Chapter 12 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena The Gaseous State

Chapter 12

Introduction to General, Organic, and Biochemistry 10eJohn Wiley & Sons, Inc

Morris Hein, Scott Pattison, and Susan Arena

The Gaseous State of Matter

The air in a hot air balloon expands When it is heated. Some of the airescapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.

Chapter Outline

Copyright 2012 John Wiley & Sons, Inc

12.1 General Properties

12.2 The Kinetic-Molecular Theory

12.3 Measurement of Pressure

12.4 Dependence of Pressure on Number of Molecules and Temperature

12.5 Boyle’s Law

12.6 Charles’ Law

12.7 Gay-Lussac’s Law

12.8 Combined Gas Laws

12.9 Dalton’s Law of Partial Pressures

12.10 Avogadro’s Law

12.11 Mole-Mass-Volume Relationships of Gases

12.12 Density of Gases

12.13 Ideal Gas Law

12.14 Gas Stoichiometry

12.15 Real Gases

Objectives for Today

Kinetic Molecular Theory of Gases Gas Measurements Boyle’s, Charles’ and Gay-Lussac’s Laws


GASES AND KINETIC MOLECULAR THEORY


General Properties

• Gases • Have an indefinite volume

Expand to fill a container• Have an indefinite shape

Take the shape of a container• Have low densities

• Have high kinetic energies


2

air

H O

d 1.2 g / L at 25 C

d 1.0 g / mL

Kinetic Molecular Theory (KMT)

Assumptions of the KMT and ideal gases include:1. Gases consist of tiny particles2. The distance between particles is large compared

with the size of the particles.3. Gas particles have no attraction for each other4. Gas particles move in straight lines in all

directions, colliding frequently with each other and with the walls of the container.


Kinetic Molecular Theory

Assumptions of the KMT (continued):5. Collisions are perfectly elastic (no energy is lost in

the collision). 6. The average kinetic energy for particles is the

same for all gases at the same temperature.

7. The average kinetic energy is directly proportional to the Kelvin temperature.


21KE = where is mass and is velocity

2mv m v

Diffusion


Effusion

• Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure.

• Graham’s law of effusion:


rate of effusion of gas A density B molar mass B= =

rate of effusion of gas B density A molar mass A

Your Turn!

Which gas will diffuse most rapidly?a. Heb.Nec. Ard.Kr


rate of effusion of gas A density B molar mass B= =

rate of effusion of gas B density A molar mass A

Measurement of Pressure

ForcePressure =

Area


Pressure depends on the•Number of gas molecules•Temperature of the gas•Volume the gas occupies

Atmospheric Pressure

• Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface.


Barometer


Pressure Conversions

• Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm


1 atm675 mm Hg = 0.888 atm

760 mm Hg

Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr.

760 torr675 mm Hg = 675 torr

760 mm Hg

Your Turn!

A pressure of 3.00 atm is equal toa. 819 torrb.3000 torrc. 2280 torrd.253 torr


Dependence of Pressure on Number of Molecules


P is proportional to n (number of molecules) at Tc (constant T) and Vc (constant V).The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.

Dependence of Pressure on Temperature


P is proportional to T at nc

(constant number of moles) and Vc.

The increased pressure is due to• more frequent collisions•higher energy collisions

Your Turn!

If you change the temperature of a sample of gas from 80°C to 25°C at constant volume, the pressure of the gas

a. will increase.b.will decrease.c. will not change


Boyle’s Law


1 1 2 2

1At and : α or c cT n V PV PV

P

What happens to V if you double P?•V decreases by half!

What happens to P if you double V?•P decreases by half!

Boyle’s Law• A sample of argon gas occupies 500.0 mL at

920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature.


1 12

2

PV

PV

1 1 2 2 PV PV

2

920. torr 500. mL = = 491 torr

937 mLP

Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL

Calculate

Set-Up

Boyle’s Law

• Another approach to the same problem:• Since volume increased from 500. mL to 937

ml, the pressure of 920. torr must decrease.• Multiply the pressure by a volume ratio that

decreases the pressure:


2

500. mL

937 mL = 920. torr = 491 torrP

Your Turn!

A 6.00 L sample of a gas at a pressure of 8.00 atm is compressed to 4.00 L at a constant temperature. What is the pressure of the gas?

a. 4.00 atmb.12.0 atmc. 24.0 atmd.48.0 atm


Your Turn!

• A 400. mL sample of a gas is at a pressure of 760. torr. If the temperature remains constant, what will be its volume at 190. torr?

• A. 100. mL• B. 400. mL• C. 25.0 mL• D. 1.60x102 mL


Charles’ Law


1 2

1 2

At and :

α or

c cP n

V VV T

T T

• The volume of an ideal gas at absolute zero (-273°C) is zero.

• Real gases condense at their boiling point so it is not possible to have a gas with zero volume.

• The gas laws are based on Kelvin temperature.

• All gas law problems must be worked in Kelvin!

Charles’ Law• A 2.0 L He balloon at 25°C is taken outside on

a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant?


1 2

1 2

V V

T T

1 22

1

rearranged gives VT

VT

2

(2.0 L)(258 K) = = 1.7 L

298 KV

Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K

Calculate

Set-Up 1 2

1 2

V V

T T

Charles’ Law

• Another approach to the same problem:• Since T decreased from 25°C to -15°C, the

volume of the 2.0L balloon must decrease.• Multiply the volume by a Kelvin temperature

ratio that decreases the volume:


2

258K

298 = 2.0L = 1.7L

KP

Your Turn

The volume of a gas always increases whena. Temperature increases and pressure

decreasesb.Temperature increases and pressure increasesc. Temperature decreases and pressure

increases d.Temperature decreases and pressure

decreases


Your Turn!

A sample of CO2 has a volume of 200. mL at 20.0

° C. What will be its volume at 40.0 °C, assuming that the pressure remains constant?

a. 18.8 mLb.100. mLc. 213 mLd.400. mL


Your Turn!

A sample of gas has a volume of 3.00 L at 10.0 °C. What will be its temperature in °C if the gas expands to 6.00 L at constant pressure?

a. 20.0°Cb.293°Cc. 566°Cd.142°C


Gay-Lussac’s Law


1 2

1 2

At and : α or c c

P PV n P T

T T

31

40oC + 273 = 313 K

At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC ,what will be the pressure of the oxygen?

Method A. Conversion FactorsStep 1. Change oC to K:

oC + 273 = K

100oC + 273 = 373 K

temperature increases pressure increases

Determine whether temperature is beingincreased or decreased.

32

Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:

= 25.6 atmP = (21.5 atm)373K313K

At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?

33

= 25.6 atmP = (21.5 atm)373K313K

A temperature ratio greater than 1 will

increase the pressure


34

Method B. Algebraic Equation

Step 1. Organize the information (remember to make units the same):

P1 = 21.5 atm T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K


35

Step 2. Write and solve the equation for the unknown:

1 22

1

P TP =

T1 2

1 2

P P =

T T


36

Step 3. Put the given information into the equation and calculate:

= 25.6 atm1 2

21

P TP =

T


P1 = 21.5 atm

T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K

(21.5 atm)(373 K) =

313 K


Kinetic Molecular Theory of Gases Gas Measurements Boyle’s, Charles’ and Gay-Lussac’s Laws



Combined Gas LawDalton’s Law of Partial PressureAvogadro’s LawDensity of Gases


Combined Gas Laws


Used for calculating the results of changes in gas conditions.

1 1 2 2

1 2

PV PV

T T

•Boyle’s Law where Tc

•Charles’ Law where Pc

1 2

1 2

V V

T T

1 1 2 2 PV PV

•Gay Lussacs’ Law where Vc1 2

1 2

P P

T T

P1 and P2 , V1 and V2 can be any units as long as they are the same. T1 and T2 must be in Kelvin.

Combined Gas Law


If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr?

STP: Standard Temperature 273K or 0°CStandard Pressure 1 atm or 760 torr

1 1 22

1 2

PVT

VT P

Knowns V1 = 500. mL T1 =273K P1= 760 torrT2 = 85°C = 358K P2= 560 torr

Set-Up

2

(760 torr)(500. mL)(358K) = 890. ml

(273K)(560 torr)V Calculate

1 1 2 2

1 2

PV PV

T T

Combined Gas Law• A sample of oxygen gas occupies 500.0 mL at

722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg.


1 2 22

1 1

T PVT

PV

1 1 2 2

1 2

PV PV

T T

2

491 torr 2530 ml 248K= =853K 580 C

722 torr 500.0 mlT

Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr V2 = 2.53 L = 2530 mL P2= 560 torr

Set-Up

Calculate

Your Turn!

A sample of gas has a volume of 8.00 L at 20.0 ° C and 700. torr. What will be its volume at STP?

a. 1.20 Lb.9.32 Lc. 53.2 Ld.6.87 L


Dalton’s Law of Partial Pressures

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

• PTotal = PA + PB + PC + ….

• Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air.


2 2 2 2.Air N O Ar CO H OP P P P P P

Collecting Gas Over Water

• Gases collected over water contain both the gas and water vapor.

• The vapor pressure of water is• constant at a given temperature• Pressure in the bottle is • equalized so that the Pinside = Patm


2 atm gas H OP P P

Your Turn!

A sample of oxygen is collected over water at 22

° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor pressure of water at 22°C is 19.8 torr.

a. 742 torrb.782 torrc. 784 torrd.750. torr


Avogadro’s Law

• Equal volumes of different gases at the same T and P contain the same number of molecules.


1 volume1 molecule

1 mol

1 volume1 molecule

1 mol

2 volumes2 molecules

2 mol

The ratio isthe same:

Mole-Mass-Volume Relationships

• Molar Volume: One mole of any gas occupies 22.4 L at STP.

• Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP.

Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm

Set-Up

Calculate

22.4 L1 mol = 22.4 L so the conversion factor is

1mol

22.4 L

1 mol

3.94 g 3.52 L= 25.1 g/mol

Your Turn!

What is the molar mass of a gas if 240. mL of the gas at STP has a mass of 0.320 grams?

a. 8.57 gb.22.4 gc. 16.8 gd.29.9 g

Density of Gases

• Calculate the density of nitrogen gas at STP.

• Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.

mass gd = =

volume L

STP

1 mold = molar mass

22.4 L

STP

28.02 g 1 mold = = 1.25g/L

1 mol 22.4 L

Your Turn!

Which of the following gases is the most dense?a. H2

b.N2

c. CO2

d.O2

Carbon dioxide fire extinguishers can be used to put out fires because CO2 is more dense than air and can be used to push oxygen away from the fuel source.


Combined Gas LawDalton’s Law of Partial PressureAvogadro’s LawDensity of Gases


Topics for Today

The Ideal Gas Law Gas Stoichiometry Real Gases & Air Pollutants

IDEAL GAS LAW

PV nRT R

L atm where = 0.0821

mol K

P is in units of AtmospheresV is in units of Litersn is in units of molesT is in units of Kelvin

Example 1: Calculate the volume of 1 mole of any gas at STP.

PV nRT R


mol K

Knowns n = 1 mole T = 273K P = 1 atm

Set-Up

Calculate

nRT

VP

L atm(1 mol)(0.0821 )(273 K)

mol K = (1 atm)

V

Molar volume!

= 22.4 L

Example 2: How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg?

Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm

Set-Up

Calculate

PV

nRT

(0.980 atm)(1.3 L) = =0.052 mol

L atm(0.0821 )(297 K)

mol K

n

PV nRT R


mol K

Example 3: Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr.

Knowns m =4.12 g V = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm

Set-Up

Calculate

g g = so =

M Mn PV RT

L atm(4.12 g)(0.0821 )(296 K)

mol KM = =107 g/mol(0.988 atm)(0.943 L)

R T

P V

g M =

Your Turn!

What is the molar mass of a gas if 40.0 L of the gas has a mass of 36.0 g at 740. torr and 30.0 °

C?a. 33.1 gb.23.0 gc. 56.0 gd.333 g

STOICHIOMETRY & GASES

•Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L.

•Use the Ideal Gas Law if the conditions are not at STP.

Example 4: Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm.

P4(s) + 6H2(g) 4PH3(g)

2 4.0 L H2 1 mol H

22.4L

4

2

1 mol P

6 mol H

4= 0.030 mol P

Knowns V = 4.0 L T = 273 K P = 1 atm

Solution Map L H2 mol H2 mol P4

Calculate

Example 5: What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C2H6?

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)

Solution Map m C2H6 mol C2H6 mol O2 volume O2

Knowns m = 3.2 g C2H6 T = 25°C = 298K P = 1 atm

2 63.2g C H 2 6

2 6

1 mol C H

30.08g C H

2

2 6

7 mol O

2 mol C H

2= 0.37mol OCalculate

L atm(0.37 mol)(0.0821 )(298 K)

mol K = = 9.1 L(1 atm)

V

Your Turn!

How many moles of oxygen gas are used up during the reaction with 18.0 L of CH4 gas measured at STP?

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

a. 1.61 molesb. 2.49 molesc. 18.0 molesd. 36.0 moles

Example 6: Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm.

N2(g) + 3H2(g) 2NH3(g)

2 9.0 L H2

2

1 L N

3 L H

2= 3.0 L N

Knowns V = 9.0 L T = 450K P = 5.00 atm

Solution Map Assume T and P for both gases are the same.Use volume ratio instead of mole ratio!L H2 L N2

Calculate

Your Turn!

What volume of sulfur dioxide gas will react when 12.0 L of oxygen is consumed at constant temperature and pressure?

2 SO2 + O2 2 SO3

a. 6.00 Lb. 12.0 Lc. 24.0 Ld. 60.0 L

REAL GASES & AIR POLLUTANTS

• Most real gases behave like ideal gases under ordinary temperature and pressure conditions.

• Conditions where real gases don’t behave ideally:• At high P because the distance between particles is

too small and the molecules are too crowded together.

• At low T because gas molecules begin to attract each other.

• High P and low T are used to condense gases.

• Stratospheric Ozone

• Tropospheric Ozone

• Oxides of Nitrogen

• Acid Rain

• Greenhouse Gases

Topics for Today

The Ideal Gas Law Gas Stoichiometry Real Gases & Air Pollutants

Documents

Chapter 12 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena The Gaseous State