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 _____________. Quadruple

BOYLE’S LAW Formula: P 1 V 1 = P 2 V 2 Remember, if pressure increases from P 1 to P 2, then volume must _________ from V 1 to V 2. So, if pressure decreases from P 1 to P 2, then volume must _________ from V 1 to V 2. 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 (P 2 ) be?

BOYLE’S LAW EXAMPLE ANSWER Formula: P 1 V 1 = P 2 V 2 What do we know? P 1 = 210 kPa V 1 = 4.0 L P 2 = ______ V 2 = 2.5 L (210kPa)(4.0L) = (P 2 )(2.5L) P 2 = 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: V 1 = V 2 T 1 T 2 Remember, if pressure increases from P 1 to P 2, then volume must _________ from V 1 to V 2. So, if pressure decreases from P 1 to P 2, then volume must _________ from V 1 to V 2. Increase Decrease

CHARLES’ LAW EXAMPLE A gas sample at 40.0 o C occupies a volume of 2.32L. If the temperature is raised to 75.0 o C, what will its volume be? (always assume the 3 rd variable is constant if it is not mentioned) (K = o C + 273) What is absolute zero? Why do we use Kelvin instead of Celsius degrees when working with gases?

Temperature must be in Kelvin! K = o C + 273

CHARLES’ LAW EXAMPLE ANSWER Formula: V 1 = V 2 T 1 T 2 What do we know? V 1 = 2.32 L T 1 = 40.0 o C V 2 = ______ T 2 = 75.0 o C We must convert from degrees Celsius to Kelvin (K = o C + 273) T 1 = 40.0 o C = 313 K T 1 = 75.0 o C = 348 K 2.32 L = V K 348 K V 2 = 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: P 1 = P 2 T 1 T 2 Remember, if pressure increases from P 1 to P 2, then temperature must _________ from T 1 to T 2. So, if pressure decreases from P 1 to P 2, then temperature must _________ from T 1 to T 2. Increase Decrease

GAY-LUSSAC’S LAW GRAPH Temperature (K) Pressure (atm)

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 22 o C. If the temperature raises to 60 o C, what will be the pressure in the gas tank?

Temperature must be in Kelvin! K = o C + 273

GAY-LUSSAC’S LAW EXAMPLE ANSWER Formula: P 1 = P 2 T 1 T 2 What do we know? P 1 = 3.20 atm T 1 = 22.0 o C P 2 = ______ T 2 = 60.0 o C We must convert from degrees Celsius to Kelvin (K = o C + 273) T 1 = 22.0 o C = 295 K T 1 = 60.0 o C = 333 K 3.20 atm = P K 333 K P 2 = 3.61 atm