Chapter 16 Section 3

Pressure!

Definition of pressure is:Amount of force exerted on an areaP = F/AUnit in SI: Pascal (Pa)One Pascal is one Newton force on an area

1 square meter

Gases exert pressure due to KE of their particles

Boyle’s Law Robert Boyle (1627-1691) Studied relationship between volume

and pressure of gases Found: as volume goes up, pressure

goes downProduct of pressure and volume of a gas is

a constant if temperature does not change As an equation: PV=constant http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/

frglab.html

Boyle’s Law

More commonly, we compare pressure and volume of the same gas

Since PV is constant, we can say:P1V1 = P2V2 for any gas

To see this in action, let’s do Math Skills Activity p 505!

Math Skills Activity!

A balloon has a volume of 10.0 L at a pressure of 101 kPa. What will the balloon’s new volume be if the pressure changes to 43 kPa?

What we know:P1 = 101 kPa

V1 = 10.0 L

P2 = 43 kPa

V2 = ?

What equation?P1V1 = P2V2

So V2 = P1V1

P2

Put in the values!V2 = (101 kPa)(10.0L)

(43 kPa) Do the math!

V2 = 23.488 L Check units and math!

Another Problem:

A volume of helium occupies 11.0 L at 98.0 kPa. What is the new volume if the pressure increases to 101.0 kPa?

What do we know?P1 = 98.0 kPa

V1 = 11.0 L

P2 = 101.0 kPa

V2 = ?

What equation?P1V1 = P2V2

So V2 = P1V1

P2

Put in the numbers!V2 = (98.0 kPa)(11.0 L)

(101.0 kPa)

Do the math!V2 = 10.67 L

Check your math and units.

Charles’ Law Jaques Charles (1746-1823) Studied relationship of temperature and

volume with constant pressure As temperature goes up, volume

increases Volume/Temperature is a constant As an equation: K = V/T This relationship is the basis for the

Kelvin temperature scale

Charles’ Law

More commonly, we compare one situation to another for a gas

V1 = V2 With constant pressure

T1 T2

Temperatures must be used in Kelvins!

Charles’ Law Problem

If a balloon has a volume of 2.0 L at 25oC, what is the volume of the balloon at 3oC?

What do we know?T1 = 25oC = 298 K

V1 = 2.0 L

T2 = 3oC = 276 K

V2 = ?

Charles’ Law Problem What equation?

V1 = V2

T1 T2

So V2 = V1T2

T1

Put in the numbers! Remember, temperature must be in Kelvins!V2 = (2.0L)(276 K)

(298 K) Solve!

V2 = 1.85 L Check for math and units.

Earth’s Atmosphere

Earth’s atmosphere divided into 5 layers

Earth’s Atmosphere

Exosphere: outer layer Thermosphere Mesosphere Stratosphere Troposphere: layer we live in!

Exosphere

Very thin outer portion of atmosphere Negligible pressure

Thermosphere Extends from about 90 km to about 500 km

above Earth Very low pressure Very high temperatures (1000oC) in upper

regions Gases here broken into atoms by energy

from sun Much of X-ray and UV radiation absorbed

here Aurora occurs in this layer

Mesosphere

About 50 to 85 km above Earth’s surface

Temperatures get very cold (-25oC to -90oC)

Air is mixed by currents Most meteors burn up in this layer

Stratosphere

Altitude of the bottom of this layer varies with latitude and season16 km at equator8 km at poles

Ozone heats this layer as it absorbs UV from sun

Very few clouds, very dry air

Troposphere

Layer we live in! From surface up to 7 to 20 km altitude Almost all weather occurs in this layer Almost all dust particles and water vapor

in this layer This layer heated from below, so

warmer at lower altitudes than higher

Gases are fun!