PressureForce acting on a unit area of a surface

Standard Atmospheric PressureBarometric Pressure

• Air has weight and therefore exerts pressure.

• A standard Atmosphere (1atm) is the average pressure at sea level.

• What happens to air pressure as you increase your elevation?

• How do we measure pressure?

Kinetic Molecular Theory

• Particles in an ideal gas…– have elastic collisions. – are in constant, random, straight-line motion.– don’t attract or repel each other.– have an avg. KE directly related to Kelvin temperature.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Characteristics of Ideal Gases

Gases expand to fill any container.

Gases are fluids (like liquids).

Gases have very low densities.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Characteristics of Gases

• Gases can be compressed.

• Gases undergo diffusion.– random motion

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Properties of Gases

V = volume of the gas (liters, L)

T = temperature (Kelvin, K)

P = pressure (atmospheres, atm)

n = amount (moles, mol)

Gas properties can be modeled using math.Model depends on:

Pressure - Temperature - Volume Relationship

P T V P T V

Gay-Lussac’s P T

Charles V T

P T

V

P T

V P T V P T V

Boyle’s P 1V ___

Boyle’s LawP1V1 = P2V2

(Temperature is held constant)

• As the pressure on a gas increases

• As the pressure on a gas increases -

the volume decreases

• Pressure and volume are inversely related

1 atm

4 Liters

2 atm

2 Liters

Pressure vs. Volume for a Fixed Amount of Gas

(Constant Temperature)

0 100 200 300 400 500

Pressure Volume PV

(Kpa) (mL)

100 500 50,000

150 333 49,950

200 250 50,000

250 200 50,000

300 166 49,800

350 143 50,500

400 125 50,000

450 110 49,500

Vol

ume

(mL)

100

200

300

400

500

600

Pressure (KPa)

Boyle’s Law Illustrated

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 404

Pressure Units

• Atmosphere• Feet of water

• mm Hg• cm Hg• Torr• Barr

• mbarr• kPa• Pa

PV Calculation (Boyle’s Law)

A quantity of gas has a volume of 120 dm3 when confined under a pressure of 93.3 kPa at a temperature of 20 oC. At what pressure will the volume of the gas be 30 dm3 at20 oC?

P1 x V1 = P2 x V2

(93.3 kPa) x (120 dm3) = (P2) x (30 dm3)

P2 = 373.2 kPa

Volume and Pressure

Two-liter flask

The molecules arecloser together; thedensity is doubled.

The average molecules hits the wall twice as often. The total number of impacts with the wall is doubled and the pressure is doubled.

One-liter flask

Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 101

Charles' Law

This means, for example, that Volume goes up as Temperature goes up.

Jacques Charles(1746 - 1823)

Isolated boron and studied gases.Balloonist.

A hot air balloon is a good example of Charles's law.

VV and TT are directly related.

T1 T2

V1 V2=

(Pressure is held constant)

• Raising the temperature of a gas increases the pressure if the volume is held constant.

• The molecules hit the walls harder.

• The only way to increase the temperature at constant pressure is to increase the volume.

Temperature

VT Calculation (Charles’ Law)

At constant pressure, the volume of a gas is increased from150 dm3 to 300 dm3 by heating it. If the original temperatureof the gas was 20 oC, what will its final temperature be (oC)?

T1 = 20 oC + 273 = 293 KT2 = X KV1 = 150 dm3

V2 = 300 dm3

150 dm3

293 K= 300 dm3

T2

T2 = 586 K

oC = 586 K - 273

T2 = 313 oC

Temperature and the Pressure of a Gas

High in the mountains, Richard checked the pressure of his car tires and observed that they has 202.5 kPa of pressure. That morning, the temperature was -19 oC. Richard then drove all day, traveling through the desert in the afternoon. The temperature of the tires increased to 75 oC because of the hot roads. What was the new tire pressure? Assume the volume remained constant. What is the percent increase in pressure?

P1 = 202.5 kPa P2 = X kPaT1 = -19 oC + 273 = 254 KT2 = 75 oC + 273 = 348 K

202.5 kPa 254 K

= P2

348 K

P2 = 277 kPa

% increase = 277 kPa - 202.5 kPa x 100 % 202.5 kPa

or 37% increase

The pressure and absolute temperature (K) of a gas are directly related – at constant mass & volume

P

T

Gay-Lussac’s Law

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Temperature (K)

Pressure(torr)

P/T(torr/K)

248 691.6 2.79

273 760.0 2.78

298 828.4 2.78

373 1,041.2 2.79

The Combined Gas Law

When measured at STP, a quantity of gas has a volume of 500 dm3. What volume will it occupy at 0 oC and 93.3 kPa?

P1 = 101.3 kPaT1 = 273 KV1 = 500 dm3

P2 = 93.3 kPaT2 = 0 oC + 273 = 273 KV2 = X dm3

(101.3 kPa) x (500 dm3) = (93.3 kPa) x (V2)

273 K 273 K

V2 = 542.9 dm3

1 1 2 2

1 2

PV PV

T T

(101.3) x (500) = (93.3) x (V2)

Standard Temperature and Pressure

• 0o C

• 1 atm = 760 mmHg = 760 torr = 101.3 kPa

Collecting Gasses over a liquid (such as water)

• Some water vapor will be mixed with the collected gas and will exert pressure.

• The partial pressure of the water vapor for each temperature is constant.

• So, take the pressure of the gas and subtract the partial pressure of water at the given temperature.