?

V1 = 2.50L V2 = ?T1 = 96.2°C = 369.2K T2 = 293K


V = 2.50LP = 101.3kPa T = 0°C = 273Kn = ?m = ? 0.112 mol O2 32.0 g O2

1 mol O2 = 3.58 g O2 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?

V1 = 2.50L V2 = ?T1 = 96.2°C = 369.2K T2 = 293K

molK

KmolkPaL

LkPa

RT

PVn 112.0

)273)(31.8(

)50.2)(3.101(


V = 2.50LP = 101.3kPa T = 0°C = 273Kn = ?m = ? 0.112 mol O2 32.0 g O2

1 mol O2 = 3.58 g O2 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?

V1 = 2.50L V2 = ?T1 = 96.2°C = 369.2K T2 = 293K

V2 = 1.99L

)293(

)(

)2.369(

)50.2(

2

2

2

1

1

K

V

K

L

T

V

T

V

molK

KmolkPaL

LkPa

RT

PVn 112.0

)273)(31.8(

)50.2)(3.101(

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?

V1 = 5.0L V2 = ?P1 = 101.3kPa P2 = 200kPaT1 = 273K T2 = 248K

V2 = 2.3L38. 6.00 g of NH4

+ are introduced into a 0.50 L flask at –200°C. What is the pressure of the gas in kPa?

Hint: You’ll need to first convert 6.00 g of NH4+ to moles. How?

Periodic Table, of course!

V = 0.50L 6.00 g NH4+ 1 mol NH4

+ 18.0 g NH4

+ = 0.33 molT = 73KP = ?

)248(

)200)((

)273(

)3.101)(0.5(

2

2

22

1

11

K

kPaV

K

kPaL

T

PV

T

PV

kPaL

KKmol

kPaLmol

V

nRTP 404

)5.0(

)73)(31.8)(33.0(

Chemistry – Ch13,14 Review + phase diagrams1. 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 27C. What will be the pressure if the temp is lowered to –173C?

13. Five liters of air at –50C is warmed to 100C. What is the new volume if the pressure remains constant?

14. A 5.0L sample of air at a temp of –50C has a pressure of 107 kPa. What is the new pressure if the temp is raised to 100C and the volume expands to 7.0L?

15. A 5.00L flask at 25C contains 0.200 mol of Cl2. What is the pressure of the flask?

16. What volume will 12.0g of oxygen gas occupy at 25C and a pressure of 52.7 kPa?

1) Determine: a) the critical temp for waterb) 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

Pressure(kPa)

101.3

22,100

3741000

SolidLiquid

Gas

Temperature (C°)


chapter 13 - states of matter kinetic theory and the nature of glass all matter is composed of tiny...

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