Chapter 19 Chapter 19 Thermal Properties of MatterThermal Properties of Matter

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Objectives: After finishing this Objectives: After finishing this unit, you should be able to:unit, you should be 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 StateThe 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 Law, Charles’ Law, and Gay-Lusac’s Law can be combined into a single formula for an ideal gas that changes from State 1 to another State 2.

BoyleBoyle’’s Laws Law, , CharlesCharles’’ LawLaw, and , and GayGay--LusacLusac’’ss LawLaw 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 PVm 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 28 psipsi in the morning at in the morning at 202000CC. . After driving for hours the temperature of air After driving for hours the temperature of air inside the tire is inside the tire is 303000CC. What will the gauge . What will the gauge read? (Assume read? (Assume 1 1 atmatm = 14.7 = 14.7 psipsi.).)

TT11 == 20 + 273 = 293 K20 + 273 = 293 K

TT22 == 30 + 273 = 303 K30 + 273 = 303 K

PPabsabs = = PPgaugegauge + 1 + 1 atmatm;; PP11 = = 28 + 14.7 = 42.7 28 + 14.7 = 42.7 psipsi

1 1 2 2

1 1 2 2

PV PVm T m T

Same air in tires: Same air in tires: mm11 = m= m22

Same volume of air: Same volume of air: VV11 = V= V22

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

1 1 2 2

1 1 2 2

PV PVm 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 PT T

1 22

1

(42.7 psi)(303 K)293 K

PTPT

PP22 = = 44.2 44.2 psipsi

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

PP22 = = 44.2 44.2 psipsi -- 14.7 14.7 psipsi ;; P2 = 29.5 psiP2 = 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, letTo understand relative scales, let’’s ignore electrons and 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 MassThe 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 carbonof carbon--12. (12. (6.023 x 106.023 x 102323 particlesparticles))

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 particlesparticles

Molecular Mass in grams/moleMolecular Mass in grams/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

H2 O = 18.0 g/mol

CO2 = 44.0 g/mol

Each mole has 6.23 x 1023 molecules

Moles and Number of MoleculesMoles and Number of Molecules

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

NnN

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 molecules6.023 x 10 molecules molA

NnN

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:

mnM

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 Oof oxygen gas O22 ? (M = 32 g/mol)? (M = 32 g/mol)

200 g32 g/mol

mnM

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 g/mol)?single atom of boron (M = 11 g/mol)?

We are given both a number We are given both a number N = 1N = 1 and a and a molecular mass molecular mass M = 11 g/molM = 11 g/mol. Recall that:. Recall that:

mnM

A

NnN

A

m NM N

23

(1)(11 g/mol)6.023 x 10 atoms/molA

NMmN

m = 1.83 x 10-23 gm = 1.83 x 10-23 g

Ideal Gas LawIdeal Gas Law

Substituting moles Substituting moles n n for for mass mass mm, we know that:, we know that:

1 1 2 2

1 1 2 2

PV P Vn T n T

In other words, the ratio In other words, the ratio PV/PV/nTnT 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 Constant RThe Universal Gas Constant R

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

PV RnT

PV nRT

Evaluate for one mole of gas at 1 Evaluate for one mole of gas at 1 atmatm, 273 K, 22.4 L., 273 K, 22.4 L.-3 3(101,300 Pa)(22.4 x 10 m )

(1 mol)(273 K)PVRnT

R = 8.314 J/mol·KR = 8.314 J/mol·K

Example 5:Example 5: Two hundred grams of oxygen Two hundred grams of oxygen ((M =M = 32 g/mol32 g/mol) fills a ) fills a 22--LL tank at a tank at a temperature of temperature of 252500CC. What is the absolute . What is the absolute pressure pressure P P ofof the gas?the gas?

VV = 2 L= 2 Lt t = 25= 2500CC

m m = 200= 200 ggO2

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

VV = 2 L = 2 x 10= 2 L = 2 x 10--33 mm33

PV nRTmnM

mPV RTM

-3 3

(200 g)(8.314 J/mol K)(298 K)(32 g/mol)(2 x 10 m )

mRTPMV

P = 7.74 MPaP = 7.74 MPa

Example 6:Example 6: How many grams of nitrogen How many grams of nitrogen gas (M = 28 g/mol) will occupy a volume of gas (M = 28 g/mol) will occupy a volume of 2.4 m2.4 m33 if the absolute pressure is 220 kPa if the absolute pressure is 220 kPa and the temperature is 300 K?and the temperature is 300 K?

VV = 2.4 m= 2.4 m33

T T = 300 K= 300 KP P = 220= 220 kPakPa

N2mPV RTM

3(220,000 Pa)(2.4 m )(28 g/mol)(8.314 J/mol K)(300 K)

PVMmRT

mm = 5930 g or= 5930 g or m = 5.93 kgm = 5.93 kg

Summary of FormulasSummary of Formulas

1 1 2 2

1 1 2 2

PV PVm T m T

1 1 2 2

1 1 2 2

PV P Vn T n T

A

NnN

mnM

PV RnT

PV nRT

CONCLUSION: Chapter 19CONCLUSION: Chapter 19 Thermal Properties of MatterThermal Properties of Matter