CH110 Chpt 7 Gases
CH110 Chapter 8: Gases
Kinetic Molecular Theory
Pressure
Gas Laws
CH110 Chpt 7 Gases
Solid
Liquid
Vapor
Slow moving, dense,Fixed shape
Moderate movement,Dense,Takes shape of container
Fast moving, Low density,Expands to fill container
Density Shape Compressibility
Small compressibility,
Very small heat expansion
Large compressibility,
Expands w/ heat
Smallcompressibility,
Small heat expansion
CH110 Chpt 7 Gases
1. All gases are made up of tiny particles moving in • straight lines • in all directions • at various speeds.
Kinetic molecular theory of Gases
Model to explain behavior of gases
Vapor
CH110 Chpt 7 Gases
3. V of a gas = V of container
V of a gas is mostly empty space.
2. Particles far apart have no effect on each other. (Don’t attract or repel)
Kinetic molecular theory
CH110 Chpt 7 Gases
Kinetic molecular theory
4. The ave KE as the T
• The average KE is the same for all gases at the same T.
TKE
(K.E. a T)
CH110 Chpt 7 Gases
5. Gas molecules exert pressure as they collide with container walls
The > the # of collisions (per unit time), the > the pressure
Kinetic molecular theory
CH110 Chpt 7 Gases
6. E is conservedwhen colliding with each other or container walls.
For an Ideal Gas Collisions are perfectly elastic & no E is gained or lost. (Like billiard balls exchanging E.)
CH110 Chpt 7 Gases
6. E is conservedwhen colliding with each other or container walls.
For an Ideal Gas Collisions are perfectly elastic & no E is gained or lost. (Like billiard balls exchanging E.)
CH110 Chpt 7 Gases
Pressure= Force per unit of Area. Force
AreaP = Force
Area
In the atmosphere, molecules of air (N2, O2, Ar, H2O, etc..) are constantly bouncing
off us.
CH110 Chpt 7 Gases
We live at the bottom of an ocean of air
Atmospheric Pressure
Atmosphere:A sea of colorless, odorless gases surrounding the earth
CH110 Chpt 7 Gases
PressureAt higher elevations, there is less air so the P is less.
CH110 Chpt 7 Gases
Boiling Point = Temp where molecules
overcome atmospheric Pressure
Sea Level
760 torrDenver (5280’)
630 torrMt. Evans,CO(14,000’)
Mt. Everest(20,000’)
467 torr
270 torr H2O
= 100 oC
= 95 oC
= 87 oC
= 73 oC
CH110 Chpt 7 Gases
Measuring PressureAttempts to
pump water out of flooded
mines often failed because
H2O can’t be
lifted more than 34 feet.
CH110 Chpt 7 Gases
Measuring PressureTorricelli believed reason was that P of atmosphere could not hold anything heavier than a 34’ column of water.
CH110 Chpt 7 Gases
Like drinking from a straw.
What causes the liquid to move up the straw to your mouth ?
Atmospheric Pressure
CH110 Chpt 7 Gases
34’ columnof water
1 Atm
The atmosphere
would support a column of
H2O> 34 feet high.
Measuring Pressure
CH110 Chpt 7 Gases
Torricelli BarometerPressure of the atmosphere supports acolumn of Hg 760 mm high.
1 atm
1 atm =760 mm Hg760 torr29.92 in Hg14.7 lb/in2
101,325 Pa
vacuum
Mercury used because it’s so dense.
CH110 Chpt 7 Gases
Blood pressure (systolic over diastolic):most often in mm Hg. (ex. 120/80)
Meteorologists refer to pressure systems in mm or inches of Hg. ex. 30.01 in
CH110 Chpt 7 Gases
STPStandard Temperature & Pressure
1 atm
1 atm =760 mm Hg760 torr29.92 in Hg14.7 lb/in2
101,325 Pa
0oC
273K
CH110 Chpt 7 Gases
Gas lawsLaws that show relationships between volume and properties of gasesBoyle’s LawCharles’ LawGay-Lussac’s Law
Avogadro’s LawDalton’s Law
CombinedGas Law
CH110 Chpt 7 Gases
V is inversely proportional to P when T is constant.
If P goes down V goes up
P
V
P V
P
V
Boyle’s law: V vs P
CH110 Chpt 7 Gases
P1 = 1 Atm
V1 = 1 L
P1V1 = P2V2 V2 =
P1V1 = V2
P2
1atm (1L) =
0.5 atm2 L
Boyle’s law: V vs P
2 L
P2 = 0.5 Atm
CH110 Chpt 7 Gases
1 L
Boyle’s law: V vs P2 L
Drive to top of mountain - ears start popping.
Breathing at high altitudes is more difficult because the pressure of O2 is less.
CH110 Chpt 7 Gases
It all “Boyle’s” down to Breathing in and out.
Boyle’s law
CH110 Chpt 7 Gases
Charles’s law: V vs TThe volume of a gas is directly proportional to the absolute temperature (K).
T V
P
If T goes up V goes up
CH110 Chpt 7 Gases
V1 = 125 mL
T1 = 273 K
Charles’s law: V vs T V1 = V2
T1 T2
V1 = V2
T1 T2
V2 =
T2 = 546 K
250 mL
(546K)125 mL = 273 K
T2V1 = V2
T1
CH110 Chpt 7 Gases
Gay-Lussac’s Law (PT)Pressure of a gas is directly proportional to
Absolute Temp (K) when Volume is constant
P1 = P2
T1 T2
P1 = P2
T1 T2
P T
V
If P goes up T goes up
CH110 Chpt 7 Gases
Boyle’s
Gay-Lussac’s
Charles’
PT
VV
T VP
TP
VGas LawsP1V1 = P2V2
V1 = V2
T1 T2
P1 = P2
T1 T2
CH110 Chpt 7 Gases
Boyle’s
Gay-Lussac’s
Charles’
CombinedGas Law
PT
VV
T VP
TP
VGas Laws
P1V1
T1
= P2V2
T2
CH110 Chpt 7 Gases
A 10 m3 balloon contains helium on the ground where the temperature is 27ºC and the pressure is 740 torr. Find the volume at an altitude of 5300 m if pressure is 370 mm Hg and temperature is -33 ºC.
P1 = 740 mm
T1 = 27 + 273 = 300 K
V1 = 10 m3
P2 = 370 mm
T2 = -33 + 273 = 240 K
V2 = ?
= 16 m3V2 = (240 K)(740 mm)(10 m3 )
(370 mm) (300 K)
P1V1
T1
= P2V2
T2
T2P1V1
P2 T1
= V2
Combined Gas Law
CH110 Chpt 7 Gases
Avogadro’s lawThe volume of a gas is directly
proportional to the number of molecules
V1 = V2
n1 n2
V1 = V2
n1 n2
More moles of a gas, takes up more space.
CH110 Chpt 7 Gases
At Standard Temperature & Pressure (STP)
V of 1 mole of gas = 22.4 liters
Equal volumes of gas (at same T and P)
contain equal numbers of molecules.
Avogadro’s law
At T = 273 K (0ºC) P = 1 atm (760 mm)
1 mol He
4 g He
22.4 L
1 mol He
4 g He
22.4 L
1 mol N2
28 g N2
22.4 L
1 mol N2
28 g N2
22.4 L
1 mol CO2
44 g CO2
22.4 L
1 mol CO2
44 g CO2
22.4 L
CH110 Chpt 7 Gases
Standard conditions (STP)When 2.00 mol of liquid H2O is vaporized,
how may liters of gas will there be?
2 mole H2O
1
22.4 liters
1 mole H2O= 44.8 L
22.4 liters
1 mole H2O1 mole H2O
22.4 liters
CH110 Chpt 7 Gases
Dalton’s law of Partial Pressures
The total pressure of a gas mix = sum of the partial pressures of each gas.
Pair = PN2 + PO2 + PAr + PCO2 + PH2O
PT = P1 + P2 + P3 + .....
Each gas acts independently of the others.
Example: Air
CH110 Chpt 7 Gases
Pair = PN2 + PO2 + PCO2 + PH2O = 760 mm
Dalton’s law of Partial Pressures
Typical values for Atmospheric air at 0 ºC (excluding argon):
(594.0mm)+(160mm) +(5.7mm)=+(0.3mm) 760mm
As T of air increases, more H2O enters mix.
example: at 20 ºC, the PH2O = 18 mm Ptotal (760 mm) can’t change, so other
gases get diluted to make room for the water.
CH110 Chpt 7 Gases
Air moving over warm water
has more water in it.
Low pressure
is often associated with this air.
Typhoons and hurricanes
are associated with very warm, moist air.
Pair = PN2 + PO2 + PCO2 + PH2O = 760 mm
CH110 Chpt 7 Gases
Blood Gases
PCO2 ~ 40 mm Hg
Normal PO2 in the air =160 mm.
If drops
< 100 mm,
can’t diffuse into the blood.
Arterial Blood Gases (ABGs)
PBG = PO2 + PCO2
PO2 ~ 100 mm Hg
PCO2 ~ 46 mm Hg
Venous Blood Gases (VBGs)
PO2 ~ 40 mm Hg
CH110 Chpt 7 Gases
We only use about 25% of the Oxygen we inhale.
The rest is exhaled along with the Nitrogen and some carbon dioxide.
THIS IS WHY CPR WORKS !!!
CH110 Chpt 7 Gases
Bernoulli's Principle
Faster moving gases exert less pressure than slow moving gases.
Fast moving Gases Low P
Slow moving Gases
High P
CH110 Chpt 7 Gases
Bernoulli's PrincipleSlow moving
Gases
Fast moving Gases
High P
Low P