Chemistry I Honors

Chemistry I Honors

Unit 9: Gases

Objectives #1-4: Introduction to the Kinetic Theory of Gases

I. The Kinetic TheoryA. Assumptions of the Kinetic Theory

Gases are composed of tiny particles that are arranged far apart from each other

Gases are composed of individual atoms such as in the element neon or molecules such as in the element oxygen


Collisions that may occur between gas particles are elastic with no net loss of energy

Gas particles are in constant, random motion

No attractive forces exist between gas particles

The temperature of a gas depends on the average kinetic energy of the particles


II. The Kinetic Theory and its Implications on the Properties of GasesThe temperature of a gas is directly related to its kinetic energyThe temperature at which no kinetic energy is present is called absolute zero; this temperature is 0 on the Kelvin scale and -273o on the Celsius scaleGases are able to expand freely due to their random motion and the lack of attractive forces between the particles of a gas


The density of gases is generally less than other substances due to the ability of gases to move freely

Gases can be compressed due to the great distances between gas particles

The ability of gases to spread out spontaneously or diffuse is due to the rapid motion of gas particles; particles with less mass and faster velocities spread out at a faster rate than heavier, slower gas particles (This is why you smell cookies baking! )


Due to their small size, gas particles can be made to pass through small openings; this characteristic of effusion depends on the velocity and molar mass of the gas particles present; the rate of effusion is directly proportional to the velocity of the gas particles and inversely proportional to the molar mass of the gas particles

When gas particles collide against a surface they exhibit pressure; which is defined as the force per unit area

Diffusion v Effusion

Objectives #5-8: Relationships Among Gas Characteristics

Common Units of Pressure:Unit Symbol Standard

ValuePascal/Kilopascal Pa/kPa 101325/101.3

Millimeters of Hg mm Hg 760

Atmospheres atm 1

Torr torr 760

Objectives #5-8: Relationships among Gas Characteristics

Pressure Conversion Examples:• Convert 2.5 atm to mmHg

2.5 atm x 760 mmHg = 1900 mmHg

1 atm• Convert 300. Pa to atm

300. Pa x 1 atm_____ = .00296 atm 101325 Pa


Dalton’s Law of Partial PressureThe total pressure of a mixture of gases is equal

to the sum of the partial pressures of each of the individual gases in the mixture

PT = P1 + P2 + P3 + ….This concept must be kept in mind when gases are

collected over water in the laboratory; the vapor pressure of water must be subtracted from the measured pressure of the gas in order to obtain the true pressure of the gas being collected

Pgas = Patm - Pwater

Interpreting Vapor Pressure Charts


The vapor pressure due to water increases with increasing temperature (see attached chart)

Molar Volumeat standard temperature and pressure

(STP), which are defined as O o C and 1 atm, 1 mole of any gas occupies 22.4 L; this is referred to as molar volume

Relationships Among Gas Characteristics


Relationships Among Gas CharacteristicsA. Amount of Gas vs. Pressure

(assumes temperature and volume are held constant)

What is happening at the particle level:

How can you tell the can is full?Gases expand to fit the space, so increasing the particles makes the can heavier !


As the amount of gas in a container increases, the pressure increases

This illustrates a direct relationship between the amt. of the gas and the pressure of the gas

Empty can = less pressure, so the can is lighter weight


B. Pressure vs. Volume of Gases (assumes constant temperature)

What is happening at the particle level: Why are balloons “over-filled” with helium?


As the pressure of a fixed amount of gas increases, its volume decreases

This illustrates an inverse relationship


C. Temperature vs. Volume (assumes constant pressure)

What is happening at the particle level: Why does tires tend to

look like they are low in air on a cold winter morning?Cold night temperatures result in slower movement of the gas particles, so the volume of the gas SEEMS to decrease. As the car moves, friction of the tire on the road increases the temp of the air in the tire.


As the Kelvin temperature of a fixed amount of gas increases, its volume increases

This illustrates a direct relationship:


D.Temperature vs. Pressure (assuming constant volume)

What is happening at the particle level: Keeping & transporting

unopened pop cans in the car in the summer is NOT advisable. Why ?

Increasing the temperature in the car increases the pressure in the can. If the pressure of the gas is greater than the pressure exerted by the wall of the can, the can will explode!


As the Kelvin temperature of a fixed amount of gas increases, its pressure increases

This illustrates a direct relationship


Ideal vs. Real GasesIdeal gases always follow the kinetic

theory under any conditionsReal gas particles do have

attractive forces among each otherReal gases no longer act as ideal

gases under conditions of high pressure and extremely low temperature

Objective #9 Solving Problems Involving the Gas Laws

*The Gas Laws are mathematical formulas based on the relationships discussed in the previous section of notes…

Combined Gas Law: P1V1 = P2V2 T1 T2

**Note that all temps must be in KELVINS!!Ideal Gas Law: (P)(V) = (n)(R)(T)

I. Combined Gas Law1. To what pressure must a gas be compressed

to get it into a 9.00 L tank if it occupies 90.0 L at 1.00 atm?

Step 1: Can a variable be eliminated? Yes; temperature is not a factor…Step 2: Substitute & Solve… P1V1 = P2V2

P2 = P1V1 = (1.00 atm) (90.0 L) V2 9.00 L P2 = 10.0 atm

2. A container with a movable piston contains .89 L of methane gas at 100.50C. If the temperature of the gas drops to 11.3oC, what is the new volume of the gas?

Step 1: Convert temps!! K = C + 273 = 100.5oC + 273 = 374 K, AND K = C + 273 = 11.3oC + 273 = 284 K

Step 2: Can a variable be eliminated? Yes; pressure is not a factor…Step 3: Substitute and solve… V1 = V2 T1 T2

(V1)(T2) = (T1)(V2) V1T2 = V2

T1

(.89 L) (284 K) = V2

374 K .676 L = V2

3. A sample of gas occupies a volume of 5.0 L at a pressure of 650. torr and a temperature of 24oC . We want to put the gas in a 100. ml container that can only withstand a pressure of 3.0 atm. What temperature must be maintained so that the container does not explode?

Step 1: Standardize units!!5.0 L = 5000 ml650. torr = .855 atm24oC = 297 K

Step 2: Can a variable be eliminated? No, all components are used…Step 3: Substitute and solve… P1V1 = P2V2

T1 T2

P1V1T2 = T1P2V2

T2 = T1P2V2 = (297 K)(3.0 atm)(100. ml) P1V1 (.855 atm) (5000 ml) T2 = 21 K

4. A sample of gas occupies 2.00 L at STP. What volume will it occupy at 27oC and 200. mm Hg?

P1V1 = P2V2 T1 T2

Can a variable be eliminated? (P1)(V1)(T2) = (T1)(P2)(V2) P1V1T2 = V2

T1P2

(760 mm Hg) (2.00 L) (300 K) = V2

(273 K) (200. mm Hg) 8.35 L = V2

II. Additional Problems Recall that at STP conditions, 1 mole

of any gas = 22.4 LExamples:1. Calculate the volume of .55000

moles of gas at STP..55000 moles x 22.4 L = 12.320 L 1 mole

2. Calculate the moles of gas contained in 350 L of gas at STP.

350 L x 1 mole = 16 moles 22.4 L3. Calculate the mass in grams of 3.50

L of chlorine gas. 3.50 L x 71.0 g = 11.1 g 22.4 L

Objective #11: The Ideal Gas Law and Gas Stoichiometry

Ideal Gas Law: (P)(V)=(n)(R)(T)R = ideal gas constant, so…

R = (P)(V) = (1 atm) (22.4 L) = .0821 L . atm.

(n)(t) (1 mole) (273K) mole.K

Examples:1. What is the temperature of a .65 L sample

of fluorine gas at 620. torr which contains 1.3 mol fluorine?

T = PV (620 torr/760 torr/atm = 0.82 atm ) nR = (0.82atm ) (.65 L) (1.3 mol) (.0821 L.atm / mol.K) = 5.0 K

2. A 25.0 gram sample of argon gas is placed inside a container with a volume of 10.0 L at a temperature of 65oC. What is the pressure of argon at this temperature?

PV=nRTP = nRT V = (25.0 g / 39.9 g) (.0821) (338 K) 10.0 L = 1.74 atm.

Gas StoichiometryThere are two types of gas

stochiometry problems:1) at STP (Standard Temp & Pressure)2) non STP

Gas Stoichmetry: Examples…

1. Calculate the volume of hydrogen gas that can be produced from the reaction of 5.00 g of zinc reacted in an excess of hydrochloric acid. Assume STP conditions.

Zn + 2 HCl H2 + ZnCl2

5.00 g Zn x 1 mole Zn x 1 mole H2 x 22.4 L H2

65.4 g Zn 1 mole Zn 1 mole H2 = 1.71 L H2

2. Calculate the volume in liters of oxygen gas that can be produced from the decomposition of 3.50 X 1024 formula units of potassium chlorate. Assume STP conditions.

2 KClO3 2 KCl + 3 O2

3.50 x 1024 formula units KClO3 x 1 mole KClO3 x 3 mole O2 x 22.4 L6.02 X 1023 f. units 2 mole KClO3 1 mole

= 195 L

3. Calculate the volume of hydrogen produced at 1.50 atm and 19oC by the reaction of 26.5 g of calcium metal with excess water. (Ignore the vapor pressure of water)

Ca + 2H2O --› H2 + Ca(OH)2

*use amount of given reactant and stoichiometry to determine moles of gas desired in problem:

26.5 g Ca X 1 mole Ca / 40.1 g Ca X1 mole H2 / 1 mole Ca = .661 mole H2

*use moles of gas found in ideal gas law to calculate volume of gas:

PV=nRTV = nRT / P = (.661 moles) (.0821) (292 K) 1.50 L = 10.6 L

4. Calculate the volume of chlorine gas produced at 1.25 atm at 25oC from the reaction of 5.00 g of sodium chloride and an excess of fluorine.

2 NaCl + F2 2 NaF + Cl2

First, find the moles of Cl2…

5.00 g NaCl x 1 mole NaCl x 1 mole Cl2 58.5 g NaCl 2 mole NaCl = .0427 mole Cl2

Now, use the ideal gas relationship…

PV = nRTV = nRT P = (.0427 mole) (.0821) (298 K) 1.25 atm = .836 L

Chemistry I Honors

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