The KineticMolecular Theory

1
The Kinetic-Molecular Theory

• Physical Characteristics of Gases

2
The Kinetic-Molecular Theory of Matter

• Particles of matter are always in motion
• Ideal gas an imaginary gas that perfectly fits
  the kinetic-molecular theory

3
5 Assumptions of the Kinetic-Molecular Theory of
Gases

• Gas particles are far apart relative to their
  size.
• Collisions of particles and walls of a container
  are elastic collisions. (no loss of KE)
• Gas particles are in continuous, rapid, random
  motion.
• There are no forces of attraction or repulsion
  between gas particles.
• Average kinetic energy depends upon the
  temperature of the gas.

4
Physical Properties of Gases explained by the
Kinetic-Molecular Theory

• Expansion completely fill container
• Fluidity flow
• Low density particles are far apart (1/1000 the
  density of the gass liquid or solid)
• Compressibility
• Diffusion gases spread out and mix with one
  another (spontaneous mixing)
• Effusion gas particles pass through a tiny
  opening.

5
Comparing Relative Volumes for a Solid, Liquid,
and Gas
Gas 1 mol oxygen (32 g) occupies 22 400 mL
Solid 1 mol aluminum (27 g) occupies 10 mL
Liquid 1 mol water (18 g) occupies 18 mL
6
Pressure

• Gases exert pressure
• Pressure is defined as the force per unit area on
  a surface
• Gases have 4 measurable quantities
• Volume
• Temperature
• Number of molecules
• pressure

7
Units of Pressure
8
Atmospheric Pressure

• Atmospheric pressure is the pressure exerted by
  air pressing down on the Earths surface.
• Atmospheric pressure is measured by a barometer
  (open) and/or a manometer (closed).

9
Standard Temperature and Pressure

• Standard temperature and pressure (STP) is 1 atm
  and 0? C (273? K).
• ?K 273 ?C

10
Problems on Atmospheric Pressure

• (p. 312) The average atmospheric pressure in
  Denver is 0.830 atm. Express this pressure in
  (a) mm Hg, (b) kPa.
• Given P 0.830 atm, unknown mm Hg
• 0.830 atm x 760 mm Hg 631 mm Hg
• 1 atm
• Given P 0.830 atm, unknown kPa
• 0.830 atm x 101.3 kPa 84.1 kPa
• 1 atm

11
Gas Laws

• Gas laws are simple mathematical relationships
  between the volume, temperature, pressure, and
  amount of gas.

12
Boyles Law Pressure Volume Relationship

• Pressure and volume are inversely proportional
  when temperature remains constant.
• P1V1 P2V2 P is pressure
• V is volume

13
Boyles Law sample problem (p. 315)A sample of
oxygen gas has a volume of 150. mL when its
pressure is 0.947 atm. What will the volume of
the gas be at a pressure of 0.987 atm if the
temperature remains constant?

• Given V1 150 mL P2 0.987 atm
• P2 0.947 atm
• Unknown V2
• P1V1 P2V2 rearranges to V2 P1V1
• P2
• V2 P1V1 (0.947 atm)(150 mL) 144 mL O2
• P2 0.987 atm

14
Charless Law Volume Temperature Relationship

• Volume is directly related to temperature when
  pressure remains constant.
• Temperature must be in Kelvin.
• ?K 273 ?C
• v1 v2 V is volume
• T1 T2 T is temperature

15
Charless Law sample problem (p. 319)A sample of
neon gas occupies a volume of 752 mL at 25?C.
What volume will the gas occupy at 50?C if the
pressure remains constant?

• Given V1 of Ne 752 mL
• T1 of Ne 25?C 273 298 K.
• T2 of Ne 50?C 273 323 K.
• Celsius temperatures must always be converted to
  kelvins. This is a very important step in
  working the problems in this unit.
• Unknown V2 of Ne in mL

16
Continuing Charless Law sample problem (p.
319)A sample of neon gas occupies a volume of
752 mL at 25?C. What volume will the gas occupy
at 50?C if the pressure remains constant?

• Because the gas remains at constant pressure, an
  increase in temperature will cause an increase in
  volume. To obtain V2, rearrange the equation for
  Charless Law.
• V2 V1T2
• T1
• V2 V1T2 (752 mL Ne)(323 K) 815 mL Ne
• T1 298 K

17
Gay-Lussacs Law Pressure Temperature
Relationship

• The pressure of a fixed mass of gas at constant
  volume varies directly with the Kelvin
  temperature.
• P1 P2
• T1 T2
• When values are known for 3 of the 4 quantities,
  the 4th value can be calculated.

18
Gay-Lussacs Law Sample Problem (p. 320)The gas
in an aerosol can is at a pressure of 3.00 atm at
25?C. Directions on the can warn the user not to
keep the can is a place where the temperature
exceeds 52?C. What would the gas pressure in the
can be at 52?C.

• Given P1 of gas 3.00 atm
• T1 of gas 25?C 273 298 K
• T2 of gas 52?C 273 325 K
• Unknown P2 of gas in atm
• Because the gaseous contents remain at the
  constant volume of the can, an increase in
  temperature will cause an increase in pressure.
  Rearrange Gay-Lussacs Law to obtain P2.

19
Gay-Lussacs Law Sample Problem (p. 320)The gas
in an aerosol can is at a pressure of 3.00 atm at
25?C. Directions on the can warn the user not to
keep the can is a place where the temperature
exceeds 52?C. What would the gas pressure in the
can be at 52?C.

• P1 P2 rearranges to P1T2
• T1 T2 T1
• P2 P1T2 (3.00 atm)(325 K) 3.27 atm
• T1 298 K

P2


20
The Combined Gas Law

• No variable remains constant
• The combined gas law expresses the relationship
  among pressure, volume, and temperature.
• P1V1 P2V2
• T1 T2

21
The Combined Gas Law sample problem (p.321)A
helium-filled balloon has a volume of 50.0 L at
25?C and 1.08 atm. What volume will it have at
0.885 atm and 10 ?C ?

• Given V1 of He 50.0 L
• T1 25?C 273 298K T2 10?C 273
• P1 1.08 atm 283K
• P2 0.855 atm
• Unknown V2 in liters (L)

22
The Combined Gas Law sample problem (p.321)A
helium-filled balloon has a volume of 50.0 L at
25?C and 1.08 atm. What volume will it have at
0.885 atm and 10 ?C ?

• P1V1 P2V2 rearranges V2 P1V1T2
• T1 T2 P2T1
• V2 (1.08 atm)(50.0 L)(283K) 60.0 L He
• (0.855 atm)(298K)

23
Daltons Law of Partial Pressures

• The total pressure of a mixture of gases is
  equal to the sum of the partial pressures of the
  component gases.
• PT P1 P2 P3 . . .

24
Sample problem (p 324)Dalton Partial
Pressures

• 2KCLO3 ? 2KCl 3O2?
• What is the partial pressure of the oxygen
  collected? Barometric pressure 731.0 torr,
  Temperature 20?C
• PT Patm 731.0 torr
• PH2O 17.5 torr from Table 8-A
• Unknown PO2
• PT PH2O PO2
• PO2 PT - PH2O 731.0 torr 17.5 torr
  713.5 torr

25
Avogadros Law

• Equal volumes of gases at the same temperature
  and pressure contain the same number of
  molecules.
• n K
• V
• 2 H2(g) O2(g) ? 2 H2O(g)
• 2 molecules 1 molecule 2 molecules
• 2 mol 1 mol 2
  mol
• 2 volumes 1 volume 2 volume

26
Standard Molecular Volume of a Gas

• The volume occupied by one mole of a gas at
  standard temperature and pressure (STP) 22.4 L

27
The Ideal Gas Law

• The Ideal Gas Law is the mathematical
  relationship among pressure, volume, temperature,
  and the number of moles of a gas.
• V nRT or PV nRT
• P
• R is the ideal gas constant, R 0.0821

Latm molK
(Or, see your reference tables for the constant
in different units.) P pressure, T
temperature , n number of moles
28
The Ideal Gas Law sample problem What is the
pressure in atmospheres exerted by a 0.500mol
sample of nitrogen gas in a 10.0L container at
298K?

• Given V of N2 10.0L Unknown P of N2 in
• n 0.500mol atm
• T 298K
• P nRT (0.5000mol)(0.0821)(298K) 1.22
• 10.0L atm

29
Gas Density

• D m Density mass
• V volume
• As temperature increases, volume increases.
• As volume increases and mass remains the same,
  density decreases.