Unit 10 Gas Laws. I. Kinetic Theory Particles in an ideal gas… 1.gases are hard, small, spherical...

Unit 10Unit 10Gas LawsGas Laws

I. Kinetic TheoryI. Kinetic Theory Particles in an ideal gas…

1.gases are hard, small, spherical particles

2.don’t attract or repel each other. 3.are in constant, random,

straight-line motion.4.indefinite shape and volume.5.have “perfectly” elastic

collisions.

A. Graham’s LawA. Graham’s Law• DiffusionDiffusion

– The tendency of molecules to move toward areas of lower concentration.•Ex: air leaving tire when valve is opened

• EffusionEffusion– Passing of gas molecules through a tiny opening in a container

A. Graham’s LawA. Graham’s Law

Which one is Diffusion and which one is Effusion?

DiffusionEffusion

Tiny opening

II. Factors Affecting Gas II. Factors Affecting Gas PressurePressure

A. Amount of GasAdd gas - ↑ pressureRemove gas - ↓ pressureEx: pumping up a tire adding air to a balloon aerosol cans

B. VolumeReduce volume - ↑ pressureIncrease volume - ↓ pressure Ex: piston in a car

II. Factors Affecting Gas II. Factors Affecting Gas Pressure Pressure

C. TemperatureIncrease Temp. - ↑ pressureDecrease Temp. - ↓

pressureEx: Helium balloon on

cold/hot day, bag of chips

II. Factors Affecting Gas II. Factors Affecting Gas Pressure Pressure

Gas Gas PressurePressure-- collision of gas collision of gas molecules with the walls of the molecules with the walls of the containercontainer

Atmospheric Pressure-Atmospheric Pressure- collision collision of air molecules with objectsof air molecules with objects

Atmospheric pressure is measured with a barometer.

Increase altitude – decrease pressureEx. Mt. Everest – atmospheric pressure is 253 mm Hg

Vacuum- empty space with no particles and no pressure Ex: space

Gas Pressure (Cont.) -- 3 ways to measure pressure:

»atm (atmosphere)»mm Hg»kPa (kilopascals)

U-tube Manometer

III. Variables that describe a III. Variables that describe a gasgas

VariableVariabless UnitsUnits

Pressure (P) – kPa, mm Hg, atm

Volume (V) – L , mL , cm3

Temp (T) – °C , K (convert to Kelvin)

K = °C + 273

Mole (n) - mol

Draw on the Left Side of Draw on the Left Side of Your SpiralYour Spiral

Pressure

Volume

Temperature

kPa

Mole

How pressure units are How pressure units are related:related:

1 atm = 760 mm Hg = 101.3 kPa

How can we make these into conversion factors?

1 atm 101.3 kPa760 mm Hg 1 atm

Guided Problem:Guided Problem:1. Convert 385 mm Hg to kPa

385 mm Hg

2. Convert 33.7 kPa to atm 33.7 kPa

x 101.3 kPa= 51.3 kPa

760 mm Hg

x = .33 atm101.3 kPa

1 atm

Standard Temperature and Pressure

Standard pressure – 1 atm, 760 mmHg, or 101.3 kPa

Standard temp. – 0° C or 273K

STP

Gases (cont.)Gases (cont.) Kelvin Temperature scale is directly

proportional to the average kinetic energy

A. Boyle’s Law

IV. Gas LawsIV. Gas Laws

P

V

• The pressure and volume of a gas are inversely related

-at constant mass & temp

•P1 × V1 = P2 × V2

10. The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temp remains constant?

P1 = 105 kPa P2 = 40.5 kPa V1 = 2.5 L V2 = ?

P1 × V1 = P2 × V2

(105) (2.5) = (40.5)(V2) 262.5 = 40.5 (V2) 6.48 L = V2

Example Problems Example Problems pg 335 # 10 &11pg 335 # 10 &11

11. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temp remains constant? P1 = 205 kPa P2 = ?

V1 = 4.0 L V2 = 12.0 L

P1 × V1 = P2 × V2

(205) (4.0) = (P2)(12)

820 = (P2) 12

68.3 L = P2

Example Problems Example Problems pg 335 # 10 &11pg 335 # 10 &11

B. Charles’ LawB. Charles’ Law

• The volume and temperature (in Kelvin) of a gas are directly related – at constant mass & pressure

V

T

• V1 = V2

***Temp must be in KelvinK = °C + 273

T1T2

Example Problems Example Problems pg. 337 # 12 & 13pg. 337 # 12 & 13

12. If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change?

V1= 6.8L V2 = ?

T1 = 325°C = 598 K T2 = 25°C = 298 K

6.8 = V2

598 298598 × V2 = 2026.4

598 598 V2 = 3.39 L

13. Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?

V1= 5.0L V2 = ?

T1 = -50°C = 223 K T2 = 100°C = 373 K

5 = V2

223 373(223) V2 = 1865 223 223

V2 = 8.36 L

Example Problems Example Problems pg. 337 # 12 & 13pg. 337 # 12 & 13

P

T

C. Gay-Lussac’s LawC. Gay-Lussac’s Law• The pressure and absolute

temperature (K) of a gas are directly related – at constant mass & volume

P1 = P2

T1 T2

***Temp must be in KelvinK = °C + 273

Example ProblemsExample Problems1. The gas left in a used aerosol can is at

a pressure of 103 kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C?

P1= 103 kPa P2 = ?

T1 = 25°C = 298 K T2 = 928°C = 1201 K

103 = P2

298 1201298 × P2 = 123,703

P2 = 415 kPa

Example Problem Example Problem pg. 338 # 14pg. 338 # 14

14. A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change?

P1= 6.58 kPa P2 = ?

T1 = 539 K T2 = 211 K

6.58 = P2

539 211539 × P2 = 1388

539 539 P2 = 2.58 kPa

D. Combined Gas LawD. Combined Gas Law

P1V1

T1

=P2V2

T2

Combines the 3 gas laws as follows:

•The other laws can be obtained from this law by holding one quantity (P,V or T) constant.

•Use this law also when none of the variables are constant.

How to remember each Law!How to remember each Law!

PP VV

TT

Gay-Lussac

Boyles

Charles

Cartesian Divers

Balloon and flask Demo

Fizz Keepers

E. Ideal Gas LawE. Ideal Gas Law• The 4th variable that considers the

amount of gas in the system

P1V1

T1 n=

P2V2

T2 n

• Equal volumes of gases contain equal numbers of moles (varies directly).

E. Ideal Gas LawE. Ideal Gas Law

You don’t need to memorize this value!

•You can calculate the # of n of gas at standard values for P, V, and T

PVTn

= R (1 atm)(22.4L)(273K)(1 mol) = R

UNIVERSAL GAS CONSTANT

R= 0.0821 atm∙L/mol∙K

E. Ideal Gas LawE. Ideal Gas Law

P= pressure in atmV = volume in litersn = number of molesR= 0.0821 atm∙L/mol∙KT = temperature in Kelvin

PV=nRT

E. Example ProblemsE. Example Problems1. At what temperature will 5.00g of Cl2

exert a pressure of 900 mm Hg at a volume of 750 mL?

2. Find the number of grams of CO2 that exert a pressure of 785 mm Hg at a volume of 32.5 L and a temperature of 32 degrees Celsius.

3. What volume will 454 g of H2 occupy at 1.05 atm and 25°C.

F. Dalton’s Partial Pressure F. Dalton’s Partial Pressure LawLaw

• The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + P3 + ...

F. Dalton’s LawF. Dalton’s Law• Example problem:

1. Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2

) if the total pressure is

101.3 kPa. And the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10 kPa, 0.040 kPa, and 0.94 kPa.

PO2 = Ptotal – (PN2

+ PCO2 + Pothers)

= 101.3 kPa – (79.10 kPa + 0.040 kPa + 0.94 kPa)

= 21.22 kPa

F. Dalton’s Law F. Dalton’s Law 2. A container holds three gases : oxygen ,

carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm respectively. What is the total pressure of the container?

3. A gas mixture contains oxygen, nitrogen and carbon dioxide. The total pressure is 50.0 kPa. If the carbon dioxide has a partial pressure of 21 kPa and the nitrogen has a partial pressure of 15 kPa, what is the partial pressure of the oxygen?

4. A container contains two gases – helium and argon, at a total pressure of 4.00 atm. Calculate the partial pressure of helium if the partial pressure of the argon is 1.5 atm.