Chapter 19 Chapter 19 Thermal Properties of Thermal Properties of

MatterMatterA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: After finishing Objectives: After finishing this unit, you should be this unit, you should be able to:able to:

• Write and apply relationships among pressure, volume, temperature, and quantity of matter for ideal gases undergoing changes of state.

• Write and apply the general gas law for a particular state of an ideal gas.

• Define and apply concepts involving molecular mass, moles, and Avogadro’s number.

Thermodynamic StateThermodynamic State

The The thermodynamic statethermodynamic state of a gas of a gas is defined by four coordinates:is defined by four coordinates:

• Absolute pressure, Absolute pressure, PP

• Absolute temperature, Absolute temperature, TT

• Volume, Volume, VV

• Mass Mass mm or quantity of matter or quantity of matter nn

Gas Laws Between StatesGas Laws Between States

Boyle’s LawBoyle’s Law, , Charles’ LawCharles’ Law, and , and Gay-Lusac’s LawGay-Lusac’s Law can be combined into a single formula for an ideal can be combined into a single formula for an ideal gas that changes from gas that changes from State 1State 1 to another to another State 2State 2..

Boyle’s LawBoyle’s Law, , Charles’ LawCharles’ Law, and , and Gay-Lusac’s LawGay-Lusac’s Law can be combined into a single formula for an ideal can be combined into a single formula for an ideal gas that changes from gas that changes from State 1State 1 to another to another State 2State 2..

P1, V1

T1 m1

P2, V2

T2 m2

Any Factor that remains constant

divides out

Any Factor that remains constant

divides out

1 1 2 2

1 1 2 2

PV PV

m T m T1 1 2 2

1 1 2 2

PV PV

m T m T

State State 11

State State 22

Example 1:Example 1: An auto tire has an gauge An auto tire has an gauge pressure of pressure of 28 psi28 psi in the morning at in the morning at 202000CC. After driving for hours the . After driving for hours the temperature of air inside the tire is temperature of air inside the tire is 303000CC. What will the gauge read? . What will the gauge read? (Assume (Assume 1 atm = 14.7 psi1 atm = 14.7 psi.).)

TT11 = = 20 + 273 = 20 + 273 = 293 K293 K

TT22 = = 30 + 273 = 30 + 273 = 303 K303 K

PPabsabs = P = Pgaugegauge + 1 atm; + 1 atm; PP11 = = 28 + 14.7 = 42.7 28 + 14.7 = 42.7 psipsi

1 1 2 2

1 1 2 2

PV PV

m T m T1 1 2 2

1 1 2 2

PV PV

m T m T

Same air in tires: Same air in tires: mm11 = = mm22Same volume of air: Same volume of air: VV11 = =

VV22

Example 1:Example 1: What will the gauge read? What will the gauge read?

1 1 2 2

1 1 2 2

PV PV

m T m T1 1 2 2

1 1 2 2

PV PV

m T m T

Given: Given: TT11 = = 293 K293 K; T; T22 = = 303 K303 K; P; P11 = = 42.7 42.7

psipsi

1 2

1 2

P P

T T1 2

1 2

P P

T T

1 22

1

(42.7 psi)(303 K)

293 K

PTP

T PP22 = = 44.2 44.2

psipsi

Gauge pressure is 14.7 psi less than this value:Gauge pressure is 14.7 psi less than this value:

PP22 = = 44.2 psi - 14.7 psi ;44.2 psi - 14.7 psi ; P2 = 29.5 psi

P2 = 29.5 psi

The Composition of MatterThe Composition of MatterWhen dealing with gases, it is much more When dealing with gases, it is much more

convenient to work with convenient to work with relativerelative masses of atoms. masses of atoms.

Atoms contain Atoms contain protonsprotons and and neutronsneutrons , which are , which are close to the same mass, surrounded by close to the same mass, surrounded by electronselectrons

which are almost negligible by comparison.which are almost negligible by comparison.

Building blocks of

atoms.electron

proton neutron

Helium atom

Relative MassesRelative MassesTo understand relative scales, let’s ignore electrons and To understand relative scales, let’s ignore electrons and

compare atoms by total number of nuclear particles.compare atoms by total number of nuclear particles.

Oxygen, O 16 particles

Carbon, C 12 particles

Lithium, Li 7 particles

Helium, He 4 particles

Hydrogen, H 1 particle

Atomic masses of a few elements:

Atomic MassAtomic MassThe The atomic massatomic mass of an element is the mass of an of an element is the mass of an atom of the element compared with the mass of an atom of the element compared with the mass of an atom of carbon taken as atom of carbon taken as 1212 atomic mass units ( atomic mass units (uu).).

Carbon, C = 12.0 u

Nitrogen, N = 14.0 u

Neon, Ne = 20.0 u

Copper, Cu = 64.0 u

Hydrogen, H = 1.0 u

Helium, He = 4.0 u

Lithium, Li = 7.0 u

Beryllium, Be = 9.0 u

Molecular MassMolecular Mass

The The molecular massmolecular mass M M is the sum of the atomic is the sum of the atomic masses of all the atoms making up the molecule.masses of all the atoms making up the molecule.

Consider Carbon Dioxide (CO2)

1 C = 1 x 12 u = 12 u

2 O = 2 x 16 u = 32 u

CO2 = 44 u

The molecule has one carbon atom and two oxygen atoms

Definition of a MoleDefinition of a MoleOne One molemole is that quantity of a substance that is that quantity of a substance that contains the same number of particles as there are contains the same number of particles as there are in in 12 g12 g of carbon-12. ( of carbon-12. (6.023 x 106.023 x 102323 particles particles))

1 mole of Carbon has a mass of 12 g

1 mole of Helium has a mass of 4 g

1 mole of Neon has a mass of 20 g

1 mole of Hydrogen (H2) = 1 + 1 = 2 g

1 mole of Oxygen (O2) is 16 + 16 = 32 g

6.023 x 106.023 x 102323 particles particles

Molecular Mass in Molecular Mass in grams/molegrams/mole

The unit of molecular mass M is grams per mole.

Hydrogen, H = 1.0 g/mol

Helium, He = 4.0 g/mol

Carbon, C = 12.0 g/mol

Oxygen, O = 16.0 g/mol

H2 = 2.0 g/mol

O2 = 16.0 g/mol

H2O = 18.0 g/mol

CO2 = 44.0 g/mol

Each mole has 6.23 x 1023 molecules

Moles and Number of Moles and Number of MoleculesMolecules

Finding the number of Finding the number of moles moles nn in a given in a given number of number of NN molecules: molecules: A

Nn

N

A

Nn

N

Avogadro’s number: NA = 6.023 x 1023 particles/mol

Example 2:Example 2: How many moles of any gas will How many moles of any gas will contain 20 x 10contain 20 x 102323 molecules? molecules?

23

23

20 x 10 molecules

6.023 x 10 molecules molA

Nn

N n = 3.32 moln = 3.32 mol

Moles and Molecular Mass MMoles and Molecular Mass M

Finding the number of Finding the number of moles moles nn in a given mass in a given mass m m of a substance:of a substance:

mn

M

mn

M

Molecular mass M is expressed in grams per mole.

Example 3:Example 3: How many moles are there in How many moles are there in 200 g200 g of oxygen gas O of oxygen gas O22? (M = 32 g/mol)? (M = 32 g/mol)

200 g

32 g/mol

mn

M n = 6.25 moln = 6.25 mol

Example 4:Example 4: What is the mass of a What is the mass of a single atom of boron (M = 11 single atom of boron (M = 11 g/mol)?g/mol)?We are given both a number We are given both a number N = 1N = 1 and and a molecular mass a molecular mass M = 11 g/molM = 11 g/mol. Recall . Recall that:that:

mn

M

mn

M

A

Nn

N

A

Nn

N

A

m N

M N

A

m N

M N

23

(1)(11 g/mol)

6.023 x 10 atoms/molA

NMm

N m = 1.83 x 10-23

gm = 1.83 x 10-23

g

Ideal Gas LawIdeal Gas LawSubstituting moles Substituting moles n n for mass for mass mm, we know , we know that:that:

1 1 2 2

1 1 2 2

PV PV

n T n T1 1 2 2

1 1 2 2

PV PV

n T n T

In other words, the ratio In other words, the ratio PV/nTPV/nT is a constant, and if we is a constant, and if we can find its value, we can work with a single state.can find its value, we can work with a single state.

Since a mole of any gas contains the same number of Since a mole of any gas contains the same number of molecules, it will have the same volume for any gas.molecules, it will have the same volume for any gas.

Volume of one mole of a gas:

V = 22.4 L or 22.4 x 10-3 m3

The Universal Gas The Universal Gas Constant RConstant R

The universal gas constant The universal gas constant R R is defined as follows:is defined as follows:

PVR

nT

PVR

nT PV nRTPV nRT

Evaluate for one mole of gas at 1 atm, 273 K, 22.4 L.Evaluate for one mole of gas at 1 atm, 273 K, 22.4 L.

-3 3(101,300 Pa)(22.4 x 10 m )

(1 mol)(273 K)

PVR

nT

R = 8.314 J/mol·KR = 8.314 J/mol·K

Example 5:Example 5: Two hundred grams of Two hundred grams of oxygen (oxygen (M =M = 32 g/mol32 g/mol) fills a ) fills a 2-L2-L tank tank at a temperature of at a temperature of 252500CC. What is the . What is the absolute pressure absolute pressure P P ofof the gas?the gas? VV = 2 = 2

LLt t = 25= 2500CCm m = 200= 200

gg

O2

TT = 25 = 2500 + 273 + 27300 = 298 = 298 KK

VV = 2 L = 2 x 10 = 2 L = 2 x 10-3-3 m m33

PV nRTPV nRT mn

M

mn

M m

PV RTM

m

PV RTM

-3 3

(200 g)(8.314 J/mol K)(298 K)

(32 g/mol)(2 x 10 m )

mRTP

MV

P = 7.74 MPaP = 7.74 MPa

Example 6:Example 6: How many grams of How many grams of nitrogen gas (M = 28 g/mol) will occupy nitrogen gas (M = 28 g/mol) will occupy a volume of 2.4 ma volume of 2.4 m33 if the absolute if the absolute pressure is 220 kPa and the pressure is 220 kPa and the temperature is 300 K?temperature is 300 K? VV = 2.4 = 2.4

mm33T T = 300 = 300

KKP P = 220= 220 kPakPa

N2m

PV RTM

m

PV RTM

3(220,000 Pa)(2.4 m )(28 g/mol)

(8.314 J/mol K)(300 K)

PVMm

RT

mm = 5930 g = 5930 g oror

m = 5.93 kg

m = 5.93 kg

Summary of FormulasSummary of Formulas

1 1 2 2

1 1 2 2

PV PV

m T m T1 1 2 2

1 1 2 2

PV PV

m T m T 1 1 2 2

1 1 2 2

PV PV

n T n T1 1 2 2

1 1 2 2

PV PV

n T n T

A

Nn

N

A

Nn

N m

nM

m

nM

PVR

nT

PVR

nT PV nRTPV nRT

CONCLUSION: Chapter 19CONCLUSION: Chapter 19Thermal Properties of Thermal Properties of

MatterMatter