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The Gas Laws. Chemistry Dr. May. Gaseous Matter. Indefinite volume and no fixed shape Particles move independently of each other Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids. Avogadro’s Number.

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the gas laws

The Gas Laws

Chemistry

Dr. May

gaseous matter
Gaseous Matter
  • Indefinite volume and no fixed shape
  • Particles move independently of each other
  • Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids
avogadro s number
Avogadro’s Number
  • One mole of a gas contains Avogadro’s number of molecules
  • Avogadro’s number is

6.02 x 1023 or

602,000,000,000,000,000,000,000

diatomic gas elements
Gas

Hydrogen (H2)

Nitrogen (N2)

Oxygen (O2)

Fluorine (F2)

Chlorine (Cl2)

Molar Mass

2 grams/mole

28 grams/mole

32 grams/mole

38 grams/mole

70 grams/mole

Diatomic Gas Elements
inert gas elements
Gas

Helium

Neon

Argon

Krypton

Xenon

Radon

Molar Mass

4 grams/mole

20 grams/mole

40 grams/mole

84 grams/mole

131 grams/mole

222 grams/mole

Inert Gas Elements
other important gases
Gas

Carbon Dioxide

Carbon Monoxide

Sulfur Dioxide

Methane

Ethane

Freon 14

Formula Molar Mass

CO2 44 g/mole

CO 28 g/mole

SO2 64 g/mole

CH4 16 g/mole

CH3CH3 30 g/mole

CF4 88 g/mole

Other Important Gases
one mole of oxygen gas o 2
One Mole of Oxygen Gas (O2)
  • Has a mass of 32 grams
  • Occupies 22.4 liters at STP
      • 273 Kelvins (0oC)
      • One atmosphere (101.32 kPa)(760 mm)
  • Contains 6.02 x 1023 molecules

(Avogadro’s Number)

mole of carbon dioxide co 2
Mole of Carbon Dioxide (CO2)
  • Has a mass of 44 grams
  • Occupies 22.4 liters at STP
  • Contains 6.02 x 1023 molecules
one mole of nitrogen gas n 2
One Mole of Nitrogen Gas (N2)
  • Has a mass of 28 grams
  • Occupies 22.4 liters at STP
  • Contains 6.02 x 1023 molecules
mole of hydrogen gas h 2
Mass

Volume at STP

Molecules

2.0 grams

22.4 liters

6.02 x 1023

Mole of Hydrogen Gas (H2)
standard conditions stp
Molar Volume

Standard

Temperature

Standard Pressure

22.4 liters/mole

0 oC

273 Kelvins

1 atmosphere

101.32 kilopascals

760 mm Hg

Standard Conditions (STP)
gas law unit conversions
liters  milliliters

milliliters  liters

o C  Kelvins

Kelvins  o C

mm  atm

atm  mm

atm  kPa

kPa  atm

Multiply by 1000

Divide by 1000

Add 273

Subtract 273

Divide by 760

Multiply by 760

Multiply by 101.32

Divide by 101.32

Gas Law Unit Conversions
charles law
Charles’ Law
  • At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins

V1 = V2

T1 T2

As the temperature goes up, the volume goes up

boyle s law
Boyle’s Law
  • At constant temperature, the volume of a gas is inversely proportional to the pressure.

P1V1 = P2V2

As the pressure goes up, the volume goes down

combined gas law
Combined Gas Law

P1V1 = P2V2

T1 T2

  • Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hg
  • Standard Temperature (T) is 273 K
  • Volume (V) is in liters, ml or cm3
charles law problem
Charles’ Law Problem

A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?

1. Convert oC to Kelvins

25 + 273 = 298 K

38 + 273 = 311 K

2. Insert into formula

charles law solution
Charles’ Law Solution

V1 = V2

T1 T2

V1 = 2 liters V2 = Unknown

T1 = 298 K T2 = 311 K

2 = V2

298 311

charles law solution1
Charles’ Law Solution

2 = V2

298 311

298 V2 = (2) 311

V2 = 622

298

V2 = 2.09 liters

charles law problem answer
Charles’ Law Problem Answer

A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?

V2 = 2.09 liters

boyle s law problem
Boyle’s Law Problem

A balloon has a volume of 2.0 liters at

743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?

1. Convert pressure to the same units

743  760 = .98 atm

2. Insert into formula

boyle s law solution
Boyle’s Law Solution

P1V1 = P2V2

P1 = 0.98 atm P2 = 2.5 atm

V1 = 2.0 liters V2 = unknown

0.98 (2.0) = 2.5 V2

boyle s law solution1
Boyle’s Law Solution

P1V1 = P2V2

0.98 (2.0) = 2.5 V2

V2= 0.98 (2.0)

2.5

V2 = 0.78 liters

boyle s law problem answer
Boyle’s Law Problem Answer

A balloon has a volume of 2.0 liters at

743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?

V2 = 0.78 liters

combined gas law problem
Combined Gas Law Problem

A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?

1. Convert 25 oC to Kelvins 25 + 273 = 298 K

2. Standard pressure is 101.32 kPa

3. Standard temperature is 273 K

4. Insert into formula

combined gas law solution
Combined Gas Law Solution

P1V1 = P2V2

T1 T2

P1 = 98 kPa P2 = 101.32 kPa

V1 = 2.0 liters V2 = unknown

T1 = 298 K T2 = 273 K

combined gas law solution1
Combined Gas Law Solution

P1V1 = P2V2

T1 T2

98 (2.0) = 101.32 V2

298 273

(298) (101.32) V2 = (273) (98) (2.0)

combined gas law solution2
Combined Gas Law Solution

P1V1 = P2V2

T1 T2

(298) (101.32) V2 = (273) (98) (2.0)

V2= 273 (98) (2.0)

(298) (101.32)

combined gas law solution3
Combined Gas Law Solution

P1V1 = P2V2

T1 T2

V2= 273 (98) (2.0)

(298) (101.32)

V2=53508=1.77 liters

30193

combined gas law problem answer
Combined Gas Law Problem Answer

A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?

V2 = 1.77 liters

combined gas law v 2
Combined Gas Law – V2

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

P1V1T2= V2

P2T1

the end
The End
  • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology
  • Please send suggestions and comments to [email protected]
the ideal gas law

The Ideal Gas Law

Chemistry

Dr. May

kinetic molecular theory
Kinetic Molecular Theory

Molecules of an ideal gas

  • Are dimensionless points
  • Are in constant, straight-line motion
  • Have kinetic energy proportional to their absolute temperature
  • Have elastic collisions
  • Exert no attractive or repulsive forces on each other
ideal gas law
Ideal Gas Law

PV = nRT

P = pressure in kilopascals (kPa) or atmospheres (atm)

V = volume in liters

n = moles

T = temperature in Kelvins

R = universal gas constant

ideal gas law pv nrt
Pressure (P)

Volume (V)

Moles (n)

Temperature (T)

The universal gas constant (R)

Atm or kPa

Always liters

Moles

Kelvins

0.0821 ( P in atm) or

8.3 (P in kPa)

Ideal Gas Law: PV = nRT
universal gas constant
Universal Gas Constant

R = 0.0821 if P = atmospheres

R = 8.3 if P = kilopascals

R = PV

nT

deriving r for p in atmospheres
Deriving R for P in Atmospheres

R = PV

nT

Assume n = 1 mole of gas

Standard P = 1atmosphere

Standard V = molar volume = 22.4 liters

Standard T = 273 Kelvins

r value when p is in atmospheres
R Value When P Is In Atmospheres

R = PV

nT

R = (1) (22.4)

(1) 273

R = 0.0821atm Liters

mole Kelvins

deriving r for p in kilopascals
Deriving R For P In Kilopascals

R = PV

nT

Assume n = 1 mole of gas

Standard P = 101.32 kilopascals

Standard V = molar volume = 22.4 liters

Standard T = 273 Kelvins

r value when p is in kilopascals
R Value When P Is In Kilopascals

R = PV

nT

R = (101.32) (22.4)

(1) 273

R = 8.3kPa Liters

mole Kelvins

ideal gas law pressure
Ideal Gas Law - Pressure

PV = nRT

P = nRT

V

Solves for pressure when moles, temperature, and volume are known

ideal gas law volume
Ideal Gas Law - Volume

PV = nRT

V = nRT

P

Solves for volume when moles, temperature, and pressure are known

ideal gas law temperature
Ideal Gas Law - Temperature

PV = nRT

T = PV

nR

Solves for temperature when moles, pressure, and volume are known

ideal gas law moles
Ideal Gas Law - Moles

PV = nRT

n = PV

RT

Solves for moles when pressure, temperature, and volume are known

ideal gas law problem
Ideal Gas Law Problem

What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?

  • V = 2.3 liters
  • P = 1.2 atmospheres
  • T = 25 oC = 298 Kelvins
  • R = 0.0821 since P is in atms.
  • Find moles (n), then grams
ideal gas law solution moles
Ideal Gas Law Solution (moles)

PV = nRT

1.2 (2.3) = n (0.0821) (298)

n = 1.2 ( 2.3) = 0.11 moles

(0.0821) (298)

ideal gas law solution grams
Ideal Gas Law Solution (Grams)

Grams = moles x molecular weight (MW)

  • Moles = 0.11
  • Molecular Weight of N2 = 28 g/mole
  • Grams = 0.11 x 28 = 3.1 grams
ideal gas law answer
Ideal Gas Law Answer

What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?

The answer is 0.11 moles and

3.1 grams

the end1
The End
  • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology
  • Please send suggestions and comments to [email protected]
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