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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

- 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

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 ElementsHelium

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 ElementsCarbon 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 GasesOne Mole of Oxygen Gas (O2) Contains 6.02 x 1023 molecules

- Has a mass of 32 grams
- Occupies 22.4 liters at STP
- 273 Kelvins (0oC)
- One atmosphere (101.32 kPa)(760 mm)

(Avogadro’s Number)

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 (N2)

- Has a mass of 28 grams
- Occupies 22.4 liters at STP
- Contains 6.02 x 1023 molecules

Standard

Temperature

Standard Pressure

22.4 liters/mole

0 oC

273 Kelvins

1 atmosphere

101.32 kilopascals

760 mm Hg

Standard Conditions (STP)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 ConversionsCharles’ 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

- 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

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

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 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

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

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 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

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

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 Solution

P1V1 = P2V2

T1 T2

98 (2.0) = 101.32 V2

298 273

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

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 Solution

P1V1 = P2V2

T1 T2

V2= 273 (98) (2.0)

(298) (101.32)

V2=53508=1.77 liters

30193

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

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]

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

PV = nRT

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

V = volume in liters

n = moles

T = temperature in Kelvins

R = universal gas constant

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 = nRTDeriving 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

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 = PV

nT

R = (101.32) (22.4)

(1) 273

R = 8.3kPa Liters

mole Kelvins

Ideal Gas Law - Pressure

PV = nRT

P = nRT

V

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

Ideal Gas Law - Volume

PV = nRT

V = nRT

P

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

Ideal Gas Law - Temperature

PV = nRT

T = PV

nR

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

Ideal Gas Law - Moles

PV = nRT

n = PV

RT

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

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)

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)

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

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 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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