GasesGases

There are four variables that affect a gas.

1. Pressure2. Volume3.Temperature4. Number of molecules

There are four variables that affect a gas.

1. Pressure2. Volume3.Temperature4. Number of molecules

The variablesThe variables

Pressure units - there are many units for pressure. kPa - kilopascalAtm - atmospheresmm Hg - millimeters of Hg

Volume is measured in LitersTemperature is in Kelvin K = oC + 273 or oC = K - 273If the Kelvin temperature doubles the

K.E. doubles.

Pressure units - there are many units for pressure. kPa - kilopascalAtm - atmospheresmm Hg - millimeters of Hg

Volume is measured in LitersTemperature is in Kelvin K = oC + 273 or oC = K - 273If the Kelvin temperature doubles the

K.E. doubles.

The pressure-volume relationship Boyle’s Law

The pressure-volume relationship Boyle’s Law

Pressure and volume are inversely related.

One goes up the other goes downP1 x V1 = P2 x V2

Pressure and volume are inversely related.

One goes up the other goes downP1 x V1 = P2 x V2

Boyle’s Lawsample problem

Boyle’s Lawsample problem

A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa?

103 kPa x 30.0 L = 25.0 kPa x V2

V2 = 124 L

A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa?

103 kPa x 30.0 L = 25.0 kPa x V2

V2 = 124 L

Boyle’s Lawsample problem

Boyle’s Lawsample problem

Your turnYour birthday balloon travels with you

from Lincoln to Denver. The balloons volume is 4.0 L with an atmospheric pressure of 101.3 kPa. You arrive in Denver where the atmospheric pressure is 90.0 kPa. What is the new volume of your balloon?

Your turnYour birthday balloon travels with you

from Lincoln to Denver. The balloons volume is 4.0 L with an atmospheric pressure of 101.3 kPa. You arrive in Denver where the atmospheric pressure is 90.0 kPa. What is the new volume of your balloon?

Boyle’s Lawsample problem

Boyle’s Lawsample problem

Answer101.3 kPa x 4.0 L = 90.0 kPa x V2

V2 = 4.5 L

Answer101.3 kPa x 4.0 L = 90.0 kPa x V2

V2 = 4.5 L

The temperature-volume relationship

Charles’s Law

The temperature-volume relationship

Charles’s LawVolume and temperature have a

direct relationship.One goes up the other goes up

One goes down the other goes down

V1 = V2

T1 T2

Volume and temperature have a direct relationship.

One goes up the other goes upOne goes down the other goes down

V1 = V2

T1 T2

Charles’s Lawsample problem

Charles’s Lawsample problem

A balloon inflated in a room at 24oC has a volume of 4.00 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?

4.00L = V2

297 K 331 KV2 = 4.46 L

A balloon inflated in a room at 24oC has a volume of 4.00 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?

4.00L = V2

297 K 331 KV2 = 4.46 L

Charles’s Lawsample problem

Charles’s Lawsample problem

Your turnA balloon is inflated in a room at

24oC and has a volume of 4.00 L. The balloon is placed in a freezer and then removed the volume is now 3.25 L. What was the temperature of the freezer in Celcius?

Your turnA balloon is inflated in a room at

24oC and has a volume of 4.00 L. The balloon is placed in a freezer and then removed the volume is now 3.25 L. What was the temperature of the freezer in Celcius?

Charles’s Lawsample problem

Charles’s Lawsample problem

Answer4.00 L = 3.25 L

297 K T2

= 241 K which is - 31.7 oC

Answer4.00 L = 3.25 L

297 K T2

= 241 K which is - 31.7 oC