GAS LAWS

BOYLE’S LAW

DEMO

Bell Jar and Marshmallow

- The marshmallow is getting bigger (expanding – volume increases). Why?

- How do volume and pressure relate to each other?

SHAVING CREAM DEMO

WHAT IS BOYLE’S LAW?

• The volume of a given amount of gas varies inversely with its pressure if the temperature remains constant

• Inverse Relationship – as pressure increases, the volume decrease by the same factor

• Example: If pressure doubles, the volume decreases by ½

• Example: If the pressure decreases by a factor of 4, the volume will _____________.Quadrupl

e

BOYLE’S LAW

Formula: P1V1 = P2V2

• Remember, if pressure increases from P1 to P2, then volume must _________ from V1 to V2.

• So, if pressure decreases from P1 to P2, then volume must _________ from V1 to V2.

Decrease

Increase

BOYLE’S LAW GRAPH

BOYLE’S LAW EXAMPLE

• A sample of helium gas in a balloon is compressed from 4.0L to 2.5L, at a constant temperature. If the pressure of the balloon started at 210kPa, what will the final pressure (P2) be?

BOYLE’S LAW EXAMPLE ANSWER

• Formula: P1V1 = P2V2

• What do we know? • P1 = 210 kPa

• V1 = 4.0 L

• P2 = ______

• V2 = 2.5 L

(210kPa)(4.0L) = (P2)(2.5L)

P2 = 336 kPa

CHARLES’ LAW

DEMO

Erlenmeyer Flask and a balloon

Why does the balloon expand?

How do temperature and volume relate to each other?

WHAT IS CHARLES’ LAW?

• The volume of a given mass of gas is directly proportional to its Kelvin temperature when held at constant temperature

• Direct Relationship – as temperature increases, the volume increases by the same factor

CHARLES’ LAW GRAPH

CHARLES’ LAW

Formula: V1 = V2

T1 T2

• Remember, if pressure increases from P1 to P2, then volume must _________ from V1 to V2.

• So, if pressure decreases from P1 to P2, then volume must _________ from V1 to V2.

Increase

Decrease

CHARLES’ LAW EXAMPLE

• A gas sample at 40.0oC occupies a volume of 2.32L. If the temperature is raised to 75.0oC, what will its volume be? (always assume the 3rd variable is constant if it is not mentioned)

(K = oC + 273)

What is absolute zero?

Why do we use Kelvin instead of Celsius degrees when working with gases?

Temperature must be in Kelvin!

K = oC + 273

CHARLES’ LAW EXAMPLE ANSWER

Formula: V1 = V2

T1 T2

• What do we know?• V1 = 2.32 L

• T1 = 40.0o C

• V2 = ______

• T2 = 75.0o C

We must convert from degrees Celsius to Kelvin(K = oC + 273)

• T1 = 40.0o C + 273 = 313 K

• T1 = 75.0o C + 273 = 348 K2.32 L = V2

313 K 348 K

V2 = 2.58 L

GAY-LUSSAC’S LAW

DEMO

Egg in a bottle

WHAT IS GAY-LUSSAC’S LAW

• The pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant.

• Direct Relationship – as temperature increases, the volume increases by the same factor

GAY-LUSSAC’S LAW

Formula: P1 = P2

T1 T2

• Remember, if pressure increases from P1 to P2, then temperature must _________ from T1 to T2.

• So, if pressure decreases from P1 to P2, then temperature must _________ from T1 to T2.

Increase

Decrease

GAY-LUSSAC’S LAW GRAPH

Temperature (K)

Pressure (atm)

150 1.0300 2.0450 3.0

GAY-LUSSAC’S LAW

Real World Example

Pressure cooker is sealed so that the volume is constant. Pressure increases in

the cooker as temperature increases

GAY-LUSSAC’S LAW EXAMPLE

The pressure of a gas tank is 3.20atm at 22oC. If the temperature raises to 60oC,

what will be the pressure in the gas tank?

Temperature must be in Kelvin!

K = oC + 273

GAY-LUSSAC’S LAW EXAMPLE ANSWER

Formula: P1 = P2

T1 T2

• What do we know?• P1 = 3.20 atm

• T1 = 22.0o C

• P2 = ______

• T2 = 60.0o C

We must convert from degrees Celsius to Kelvin(K = oC + 273)

• T1 = 22.0o C + 273 = 295 K

• T1 = 60.0o C + 273 = 333 K3.20 atm = P2

295 K 333 K

P2 = 3.61 atm