Title: Honors Chemistry, Chapter 10
1Chapter 10 Physical Characteristics of Gases
2Kinetic-Molecular Theory
- Kinetic-molecular theory is based on the idea
that particles of matter are always in motion. - Ideal Gas an imaginary gas that perfectly fits
all the assumptions of the kinetic-molecular
theory.
3Assumptions of Kinetic-Molecular Theory
- Gases consist of large numbers of tiny particles
that are far apart relative to their size. - Collisions between gas particles and between
particles and container walls are elastic
collisions. An elastic collision is one in which
there is no net loss of kinetic energy. - Gas particles are in continuous, rapid, random
motion. They therefore posses kinetic energy,
which is energy of motion.
4Assumptions of Kinetic-Molecular Theory
- There are no forces of attraction or repulsion
between gas particles. - The average kinetic energy of gas particles
depends on the temperature of the gas. - KE ½ mv2 where KE is kinetic energy
- m is mass and
- v is velocity
5Nature of Gases
- Expansion Gases have no definite shape or
definite volume so they expand to fill the
container. Gas particles move rapidly in all
directions (assumption 3) without significant
attraction or repulsion between them (assumption
4). - Fluidity Because the attractive forces between
gas particles are insignificant (assumption 4),
gas particles glide easily past one another.
6Nature of Gases
- Low Density The density of a gas is about
1/1000 the density of the same substance in the
liquid or solid state. - Compressibility During compression, the gas
particles, which are initially very far apart
(assumption 1), are crowded closer together.
Under the influence of pressure, the volume of a
gas can be greatly decreased.
7Nature of Gases
- Diffusion Gases spread out and mix with one
another, even without being stirred. This
spontaneous mixing of particles of two substances
caused by their random motion is called
diffusion. - Effusion is a process by which gas particles pass
through a tiny opening. Rates of effusion of
different gases are directly proportional to the
velocities of their particles.
8Real Gases
- A real-gas is a gas that does not behave
completely according to the assumptions of the
kinetic-molecular theory. - Deviations from ideal-gas behavior usually occur
at high pressures and/or low temperatures where
the attractive forces between molecules begin to
play a role.
9Chapter 10, Section 1 Review
- State the kinetic molecular theory of matter, and
describe how it explains certain properties of
matter. - List the five assumptions of the
kinetic-molecular theory of gases. - Define the terms ideal gas and real gas.
10Chapter 10, Section 1 Review continued
- Describe each of the following characteristic
properties of gases expansion, density,
fluidity, compressibility, diffusion, and
effusion. - Describe the conditions under which a real gas
deviates from ideal behavior.
11Pressure and Force
- Pressure (P) is defined as the force per unit
area on a surface. - pressure force/area
- The SI unit for force is the newton, abbreviated
N. It is the force that will increase the speed
of a one kilogram mass by 1 meter per second per
second that it is applied. - force mass x acceleration
12Example Pressure on the Feet of a Ballet Dancer
- Acceleration of gravity is 9.8 m/s/s
- What is the force of a 51 Kg dancer on the floor?
- F m x a 51 kg x 9.8 m/s/s 500 N
- What is the pressure?
- Flat footed 500 N/325 cm2 1.5 N/cm2
- Two feet tip toes 500 N /13 cm2 38.5
N/cm2 - One foot tip toes 500 N /6.5 cm2 77 N/cm2
13Units of Pressure
- A common unit of pressure is millimeters of
mercury, mm Hg. A pressure of 1 mm of Hg is now
called 1 torr in honor of Torricelli. - One atmosphere is defined as being exactly
equivalent to 760 mm Hg. - In SI units, pressure is expressed in derived
units called pascals. One pascal (Pa) is defined
as the pressure exerted by a force of one newton
(1 N) acting on an area of 1 square meter. One
atmosphere is 1.01325 x 105 Pa or 101.325 kPa.
14Units of Pressure
15Standard Temperature and Pressure
- For purposes of comparison, scientists have
agreed on standard conditions of exactly 1 atm
pressure and 0oC. These conditions are called
standard temperature and pressure and are
commonly abbreviated STP.
16Pressure Units Conversions
- Convert 0.830 atm to mm of Hg and kPa
- 0.830 atm x 760 mm of Hg/atm 631 mm
of Hg - 0.830 atm x 101.325 kPa/atm
- 84.1 kPa
17Chapter 10, Section 2 Review
- Define pressure and relate it to force.
- Describe how pressure is measured.
- Convert units of pressure.
- State the standard conditions of temperature and
pressure.
18Gas Laws
- Gas laws are simple mathematical relationships
between the volume, temperature, pressure, and
amount of a gas. - Boyles Law states that the volume of a fixed
mass of gas varies inversely with the pressure at
constant temperature.
19Illustration of Boyles Law
20Mathematical Expression of Boyles Law
- V k/P or PV k
- P1V1 k P2V2 k
- P1V1 P2V2
- P1V1 / V2 P2
21Volume-Pressure Data
22Boyles Law Example Problem
- V 150. mL of O2 at 0.947 atm.
- What is the volume at 0.987 atm (at constant
temperature)? - Formula P1V1 / P2 V2
- 0.947 atm x 150. mL / 0.987 atm 144 mL of O2
23Absolute Zero
- Absolute zero, -273.15 oC., is the lowest
temperature possible. This is assigned a value
of zero on the Kelvin scale. - To convert from Celsius to Kelvin
- K 273.15 oC.
- To convert from Kelvin to Celsius
- oC. K 273.15
24Charless Law
- Charless Law states that the volume of a fixed
mass of gas at constant pressure varies directly
with the Kelvin temperature. - V kT or V/T k
- V1 / T1 V2 / T2
25Plot of Volume vs. Temperature
26Charless Law Example Problem
- A sample of Ne gas has a volume of 752 mL at 25
oC. What volume will the gas occupy at 50 oC. if
the pressure remains constant? - Convert temperatures to Kelvin
- 25 oC. 273 298 K.
- 50 oC. 273 323 K.
27Charless Law Sample Problem
- V2 V1 x T2 / T1
- V2 752 mL Ne x 323 K. / 298 K
- 815 mL of Ne
28Gay-Lussacs Law
- Gay-Lussacs Law The pressure of a fixed mass of
gas at constant volume varies directly with the
Kelvin temperature. - P kT or P/T k
- P1/T1 P2/T2
29Gay-Lussacs Law Example Problem
- Gas in an aerosol can is 3.00 atm at 25oC.
- What is the pressure at 52oC.?
- P2 P1 x T2/T1
- Convert temperatures to Kelvin
- 25oC. 273 298 K.
- 52oC. 273 325 K.
30Gay-Lussacs Law Example Problem
P2 3.00 atm x 325 K / 298 K 3.27 atm
31Combined Gas Law
- The combined gas law expresses the relationship
between pressure, volume, and temperature of a
fixed amount of gas. - PV/T k
- Or
- P1 V1/T1 P2 V2 / T2
32Combined Gas Law Example Problem
- A helium-filled balloon has a volume of 50.0 L at
25 oC. and 1.08 atm. What volume will it have at
0.855 atm and 10 oC.? - Convert the temperatures to Kelvin
- 25 oC. 273 298 K
- 10 oC. 273 283 K.
- V2 P1 V1 T2 /( P2 T1)
33Combined Gas Law Example Problem
V2 P1 V1 T2 /( P2 T1) V2 1.08 atm x50 L
Hex283 K/(0.855atm x298 K) 60.0 L
34Daltons Law of Partial Pressures
- Daltons law of partial pressures states that 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
35Partial Pressure Example
- Oxygen gas is collected over water. The
barometric pressure was 731 torr and the
temperature was 20.0oC. What was the partial
pressure of the oxygen collected? - PT 731 torr
- Pwater 17.5 torr at 20.0oC.
36Partial Pressure Example
Poxygen PT Pwater Poxygen 731
torr 17.5 torr 713.5 torr
37Chapter 10, Section 3, Review
- Use the kinetic-molecular theory to explain the
relationship between gas volume, temperature, and
pressure. - Use Boyles law to calculate volume-pressure
changes at constant temperature. - Use Gay-Lussacs law to calculate
pressure-temperature changes at constant volume.
38Chapter 10, Section 3, Review continued
- Use the combined gas law to calculate
volume-temperature-pressure changes. - Use Daltons law of partial pressures to
calculate the partial pressures and total
pressures.