# honors chemistry, chapter 10 page 1 chapter 10 – physical characteristics of gases

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Honors Chemistry, Chapter 10Page 1

Chapter 10 – Physical Characteristics of Gases

Honors Chemistry, Chapter 10Page 2

Kinetic-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.

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Assumptions of Kinetic-Molecular Theory

1. Gases consist of large numbers of tiny particles that are far apart relative to their size.

2. 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.

3. Gas particles are in continuous, rapid, random motion. They therefore posses kinetic energy, which is energy of motion.

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Assumptions of Kinetic-Molecular Theory

4. There are no forces of attraction or repulsion between gas particles.

5. 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

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Nature 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.

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Nature 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.

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Nature 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.

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Real 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.

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Chapter 10, Section 1 Review

1. State the kinetic molecular theory of matter, and describe how it explains certain properties of matter.

2. List the five assumptions of the kinetic-molecular theory of gases.

3. Define the terms: ideal gas and real gas.

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Chapter 10, Section 1 Review continued

4. Describe each of the following characteristic properties of gases: expansion, density, fluidity, compressibility, diffusion, and effusion.

5. Describe the conditions under which a real gas deviates from “ideal” behavior.

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Pressure 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

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Example: 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

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Units 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.

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Units of PressureUnit Symbol Definition

Pascal Pa 1Pa = 1 N/m2

Millimeter of mercury

mm Hg Pressure that supports a 1 mm column of mercury

torr torr 1 torr = 1 mm Hg

1 Atmosphere Atm 760 mm Hg 101.325 kPa

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Standard 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.

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Pressure 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

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Chapter 10, Section 2 Review

1. Define pressure and relate it to force.

2. Describe how pressure is measured.

3. Convert units of pressure.

4. State the standard conditions of temperature and pressure.

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Gas Laws

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

• Boyle’s Law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.

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Illustration of Boyle’s Law

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Mathematical Expression of Boyle’s Law

V = k/P or PV = k

P1V1 = k P2V2 = k

P1V1 = P2V2

P1V1 / V2 = P2

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Volume-Pressure Data

V – mL P – atm P x V

1200 0.5 600

600 1.0 600

300 2.0 600

200 3.0 600

150 4.0 600

120 5.0 600

100 6.0 600

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Boyle’s 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

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Absolute 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

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Charles’s Law

• Charles’s 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

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Plot of Volume vs. Temperature

0200

400600800

10001200

0 200 400 600

Temperature - K.

Vo

lum

e -

mL

Volume

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Charles’s 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.

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Charles’s Law Sample Problem

V2 = V1 x T2 / T1

V2 = 752 mL Ne x 323 K. / 298 K

= 815 mL of Ne

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Gay-Lussac’s Law

Gay-Lussac’s 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

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Gay-Lussac’s 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.

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Gay-Lussac’s Law Example Problem

P2 = 3.00 atm x 325 K / 298 K

= 3.27 atm

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Combined 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

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Combined 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)

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Combined 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

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Dalton’s Law of Partial Pressures

Dalton’s 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 + …

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Partial 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.

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Partial Pressure Example

Poxygen = PT – Pwater

Poxygen = 731 torr – 17.5 torr

= 713.5 torr

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Chapter 10, Section 3, Review

1. Use the kinetic-molecular theory to explain the relationship between gas volume, temperature, and pressure.

2. Use Boyle’s law to calculate volume-pressure changes at constant temperature.

3. Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.

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Chapter 10, Section 3, Review continued

4. Use the combined gas law to calculate volume-temperature-pressure changes.

5. Use Dalton’s law of partial pressures to calculate the partial pressures and total pressures.