BEHAVIOR OF GASES Chapter 12 General Properties of Gases There is a lot of “free” space in a gas. The particles of gas are considered to have insignificant

BEHAVIOR OF BEHAVIOR OF GASESGASESChapter 12Chapter 12

General General Properties of Properties of

GasesGases• There is a lot of “free” space in a There is a lot of “free” space in a gas. The particles of gas are gas. The particles of gas are considered to have insignificant considered to have insignificant volume.volume.

• Gases can be expanded infinitely. Gases can be expanded infinitely. Gases occupy containers Gases occupy containers uniformly and completely.uniformly and completely.

• Gases diffuse and mix rapidly. Gases diffuse and mix rapidly. Perfectly elastic collisions.Perfectly elastic collisions.

KINETIC MOLECULAR THEORYKINETIC MOLECULAR THEORY(KMT)(KMT)

Theory used to explain gas laws. KMT Theory used to explain gas laws. KMT assumptions areassumptions are

• Gases consist of molecules in constant, Gases consist of molecules in constant, random motion.random motion.

• P arises from collisions with container walls.P arises from collisions with container walls.

• No attractive or repulsive forces between No attractive or repulsive forces between molecules. Collisions elastic.molecules. Collisions elastic.

• Volume of molecules is negligible.Volume of molecules is negligible.

THREE THREE STATES STATES

OF OF MATTERMATTER

THREE THREE STATES STATES

OF OF MATTERMATTER

Properties of Properties of GasesGases

Gas properties can be Gas properties can be modeled using math. Model modeled using math. Model depends on—depends on—

• V = volume of the gas (L)V = volume of the gas (L)• T = temperature (K)T = temperature (K)• n = amount (moles)n = amount (moles)• P = pressureP = pressure

(atmospheres) (atmospheres)

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

Can be derived from Can be derived from experiment and theory.experiment and theory.

P V = n R TP V = n R T

Boyle’s LawBoyle’s Law

If n and T are If n and T are constant, thenconstant, then

PV = (nRT) = kPV = (nRT) = kThis means, that This means, that

PP(pressure)(pressure) goes up as goes up as VV(volume)(volume) goes down. goes down.

Robert Boyle Robert Boyle (1627-1691). (1627-1691). Son of Early of Son of Early of Cork, Ireland.Cork, Ireland.

Boyle’s LawBoyle’s Law

If (nRT) are If (nRT) are constant, = to 1, constant, = to 1, then Boyles Law then Boyles Law becomes:becomes:

PP11VV1 1 = P= P22VV22

Robert Boyle Robert Boyle (1627-1691). (1627-1691). Son of Early of Son of Early of Cork, Ireland.Cork, Ireland.

Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s LawA bicycle pump is a good example of A bicycle pump is a good example of

Boyle’s law. Boyle’s law.

As the volume of the air trapped in the As the volume of the air trapped in the pump is reduced, its pressure goes up, pump is reduced, its pressure goes up, and air is forced into the tire.and air is forced into the tire.

Charles’s original balloonCharles’s original balloon

Modern long-distance balloonModern long-distance balloon

Sample problem 14.1A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa?

P1V1 = P2V2

P1 = 103 kPa P2 = 25.0 kPa

V1 = 30.0 L V2 = ?????

V1 x P1

P2

V2 =

Sample problem 14.1A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa?

30.0 L x 103 kPa

25.0 kPa

V2 =

V1 x P1

P2

V2 =

= 124 L

Question 1and 2

CPS

Charles’s Charles’s LawLaw

If n and P are If n and P are constant, thenconstant, then

V = (nR/P)T = kTV = (nR/P)T = kTV and T are directly V and T are directly

related.related. Jacques Charles (1746-Jacques Charles (1746-1823). Isolated boron 1823). Isolated boron and studied gases. and studied gases. Balloonist.Balloonist.

Charles’s LawCharles’s Law

If (nR/P) are If (nR/P) are constant, = to 1, constant, = to 1, then Charles’s then Charles’s Law becomes:Law becomes:

V1 V2=

T1 T2

Jacques Charles (1746-Jacques Charles (1746-1823). Isolated boron 1823). Isolated boron and studied gases. and studied gases. Balloonist.Balloonist.

Charles’s LawCharles’s Law

Sample problem 14.2A balloon inflated in a room at 24oC has a volume of 4.00 L. The ballon is then heated to a temperature of 58C. What is the new volume if the pressure remains constant?

V1 = V2

T1 = 24oC T2 = 58oC

V1 = 30.0 L V2 = ?????

V1 x T2

T1

V2 =

T1 T2

Sample problem 14.2A balloon inflated in a room at 24oC has a volume of 4.00 L. The ballon is then heated to a temperature of 58C. What is the new volume if the pressure remains constant?

V1 = V2

4.00 L x 331 K

297 K

V2 =

T1 T2

Change the temperature to Kelvins

= 4.46 L

Question 3

CPS

The The Combined Combined Gas LawGas Law

Bigger combines use more gasBigger combines use more gasThe combined gas law is when the only variable held constant is the amount (mol) of gas.

Sample problem 14.4The volume of a gas-filled balloon is 30.0 L at 313 K and 152 kPa pressure. What would the volume be at standard temperature and pressure?

P1V1 = P2V 2

T1 = 313 K T2 = 273 K

V1 = 30.0 L P2 = 101.3 kPa

P1 = 153 kPa V2 = ????

T1 T2

Sample problem 14.4The volume of a gas-filled balloon is 30.0 L at 313 K and 152 kPa pressure. What would the volume be at standard temperature and pressure?

P1V1 = P2V 2 T1 T2

P1V1T2 = V2 T1P2

152 kPa • 30.0L • 273 KT2 = V2

313 K • 101.3 kPa

39.5 L = V2

Question 4

CPS

IDEAL GAS LAWIDEAL GAS LAW

P V = n R TP V = n R T

R is the ideal gas constant.

R = 8.314 L•kPa/K•mol or 0.0821 L•atm/K•mol

Avogadro’s Avogadro’s HypothesisHypothesis

Equal volumes of gases at the same T Equal volumes of gases at the same T and P have the same number of and P have the same number of molecules.molecules.

V = n (RT/P) = knV = n (RT/P) = knV and n are directly related.V and n are directly related.

twice as many twice as many moleculesmolecules

Avogadro’s Avogadro’s HypothesisHypothesisAvogadro’s Avogadro’s HypothesisHypothesis

The gases in this experiment are all The gases in this experiment are all measured at the same T and P.measured at the same T and P.

Using PV = nRTUsing PV = nRTHow much NHow much N22 is req’d to fill a small room is req’d to fill a small room

with a volume of 960 cubic feet (27,000 L) with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 to P = 745 mm Hg at 25 ooC?C?

R = 0.082057 L•atm/K•molR = 0.082057 L•atm/K•molSolutionSolution1. Get all data into proper units1. Get all data into proper units V = 27,000 LV = 27,000 L T = 25 T = 25 ooC + 273 = 298 KC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) P = 745 mm Hg (1 atm/760 mm Hg)

= 0.98 atm = 0.98 atm

Using PV = nRTUsing PV = nRTHow much NHow much N22 is req’d to fill a small room with a is req’d to fill a small room with a

volume of 960 cubic feet (27,000 L) to P = 745 volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 mm Hg at 25 ooC?C?

R = 0.082057 L•atm/K•molR = 0.082057 L•atm/K•mol

SolutionSolution

2. Now calc. n = PV / RT2. Now calc. n = PV / RT

n = (0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)n =

(0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)

n = 1.1 x 10n = 1.1 x 1033 mol (or about 30 kg of gas) mol (or about 30 kg of gas)

Gases and Gases and StoichiometryStoichiometry

2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume in a flask with a volume of 2.50 L. What is the pressure of Oof 2.50 L. What is the pressure of O22 at 25 at 25 ooC? C? Of HOf H22O?O?

Bombardier beetle Bombardier beetle uses decomposition uses decomposition of hydrogen peroxide of hydrogen peroxide to defend itself.to defend itself.

Gases and StoichiometryGases and Stoichiometry2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of in a flask with a volume of 2.50 L. What is the pressure of O2.50 L. What is the pressure of O22 at 25 at 25 ooC? Of C? Of HH22O?O?

SolutionSolution

Strategy: Strategy: Calculate moles of HCalculate moles of H22OO22 and then and then

moles of Omoles of O22 and H and H22O. O.

Finally, calc. P from n, R, T, and V.Finally, calc. P from n, R, T, and V.


Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of in a flask with a volume of

2.50 L. What is the pressure of O2.50 L. What is the pressure of O22 at 25 at 25 ooC? Of C? Of

HH22O?O?

SolutionSolution1.1 g H2O2 •

1 mol34.0 g

0.032 mol1.1 g H2O2 • 1 mol34.0 g

0.032 mol

0.032 mol H2O2 • 1 mol O2

2 mol H2O2= 0.016 mol O20.032 mol H2O2 •

1 mol O22 mol H2O2

= 0.016 mol O2


Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of 2.50 L. What is in a flask with a volume of 2.50 L. What is the pressure of Othe pressure of O22 at 25 at 25 ooC? Of HC? Of H22O?O?

SolutionSolution

P of O2 = nRT/V

= (0.016 mol)(0.0821 L • atm/K •mol)(298 K)

2.50 L

P of O2 = nRT/V

= (0.016 mol)(0.0821 L • atm/K •mol)(298 K)

2.50 L

P of OP of O22 = 0.16 atm = 0.16 atm

Gases and StoichiometryGases and Stoichiometry

What is P of HWhat is P of H22O? Could calculate as above. O? Could calculate as above. But recall Avogadro’s hypothesis. But recall Avogadro’s hypothesis.

V V n at same T and Pn at same T and P

P P n at same T and Vn at same T and V

There are 2 times as many moles of HThere are 2 times as many moles of H22O as O as moles of Omoles of O22. P is proportional to n. . P is proportional to n. Therefore, P of HTherefore, P of H22O is twice that of OO is twice that of O22..

P of HP of H22O = 0.32 atmO = 0.32 atm


Question 1 Exam View

CPS

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

What is the total pressure in the flask?What is the total pressure in the flask?

PPtotaltotal in gas mixture = P in gas mixture = PAA + P + PBB + ... + ...Therefore, Therefore,

PPtotaltotal = P(H = P(H22O) + P(OO) + P(O22) = 0.48 atm) = 0.48 atm

Dalton’s Law: total P is sum of Dalton’s Law: total P is sum of PARTIALPARTIAL pressures. pressures.


0.32 atm 0.32 atm 0.16 atm 0.16 atm

Sample problem 14.6Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.30 kPa of total pressure if the partial pressure of nitrogen, carbon dioxide, and other gases arte 79.10 kPa, 0.040 kPa, and 0.94 kPa, respectively?

Ptotal = PO2 + PN2 + PCO2 + Pother

Rearrange Dalton’s Law to solve for Po2

PO2 = Ptotal - (PN2 + PCO2 + Pother)

= 101.30 Kpa – (79.10 kPa + 0.040 kPa + 0.94 kPa)

= 21.22 kPa

Question 1 Exam View

CPS

Dalton’s Dalton’s LawLaw

John DaltonJohn Dalton1766-18441766-1844

Kinetic Molecular Kinetic Molecular TheoryTheory

Because we assume molecules are in Because we assume molecules are in motion, they have a kinetic energy.motion, they have a kinetic energy.

KE = (1/2)(mass)(speed)KE = (1/2)(mass)(speed)22

At the same T, all gases At the same T, all gases have the same average KE.have the same average KE.

At the same T, all gases At the same T, all gases have the same average KE.have the same average KE.

As T goes up, KE also increases — As T goes up, KE also increases — and so does speed.and so does speed.

Kinetic Molecular Kinetic Molecular TheoryTheory

At the same T, all gases have the same At the same T, all gases have the same average KE.average KE.

As T goes up, KE also increases — and As T goes up, KE also increases — and so does speed.so does speed.

Velocity of Gas MoleculesVelocity of Gas MoleculesAverage velocity decreases with increasing Average velocity decreases with increasing

mass.mass.

GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

• diffusiondiffusion is the is the gradual mixing of gradual mixing of molecules of different molecules of different gases.gases.

• effusioneffusion is the is the movement of molecules movement of molecules through a small hole through a small hole into an empty container.into an empty container.


Molecules effuse thru holes in a rubber Molecules effuse thru holes in a rubber balloon, for example, at a rate (= balloon, for example, at a rate (= moles/time) that ismoles/time) that is

• proportional to Tproportional to T• inversely proportional to M.inversely proportional to M.Therefore, He effuses more rapidly Therefore, He effuses more rapidly

than Othan O22 at same T. at same T.

HeHe


Graham’s law governs Graham’s law governs effusion and diffusion effusion and diffusion of gas molecules.of gas molecules.

Thomas Graham, 1805-1869. Thomas Graham, 1805-1869. Professor in Glasgow and London.Professor in Glasgow and London.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

M of AM of B

Rate for B

Rate for A

Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion

Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

Using KMT to Using KMT to Understand Understand Gas Gas

LawsLawsRecall that KMT assumptions areRecall that KMT assumptions are• Gases consist of molecules in constant, Gases consist of molecules in constant,

random motion.random motion.• P arises from collisions with container P arises from collisions with container

walls.walls.• No attractive or repulsive forces between No attractive or repulsive forces between

molecules. Collisions elastic.molecules. Collisions elastic.• Volume of molecules is negligible.Volume of molecules is negligible.

Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

P proportional to nP proportional to n

Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

P proportional to TP proportional to T

Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular

TheoryTheory

Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular

TheoryTheory

P proportional to 1/VP proportional to 1/V

Deviations from Deviations from Ideal Gas LawIdeal Gas Law

• Real molecules Real molecules

have have volumevolume..• There are There are

intermolecular intermolecular forcesforces..– Otherwise a gas Otherwise a gas

could not could not become a liquid.become a liquid.

Fig. 12.20Fig. 12.20

Documents

BEHAVIOR OF GASES Chapter 12 General Properties of Gases There is a lot of “free” space in a gas. The particles of gas are considered to have insignificant