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The Gas Laws

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The Gas Laws

Chemistry

Dr. May

- 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

- 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

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

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

Gas

Carbon Dioxide

Carbon Monoxide

Sulfur Dioxide

Methane

Ethane

Freon 14

Formula Molar Mass

CO244 g/mole

CO28 g/mole

SO264 g/mole

CH416 g/mole

CH3CH330 g/mole

CF488 g/mole

- 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)

- Has a mass of 44 grams
- Occupies 22.4 liters at STP
- Contains 6.02 x 1023 molecules

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

Mass

Volume at STP

Molecules

2.0 grams

22.4 liters

6.02 x 1023

Molar Volume

Standard

Temperature

Standard Pressure

22.4 liters/mole

0 oC

273 Kelvins

1 atmosphere

101.32 kilopascals

760 mm Hg

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

- 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

- 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

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

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

V1 = V2

T1 T2

V1 = 2 litersV2 = Unknown

T1 = 298 KT2 = 311 K

2 = V2

298 311

2 = V2

298 311

298 V2 = (2) 311

V2 = 622

298

V2 = 2.09 liters

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

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

P1V1 = P2V2

P1 = 0.98 atmP2 = 2.5 atm

V1 = 2.0 litersV2 = unknown

0.98 (2.0) = 2.5 V2

P1V1 = P2V2

0.98 (2.0) = 2.5 V2

V2= 0.98 (2.0)

2.5

V2 = 0.78 liters

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

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

P1V1 = P2V2

T1 T2

P1 = 98 kPaP2 = 101.32 kPa

V1 = 2.0 litersV2 = unknown

T1 = 298 KT2 =273 K

P1V1 = P2V2

T1 T2

98 (2.0) = 101.32 V2

298 273

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

P1V1 = P2V2

T1 T2

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

V2= 273 (98) (2.0)

(298) (101.32)

P1V1 = P2V2

T1 T2

V2= 273 (98) (2.0)

(298) (101.32)

V2=53508=1.77 liters

30193

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

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

P1V1T2= V2

P2T1

- 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

Chemistry

Dr. May

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

PV = nRT

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

V = volume in liters

n = moles

T = temperature in Kelvins

R = universal gas constant

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)

R = 0.0821 if P = atmospheres

R = 8.3 if P = kilopascals

R = PV

nT

R = PV

nT

Assume n = 1 mole of gas

Standard P = 1atmosphere

Standard V = molar volume = 22.4 liters

Standard T = 273 Kelvins

R = PV

nT

R = (1) (22.4)

(1) 273

R = 0.0821atm Liters

mole Kelvins

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 = PV

nT

R = (101.32) (22.4)

(1) 273

R = 8.3kPa Liters

mole Kelvins

PV = nRT

P = nRT

V

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

PV = nRT

V = nRT

P

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

PV = nRT

T = PV

nR

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

PV = nRT

n = PV

RT

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

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

PV = nRT

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

n = 1.2 ( 2.3) = 0.11 moles

(0.0821) (298)

Grams = moles x molecular weight (MW)

- Moles = 0.11
- Molecular Weight of N2 = 28 g/mole
- Grams = 0.11 x 28 = 3.1 grams

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

- 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]