Ch. 12 Behavior of Gases

Gases

• Gases expand to fill its container, unlike solids or liquids

• Easily compressible: measure of how much the volume of matter decreases under pressure

Variables that describe a gas

• Pressure (P)– Measured in kilopascals, kPa– Pressure and number of molecules are directly

related increase molecules = increase pressure

– Gases naturally move from areas of high pressure to low pressure, due to the available space to move into

Gas Pressure- collision of gas molecules with the walls of the container

Variables that describe a gas

• Volume (V)– Measured in Liters, L– Volume and pressure are inversely related

• As volume decreases, the pressure increases• Smaller container = less room for movement, therefore

molecules hit sides of container more often

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

II. Factors Affecting Gas Pressure

Variables that describe a gas

• Temperature (T)– Measured in Kelvin, K– The temperature and pressure are directly related

• Increase in temp = increase in pressure• Volume must be held constant• Molecules hit the walls harder (due to increase in K.E.)

and more frequently. Think about a tire in hot weather…

Variables that describe a gas

• Amount– Measured in moles, mol– Moles and pressure are directly related

• Increase in # of moles = increase in pressureEx: Inflating a balloon is adding more molecules.• Temperature must remain constant

II. Factors Affecting Gas Pressure

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

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

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

U-tube Manometer

III. Variables that describe a gasVariable

s UnitsPressure (P) – kPa, mm Hg, atmVolume (V) – L , mL , cm3

Temp (T) – °C , K (convert to Kelvin)

K = °C + 273Mole (n) - mol

How pressure units are 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:

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 kPa1 atm

Standard Temperature and Pressure

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

Standard temp. – 0° C or 273K

STP

Gas Laws

• Describe how gases behave• Change can be calculated• Know the math and the theory!!

Boyle’s Law (1662)

• Gas pressure is inversely related to volume (as volume increases, pressure decreases)

• Temperature is constant

P1V1= P2V2

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 pg 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 pg 335 # 10 &11

Charles’s Law (1787)

• Volume is directly proportional to temp. (increase volume, increase temp)• Pressure is constant

=

Example Problems pg. 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 K6.8 = V2

598 298598 × V2 = 2026.4598 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 pg. 337 # 12 & 13

Gay-Lussac’s Law (1802)

• Pressure and temperature are directly related(Increase pressure= Increase

temperature)• Volume is constant!

Example Problems

1. 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 = 123,703P2 = 415 kPa

Example Problem pg. 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 = 1388539 539

P2 = 2.58 kPa

Combined Gas Law

• Combines 3 gas laws: Boyle’s, Charles’, and Gay-Lussac’s • Used when it is difficult to hold any one variable (P, V, or

T) constant

=

• Can take away any variable that is constant– Take temp away = Boyle’s– Take Pressure away = Charle’s– Take Volume away = Gay-Lussac’s

How to remember each Law!

P V

T

Gay-Lussac

Boyles

Charles

Cartesian Divers

Balloon and flask Demo

Fizz Keepers

Ex: 3.0 L of Hydrogen gas has a pressure of 1.5 atm at 20oC. What would the volume be if the pressure increased to 2.5 atm at 30oC?

Ideal Gas Law• Used for gases that behave “ideally”• Allows you to solve for # of moles of a contained gas when

P, V, and T are known. • Use constant R=8.31

P(pressure)- must be in kPaV (volume)- must be in Ln (# of moles)- muse be in moles of gasR- gas constantT (Temperature)- Must be in Kelvin (oC + 273= K)

Ideal Gas Law• A gas behaves “ideally” if it conforms to the gas laws

– Gases do not usually do this– Real gases only behave this way at:

1. High temps (molecules move fast)2. Low pressure (molecules are far apart)• This is because gases will stay a gas under these conditions

– Molecules are not next to each other very long so attractive forces can’t play a role b/c molecules are moving too fast

– Ideal Gases do no exist because:1. Molecules do take up space2. There are attractive forces between molecules otherwise no liquid

would form. (Molecules slow down to become liquids)

E. 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 Law

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

PV=nRT

E. Example Problems

1. 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.

Ex: What volume will 2.0 mol of N2 occupy at 720 torr and 20oC?

Dalton’s Law of Partial Pressures

• Used for mixture of gases in a container• If you know the P exerted by each gas in a

mixture, you can calculate the total gas pressure

• It is particularly useful in calculating pressure of gases collected over water.

Ptotal = P1 + P2 + P3…*P1 represents the “partial pressure” or the contribution by the gas

F. Dalton’s Partial Pressure Law

• 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 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 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.

Graham’s Law of Effusion• Rate of effusion and diffusion are inversely proportional to the

square root of the mm of molecules – Effusion: Gas escaping through tiny holes in a container– Diffusion: movement from area of high concentration to low

concentration (ex: perfume spreading across a room)(Both depend of the mm of the molecule, which determines speed)

= • Type of Molecule is important

– Gases with lower mm effuse/diffuse faster– Ex: Helium diffuses/effuses faster than Nitrogen from a balloon b/c

Helium moves faster due to lower mm.Big = Slow small = Fast

Ex:

A. Graham’s Law

• Diffusion– The tendency of molecules to move toward

areas of lower concentration.• Ex: air leaving tire when valve is opened

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

A. Graham’s Law

Which one is Diffusion and which one is Effusion?

DiffusionEffusion

Tiny opening