Unit 5: Gas and Atmospheric Chemistry

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/se

ctions/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 kPa550 L75kPa?

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’ LAWEx. 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 = V2T1 T2

PRESSURE AND TEMPERATURE: GAY-LUSSACS LAWPressure 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 = kTP = kTP1 = 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”