UNIT 5: GAS AND ATMOSPHERIC CHEMISTRY11.2 & 11.3 Gas Laws

CONTEXT

In a gas, the particles are very far apart This means that there is a lot of empty space

between them This makes gases compressible

CONSIDER THIS…

What happens to the volume of a gas when you increase the pressure? Ex: press a syringe that is stoppered)

Why? Gases are compressible

ANIMATION

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html

BOYLE’S LAW

In the 17th, Robert Boyle described this as “the spring in the air”

As pressure on a gas increases, the volume of the gas decreases proportionally, if temperature and amount of gas (moles) remain constant.

P α 1/V

PV = k

P1V1 = P2V2

EXAMPLE PROBLEM

A 550 L weather balloon at 98 kPa is released from the ground and rises into the atmosphere. It is caught a later height and instruments indicate the air pressure is 75 kPa. What is the volume of the captured balloon?

P1=

V1=

P2=

V2=

98 kPa

550 L

75kPa

?

P1V1=P2V2

(98 kPa)(550L) = (75 kPa)(V2)V2= 719 L

REAL LIFE APPLICATIONS

Scuba Diving As they dive down underwater, the pressure

increases volume? What happens to volume as you ascend in

water?

Ears popping!

CHARLES’ LAW

Temperature and Volume

ABSOLUTE ZERO This temperature is absolute zero, the

lowest possible temperature. A theoretical temperature at which matter

has no kinetic energy (volume of 0) and therefore transmit no thermal energy.

-273.15 ˚C

TEMPERATURE SCALES

1˚C = 1 K ˚C = K - 273.15 K = ˚C + 273.15

Kelvin devised a new scale that would have no negative values

In this unit, we will ALWAYS use K

CHARLES’ LAW

A direct relationship V α TV = kTVi = VfTi Tf

As the temperature of a gas increases, the volume increases proportionally, when pressure and amount of gas (moles) remains

constant.

Temperature (in Kelvins) is directly proportional to volume

CHARLES’ LAW

Ex. A balloon with a volume of 1.0L at 25°C is cooled to -190°C. The new volume is…

V1 = V2V1= 1.0L T1 = 25 + 273 = 298KT1 T2 T2 = -190 + 273 = 83K

V2 = 1.0 L x 83 K 298K

V2 = 0.28L = 280mL

V1 = V2

T1 T2

PRESSURE AND TEMPERATURE: GAY-LUSSACS LAW

Pressure increases proportionally as temperature increases, provided volume and amount remain

constant. Pressure is directly proportional to temperature (in

Kelvins)P T A direct relationship

P = kT

P = k

T

P1 = P2

T1 T2

EXAMPLEEx. A gas cylinder with a pressure of 1000kPa at

25°C is placed in a boiling water bath. What will the new pressure reading be?

P1 = P2 P1 = 1000kPa T1 = 25 + 273 = 298K

T1 T2 P2 = ? T2 = 100 + 273 = 373K

P2 = 1000kPa x 373 K 298 KP2 = 1252 kPa

Video!

PRACTICE!!

Popsicle stick analogy

P. 435 # 2,5 P. 451 # 2-5 Worksheet – “Problems – Gas Laws”