Gas Laws Quest Study Guide - Sophomore Year

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38 Terms

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KINETIC MOLECULAR THEORY

KINETIC MOLECULAR THEORY

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there are _ Posutlates

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The postulates are:

  • gas particles are in constant motion

  • the combined volume of the particles of gas sample is negative compared to the volume of a container

  • there are no attractive forces between the particles of a gas sample (so we assume)

  • collisions between the gas particles ate completely elastic (meaning no energy is used)

  • the average kinetic energy of gas particles in a sample is directly proportional to the kelvin temperature

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Pressure =

force applied over a specific area

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Why do gas particles exert pressure?

in constant motion, the particles collide with the walls of the container

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How can we increase pressure?

  1. increase temperature

  2. decrease volume of the container

  3. add more gas particles

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Units for pressure

atm = atmosphere (of pressure)

Kpa = kiloPascal

mmHg = millimeter of Mercury (torr)

psi = pounds per square inch (tire pressure)

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STP VALUES

1 atm = 101.3 kPa = 760 mmHg(torr)

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How many atm is 252 kPa?

252 kPa x 1 atm/101.3 kPa = 2.49 atm

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BOYLE’S LAW

BOYLE’S LAW

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Boyle’s law =

the relationship between pressure and volume

inversely proportional relationship

Equation = P1V1 = P2V2

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example

P1V1 = P2V2

(103.1kPa)(1.00L) = P2 (0.35L)

divide both sides of 0.35L

P2 = 295 kPa

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CHARLES’S LAW

relationship between temperature and volume

directly proportional relationship

Equation = V1/T1 = V2/T2

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Temperature Must be in..

Kelvin (because it does not have any negative numbers)

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To convert celsius to Kelvin…

add 273 to the celsius

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Gay-Lussac’s Law

relationship between pressure and temperature

directly proportional relationship

Equation = P1/T1 = P2/T2

Temp must be in kelvin

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Combined Gas Law

looks at relationship between pressure,volume, and temperature when all 3 values are changing

Equation = P1V1/T1 = P2V2/T2

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STP =

1 atm, O degrees Celsius, 273 kPa

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The Ideal Gas Law

no change in conditions

Equation = PV = nRT

R = gas law constant

n = number of moles

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IDEAL GAS LAW CONSTANT

0.08206 L x atm/mol x K

8.314 L x kPa/mol x K

62.36 L x torr/mol x K

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To calculate the number of moles to grams it is…

number of moles (n) x molar mass/1 mole

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Avogradro’s Law

relationship between volume and moles of a gas

directly proportional relationship

Equation = V1 / n1 = V2 / n2

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n =

moles of a substance

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Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases is the sum of the partial pressures of each component

P total = Pa + Pb + Pc + …

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Notes about Dalton’s Law

most gas samples are mixtures

Air → Mixture

O2, N2, CO2, H2O, others

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The pressure due to any one of those components in the mixture is called…

partial pressure

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Partial pressure of any component =

mole fraction x P total

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Mole Fraction =

Xn

(the X is a little curvy on the top left corner and bottom right corner)

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EXAMPLE

EXAMPLE

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What fraction of the total moles of the mixtures comes from each component?

mixture of O2, H2, N2

3 moles O2, 4 moles H2, 4 moles N2

XO2 = 3/11

XN2 = 4/11

XH2 = 4/11

The total pressure of the mixture above is 122.5 KPa

122.5 KPa = P total

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Calculate the partial pressure of O2

The Partial Pressure of any component equals the mole fraction of that component times the total pressure

PO2 = XO2 x Ptotal

= 3/11 × 122.5 KPa

= 33.4 KPa

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In the mole ratio of O2 (3/11)…

the 3 is the number of moles of the component

the 11 is the moles added

(ex: 4 + 4 + 3 = 11)

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Graham’s Law of Effusion

relates the speed at which a gas moves to its molar mass

Inverse Relationship (not proportional)

Equation =

speed of Gas A/ Speed of Gas B = √M.M of B/ M.M. of A

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Heavier =

Lighter =

heavier = slower

lighter = faster

(higher molar mass = lower speed)

(lower molar mass = faster speed)

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Gases move in _ ways

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One way Gases Move: Effusion

the movement of gas particles escaping from a small hole in a container

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Second way Gases move: Diffusion

movement from an area of high concentration to low concentration

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Example of Graham’s Law

X/H2

(1/50)2 = (√2.02/x)2

1/2,500 = 2.02/x

x = 5,050 g/mol