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Chapter 17 Gases (p.300-301)

Properties of Gases, Gas Pressure, and Gas Laws

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Properties of Gases

Expand to completely fill their container.Take the shape of their container.Low density.

Much less than solid or liquid state.Compressible.Mixtures of gases are always

homogeneous.

Air Pressure

Pressure: a measure of the force applied by gas molecules on objectsThis can be felt by the

walls of a containeror on objects in air

http://www.youtube.com/watch?v=t-Iz414g-ro

Pressure

What causes gas pressure in a closed container?

Collisions between particles of a gas and the walls of the container cause the pressure in a closed container of gas.

Pressure is the result of a force distributed over an area.

Helium Filled Balloon

The helium atoms in a balloon are constantly moving. When many particles collide with the walls of a container at the same time, they produce a measurable pressure.

• The more frequent the collisions, the greater the pressure is.

• The speed of the particles and their mass also affect the pressure.

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Atmospheric Pressure

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

Factors That Affect Gas Pressure

Factors that affect the pressure of an enclosed gas are its temperature, its volume, and the number of its particles.

Raising the temperature of a gas will increase its pressure if the volume of the gas and the number of particles are constant.

Temperature

P= 60 kPa

P= 72 kPa

Effect of Temperature on Pressure in a Fixed Volume

As the temperature rises, the average kinetic energy of the particles in the air increases. With increased kinetic energy, the particles move faster and collide more often with the inner walls of the container. Faster-moving particles hit the walls with greater force. More collisions and increased force cause the pressure of the gas in the container to rise.

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf

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Pressure and Volume

Boyle’s Lawhttp://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm

Reducing the volume of a gas increases its pressure if the temperature of the gas and the number of particles are constant.

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Boyle’s Law

Boyle’s law states that

• the pressure of a gas is inversely related to its volume when T and the amount of molecules are constant.

• if volume decreases, the pressure increases.

Tro's Introductory Chemistry, Chapter 11

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When you double the pressure on a gas,the volume is cut in half (as long as the

temperature and amount of gas do not change).

In the next lab, we will test…

Robert Boyle described the relationship between the pressure and volume of a gas. The graph shows an inverse relationship between the volume of a gas and the pressure of the gas.

Boyle’s Law

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In Boyle’s law, the product P x V is constant as long as T and n do not change.

P1V1 = 8.0 atm x 2.0 L = 16 atm L

P2V2 = 4.0 atm x 4.0 L = 16 atm L

P3V3 = 2.0 atm x 8.0 L = 16 atm L

Boyle’s law can be stated as P1V1 = P2V2

(T, number of molecules (n) constant)

PV Constant in Boyle’s Law

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Solving for a Gas Law Factor

The equation for Boyle’s law can be rearranged tosolve for any factor.

P1V1 = P2V2 Boyle’s law

To solve for V2 , divide both sides by P2.

P1V1 = P2V2

P2 P2

V1 x P1 = V2

P2

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Learning Check

For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant).

1) pressure decreases2) pressure increases

Copyright © 2009 by Pearson Education, Inc.

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Solution

For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant).1) pressure decreases B2) pressure increases A

Copyright © 2009 by Pearson Education, Inc.

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Learning Check

A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is constant), is the new volume represented by A, B, or C?

Copyright © 2009 by Pearson Education, Inc.

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Solution

A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon A.

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf

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Calculations with Boyle’s Law

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

1. Set up a data table:Conditions 1 Conditions 2 P1 = 550 mmHg P2 = 2200 mmHg

V1 = 8.0 L V2 =

Calculation with Boyle’s Law

?

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2. When pressure increases, volume decreases.

Solve Boyle’s law for V2:

P1V1 = P2V2

V2 = V1 x P1

P2 V2 = 8.0 L x 550 mmHg = 2.0 L

2200 mmHg pressure ratio

decreases volume

Calculation with Boyle’s Law (continued)

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Example 1

If a sample of helium gas has a volume of 120 mLand a pressure of 850 mmHg, what is the newvolume if the pressure is changed to 425 mmHg?

1) 60 mL 2) 120 mL 3) 240 mL

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Choice 3) 240 mL P1 = 850 mmHg P2 = 425 mmHg

V1 = 120 mL V2 = ??

V2 = V1 x P1 = 120 mL x 850 mmHg = 240 mL P2 425 mmHg

Pressure ratio

increases volume

Example 1 Answer

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Example 2

A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?

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P1 = 4.0 atm P2 = 1.0 atm V1 = 6.0 L V2 = ??

V2 = V1 x P1 = 6.0 L x 4.0 atm = 24 L P2 1.0 atm

Pressure ratio

increases volume

Example 2: A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?

Check: Since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it

does.

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If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)?

A) 3.2 L B) 6.4 L C) 12.8 L

Example 3

Choice A) 3.2 L

V2 = V1 x P1

P2 V2 = 6.4 L x 0.70 atm = 3.2 L

1.40 atmVolume decreases when there is an increase in the pressure (temperature is constant.)

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Example 3: Solution

P1 = 0.7 atm P2 = 1.40 atmV1 = 6.4 L V2 = ??

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Example 4

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

1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg

Copyright © 2009 by Pearson Education, Inc.

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Example 4: Solution

Choice 1) 200. mmHg Data Table Conditions 1 Conditions 2P1 = 600. mmHg P2 = ???

V1 = 12.0 L V2 = 36.0 L

P2 = P1 x V1

V2

600. mmHg x 12.0 L = 200. mmHg 36.0 L

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Temperature and Volume

Charles’s Lawhttp://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm

Tro's Introductory Chemistry, Chapter 11

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Standard Conditions

When doing gas problems, always have temperatures in kelvin.K = °C + 273

Common reference points for comparing.Standard pressure = 1.00 atm.Standard temperature = 0 °C= 273 K.STP.

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Charles’s Law

In Charles’s Law,• the Kelvin temperature

of a gas is directly related to the volume.

• P and amount of molecules (n) are constant.

• when the temperature of a gas increases, its volume increases.

Copyright © 2009 by Pearson Education, Inc.

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Volume and Temperature

Because the hot air in theballoon is less dense than thesurrounding air, it rises.

As a gas is heated, it expands.This causes the density of thegas to decrease.

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Charles’s Law: V and T

• For two conditions, Charles’s law is writtenV1 = V2

(P and n constant)

T1 T2

• Rearranging Charles’s law to solve for V2:T2 x V1 = V2 x T1

T1 T1

V2 = V1 x T2

T1

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Learning Check

Solve Charles’s law expression for T2.

V1 = V2

T1 T2

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Solution

V1 = V2

T1 T2

Cross-multiply to give: V1T2 = V2T1

Isolate T2 by dividing through by V1:V1T2 = V2T1

V1 V1

T2 = T1 x V2 V1

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A balloon has a volume of 785 mL at 21 °C. If thetemperature drops to 0 °C, what is the new volume ofthe balloon (P constant)?

1. Set up data table:Conditions 1 Conditions 2V1 = 785 mL V2 = ?

T1 = 21 °C = 294 K T2 = 0 °C = 273 K

Be sure to use the Kelvin (K) temperature ingas calculations.

E1: Using Charles’s Law

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E1: Solution

2. Solve Charles’s law for V2:

V1 = V2

T1 T2

V2 = V1 x T2

T1

V2 = 785 mL x 273 K = 729 mL 294 K

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A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?

1) 443 °C2) 170 °C 3) - 82 °C

E2

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2. 170 °C

T2 = T1 x V2

V1

T2 = 291 K x 640 mL = 443 K 420 mL

= 443 K – 273 = 170 °C

E2: Solution

1. Set up data table:Conditions 1 Conditions 2V1 = 420 mL V2 = 640 mL

T1 = 18 °C +273= 291 K T2 = ?

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Use the gas laws to complete each sentence with

1) increases or 2) decreases.

A. Pressure _______ when V decreases.

B. When T decreases, V _______.

C. Pressure _______ when V changes from 12 L to 24 L.

D. Volume _______when T changes from 15 °C to 45 °C.

Learning Check

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Use the gas laws to complete each sentence with

1) increases or 2) decreases.

A. Pressure 1) increases when V decreases.

B. When T decreases, V 2) decreases.

C. Pressure 2) decreases when V changes from 12 L to 24 L.

D. Volume 1) increases when T changes from 15 °C to 45 °C.

Solution