The Gas Laws

Chapter 14

The Kinetic Molecular Theory

Gas particles do not attract or repel each other

Gas particles are very small with large amounts of space between them

Gas particles are in constant random motion Gas particles have elastic collisions – no

energy lost Gas particles have the same average kinetic

energy at the same temperature

Gas Pressure

Gas particles exert pressure when they collide with the walls of their containers

Temperature, volume, and the number of moles affect the pressure that a gas exerts

Pressure UnitsSI unit for pressure – Pascal (Pa)101.3 kPa = 1 atm760 mm Hg = 1 atm760 torr = 1 atm

Gas Laws

4 factors that affect gases – when one changes it changes the other factors

Volume Temperature Pressure Moles (# of particles)

Boyle’s Law

As the volume of a container of gas decreases, then the pressure of that gas increases

This is an inverse relationship (as one goes up the other goes down)

Temperature remains constant

P1V1 = P2V2

Boyle’s Law – Sample Problem

P1V1 = P2V2

P1 = 1.2 atm check to make sure your

P2 = x your units are the same

V1 = 3.5 L V2 = 6.4 L

(1.2)(3.5) = P2(6.4)

= P2 0.66 atm = P2

)4.6(

)5.3)(2.1(

Boyle’s Law – Sample Problem

P1V1 = P2V2

P1 = 7.5 atm check to make sure your

P2 = 10.3 atm your units are the same

V1 = x V2 = 2.65 L

(7.5)V1 = (10.3)(2.65)

= V1 3.6 L = V1

)5.7(

)65.2)(3.10(

Boyle’s Law

Homework – pg. 422 # 1-5

Charles’ Law As the temperature of a gas increases so

does its volume This is a direct relationship (the both change

in the same direction) Pressure remains constant

=

Temperature must be in Kelvin K = Celsius + 273

1

1

T

V

2

2

T

V

Charles’ Law – Sample Problem

V1 = 3.4 L check to make sure your

V2 = 7.8 L your units are the same

T1 = 45°C (45 + 273 = 318) T2 = x = cross multiply

= T2 729 K = T2

729 – 273 = 456°C

460°C

318

4.3

2T

7.8

(3.4)

)318)(8.7(

Gay Lussac’s Law

As the pressure of a gas increases so does its temperature

This is a direct relationship (the both change in the same direction)

Volume remains constant

=

Temperature must be in Kelvin

1

1

T

P

2

2

T

P

Gay Lussac’s Law – Sample Problem

P1 = 4.52 atm check to make sure your

P2 = x your units are the same

T1 = 22°C (22 + 273 = 295)

T2 = 315 K

= cross multiply

= P2 4.8 atm = P2

295

52.4

315

P2

(295)

)315)(52.4(

Charles’ and Gay Lussac’s Law

Homework pg. 425 # 6-8

pg. 427 # 9-13

Combined Gas Law

Combines pressure, volume, and temperature

Amount of gas (moles) is constant

Combined Gas Law - Example

A sample of nitrogen monoxide has a volume of 72.6 mL at a temperature of 16°C and a pressure of 104.1 kPa. What volume will the sample occupy at 24°C and 99.3 kPa?

P1 = 104.1 kPa T1 = 16°C

P2 = 99.3 kPa T2 = 24°C

V1 = 72.6 ml

V2 = X

Combined Gas Law - Example

297

)V)(3.99( 2

First convert your temperature to Kelvin

T1 = 16 + 273 = 289 K

T2 = 24 + 273 = 297 K

289

)6.72)(1.104(=

(104.1)(72.6)(297) = (289)(99.3)V2

)3.99 289(

297 6.72 1.104

x

xx= V2

78.2ml = V2

Combined Gas Law - Example

P1 = 98.0 kPa T1 = 25°C + 273 = 298

P2 = x T2 = 60°C + 273 = 333

V1 = 1.5 L

V2 = 3.2 L 298

)5.1)(0.98(333

)2.3)(P( 2=

)298 2.3(

333 5.1 0.98

x

xx= P2 51.3 kPa = P2

Combined Gas Law - Homework

Practice problems #19-23 pg. 430

Ideal Gas Law

Particles take up no space Particles have no intermolecular attractive

forces NO GAS IS TRULY IDEAL! PV = nRTP = pressure , V = volume, T = temperaturen = number of molesR = ideal gas constant (depends on units of pressure)

Ideal Gas Law

The value of R depends on the pressure units:

Value of R Units of Pressure

0.0821 atm

8.314 kPa

62.4 mm Hg ; torr

Volume = Liter

Temp = Kelvin

n = moles

Ideal Gas Law – Example

P = 2.5 atm

V = 3.2 L

T = 47 °C + 273 = 320 K

PV = nRT

Which R value do you use? atm = 0.0821

(2.5)(3.2) = n (0.0821)(320)

)320)(0821.0(

)2.3)(5.2(= n 0.30mol = n

Ideal Gas Law - Example

n = 0.30 mol If the molar mass of this gas is 20.2 g/mol.

How many grams of this gas do you have?

What is this gas?

0.30 mol x mol1

g2.20= 6.06 g

If the molar mass is 20.2 g/mol then the gas is:

Neon

Ideal Gas Law – ExampleP = 652 mm Hg

V = 17.5 L

T = 27 °C + 273 = 300 K

PV = nRT

Which R value do you use?

62.4

(652)(17.5) = n (62.4)(300)

)300)(4.62(

)5.17)(652(= n 0.61 mol = n

Ideal Gas Law – Homework

Practice problems # 41-45 pg. 437