Kinetic Molecular

Theory

High kinetic energy.Move rapidly in straight lines.

Are very far apart. Have essentially no attractive (or repulsive) forces.

Have very small volumes compared to the volume of the container they occupy.Can be easily compressed.

1

• Atmospheres• mmHg

• psi

• Grams• moles

• mL• L• cc

• ˚C• K• ˚F

Temperature Volume

PressureAmount

Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

2

Atmospheric pressureIs the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth.

Barometer, for measuring atmospheric pressure.

Evangelista TorricelliItalian physicist

1 atm = 760 mm Hg (exact)1 atm = 760 torr

1 atm = 14.7 lb./in.2

1 atm = 101.325 kPa

Pressure and Volume

Compressibility = ability to be squeezed into a smaller volume by the application of pressure.

Liquids and solids are relatively incompressible.

Robert Boyle-1661= quantitative study of gas compressibility.Used mercury and a J shaped tube.

4

Pressure (atm)

Volume (L)

0.2500 2.801

0.5000 1.400

0.7500 0.9333

1.000 0.6998

2.000 0.3495

3.000 0.2328

Pincreases

Vdecreases

Data: Conclusion: volume of a gas is inversely proportional to the applied pressure.

𝑉 ∝1𝑃Math follows:

𝑉=𝑐1𝑃

𝑃𝑉=𝑐

P1V1 = P2V2

final conditions P2V2 = constant

If there is a change: initial conditions

5

P1V1 = constant

Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.c

Boyle’s Law

6

Boyle’s Law

To inhale:

•The rib cage expands and the diaphragm lowers.

•This increases the volume of the lungs.

•Increasing the volume causes the pressure to decrease.

•Air is drawn into the lungs to equalize the pressure.

To exhale:

•The rib cage contracts and the diaphragm is raised.

•This decreases the volume of the lungs.

•Decreasing the volume causes the pressure to increase.

•Air is expelled out of the lungs to equalize the pressure.

7

Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of an 8.0 L sample of Freon gas after its pressure is changed from 550 mmHg to 2200 mmHg at constant T?

STEP 1 Set up a data table:

Conditions 1 Conditions 2P1 = 550 mmHg P2 = 2200 mmHg V1 = 8.0 L V2 = ?

STEP 2 Solve Boyle’s law for V2.

P1V1 = P2V2

V2 = V1 x P1

P2

STEP 3 Set up problemV2 = 8.0 L x 550 mmHg =

2.0 L 2200 mmHg

Example: Boyle’s Law

8

Boyle’s Law

P1V1 = P2V2

Charles’ Law

V1 = V2

T1 T2

Quantitative observations of gases at different temperatures.

Jacques Alexandre Charles (1787)French physicist

9

y = m x + b

V = m ˚C + b

Data shows that @ T = -273.15 ˚C, the V is 0.

0 = m (-273.15 ) + b

Therefore: b = m (273.15)

Substituting: V = m ˚C + m (273.15 )

Rearrange: V = m(˚C + 273.15)

Gives: V = m (K)

Charles’s Law:

V of a gasIs directly proportionalto the temperature in K

10

T = -273.15 ˚C

Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.c

Charles’ Law

11

Boyle’s Law

P1V1 = P2V2

Charles’ Law

V1 = V2

T1 T2

P1 = P2

T1 T2

12

PV = constant

T

Boyle’s Law

P1V1 = P2V2

Charles’ Law

V1 = V2

T1 T2

𝑃1𝑉 1

𝑇1

𝑃2𝑉 2

𝑇 2

=

Combine

13

Boyle’s Law

P1V1 = P2V2

Charles’ Law

V1 = V2

T1 T2

V1 = V2

n1 n2

Ideal Gas EquationR = 0.0821 L atm

mole KR = 62.4 L mmHg

mole K

14

P2V2 = n2RT2

P1V1 = n1RT1

STEP 2PV = nRT

PV = nRT

Summary

STEP 1Organize the data in a table of initial and final conditions.

STEP 2Use PV = nRT to find the needed gas law.

STEP 3Substitute values into the gas law eq. and solve for the variable.

A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant).

STEP 1 1 2P1 = 600. mm Hg P2 = ?

V1 = 12.0 L V2 = 36.0 L

P2V2 = P1V1 P2 = P1 x V1

V2

STEP 3P2 = 600. mm Hg x 12.0 L = 200. mm Hg

36.0 L15

P2V2 = 1

P1V1

P2V2 = n2RT2

P1V1 = n1RT1

STP(Standard

Temperature and Pressure)

Standard pressure

1 atm or 760 mm Hg

Standard temperature 0°C

or 273 K

A convenient way of looking at gases in chemical reactions.

The Molar Volume of gas @ STP = 22 4 L volume occupied by 1 mole of any gas

1 mole= 22.4 L

What is the volume, in liters, of 64.0 g of O2 gas at STP?

16

The partial pressure of a gas• Is the pressure of each gas in a mixture.• Is the pressure that gas would exert if it were by itself in the container.

Dalton’s Law of Partial Pressures• Pressure depends on the total number of gas particles, not on the types of

particles.• The total pressure exerted by gases in a mixture is the sum of the partial

pressures of those gases.

PT = P1 + P2 + P3 +.....John Dalton

17

Collecting a gas over water

Pt = Patm = Pgas + Pwv

Pgas = Patm – Pwv

The pressure of the water vapor depends on the T of the water and is tabulated.

Partial Pressure