Dr. S. M. Condren

Chapter 12

BEHAVIOR OF GASES

Dr. S. M. Condren

BEHAVIOR OF GASES

Dr. S. M. Condren

Importance of Gases

• Airbags fill with NAirbags fill with N22 gas in an accident. gas in an accident.

• Gas is generated by the decomposition of Gas is generated by the decomposition of sodium azide, NaNsodium azide, NaN33..

• 2 NaN2 NaN33 2 Na + 3 N 2 Na + 3 N22

• if bag rupturesif bag ruptures 2 Na + 2 H 2 Na + 2 H22O O 2 NaOH + H 2 NaOH + H22

Dr. S. M. Condren

THREE STATES OF MATTER

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THREE STATES OF MATTER

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General Properties of Gases

• There is a lot of “free” There is a lot of “free” space in a gas.space in a gas.

• Gases can be expanded Gases can be expanded infinitely.infinitely.

• Gases occupy containers Gases occupy containers uniformly and completely.uniformly and completely.

• Gases diffuse and mix Gases diffuse and mix rapidly.rapidly.

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Properties of Gases

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

• V = volume of the gas (L)

• T = temperature (K)

• n = amount (moles)

• P = pressure (atmospheres)

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PressurePressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

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PressureHg rises in tube until Hg rises in tube until

force of Hg (down) force of Hg (down) balances the force of balances the force of atmosphere (pushing atmosphere (pushing up). up).

P of Hg pushing down P of Hg pushing down related to related to

• Hg densityHg density• column heightcolumn height

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PressureColumn height measures P Column height measures P

of atmosphereof atmosphere• 1 standard atm1 standard atm

= 760 mm Hg= 760 mm Hg

= 29.9 inches Hg= 29.9 inches Hg

= about 34 feet of water= about 34 feet of water

SI unit is PASCAL, Pa,

where 1 atm = 101.325 kPa

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Effect of Pressure Differential

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IDEAL 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

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Boyle’s LawIf n and T are If n and T are

constant, thenconstant, then

PV = (nRT) = kPV = (nRT) = k

This means, for This means, for example, that P example, that P goes up as V goes goes up as V goes down.down.

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

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A bicycle pump is a good example of Boyle’s A bicycle pump is a good example of Boyle’s law. law.

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

Boyle’s Law

Dr. S. M. Condren

Boyle’s Law

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Charles’s LawIf n and P are If n and P are

constant, thenconstant, then

V = (nR/P)T = kTV = (nR/P)T = kT

V 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.

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Charles’s original balloon

Modern long-distance balloon

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Charles’s Law

Balloons immersed in liquid NBalloons immersed in liquid N22 (at -196 ˚C) will (at -196 ˚C) will

shrink as the air cools (and is liquefied).shrink as the air cools (and is liquefied).

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Charles’s Law

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Avogadro’s Hypothesis HypothesisEqual 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) = kn

V and n are directly related.V and n are directly related.

twice as many twice as many moleculesmolecules

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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.

2 H2 H22(g) + O(g) + O22(g) (g) 2 H2 H22O(g)O(g)

Avogadro’s Hypothesis

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Combining the Gas Laws• V proportional to 1/PV proportional to 1/P• V prop. to TV prop. to T• V prop. to nV prop. to n• Therefore, V prop. to nT/PTherefore, V prop. to nT/P• V = 22.4 L for 1.00 mol V = 22.4 L for 1.00 mol

when when Standard pressure and temperature (STP)

ST = 273 KST = 273 K

SP = 1.00 atmSP = 1.00 atm

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

R = 0.082057 L•atm/K•molSolution1. 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

memorize

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

R = 0.082057 L•atm/K•molSolution

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 22 kg of gas) mol (or about 22 kg of gas)

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Ideal Gas Constant

R = 0.082057 L*atm/mol*K

R has other values for other sets of units.

R = 82.057 mL*atm/mol*K

= 8.314 J/mol*K

= 1.987 cal/mol*K

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Gases 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 in a flask with a

volume of 2.50 L. What is the pressure of Ovolume of 2.50 L. What is the pressure of O22

at 25 at 25 ooC? Of HC? Of H22O?O?

Solution

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.

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Gases 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 in a flask with a

volume of 2.50 L. What is the pressure of Ovolume of 2.50 L. What is the pressure of O22 at at

25 25 ooC? Of HC? Of H22O?O?

Solution#mol H2O2 = 1.1g H2O2 (1mol/ 34.0g H2O2) = 0.032 mol H2O2

#mol O2 = (0.032mol H2O2)(1mol O2/2mol H2O2) = 0.016mol O2

P of O2 = nRT/V

= (0.016mol)(0.0821L*atm/K*mol)(298K)2.50L

= 0.16 atm

Dr. S. M. Condren

Gases and Stoichiometry

What is P of HWhat is P of H22O? Could calculate as O? Could calculate as

above. But recall Avogadro’s hypothesis. above. 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 H2O = 0.32 atm

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

Dr. S. M. Condren

Dalton’s Law

John DaltonJohn Dalton1766-18441766-1844

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Dalton’s Law of Partial Pressures

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 = P(H(H22O)O) + P + P(O(O22)) = 0.48 atm = 0.48 atm

Dalton’s Law: total P is sum of Dalton’s Law: total P is sum of

PARTIAL pressures. pressures.

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

0.32 atm 0.32 atm 0.16 atm 0.16 atm

Dr. S. M. Condren

Collecting Gases over Water

Dr. S. M. Condren

Example A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25oC, what is the volume of the “dry” oxygen gas at STP?Pwater = 23.8 torr at 25oC PO2 = Pbar - Pwater = (758 - 23.8) torr = 734 torrP1= PO2 = 734 torr; P2= SP = 760. torrV1= 245mL; T1= 298K; T2= 273K; V2= ?(V1P1/T1) = (V2P2/T2)V2= (V1P1T2)/(T1P2)

= (245mL)(734torr)(273K)(298K)(760.torr)

= 217mL

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Higher Higher Density airDensity air

Low Low density density heliumhelium

PV = nRT

n = PV RT

m = PMV RT

Where m => massM => molar mass

and density (d) = m/V

d = m/V = PM/RTd and M are proportional

GAS DENSITY

Dr. S. M. Condren

USING GAS DENSITYThe density of air at 15 The density of air at 15 ooC and 1.00 atm is C and 1.00 atm is

1.23 g/L. What is the molar mass of air?1.23 g/L. What is the molar mass of air?

What is air? What is air?

mass/mol = 1.23 g/0.0423 mol = 29.1 g/molmass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

79% N79% N22; M ; M 28g/mol 28g/mol 20% O20% O22; M ; M 32g/mol 32g/mol1. Calc. moles of air.1. Calc. moles of air.

V = 1.00 LV = 1.00 L P = 1.00 atmP = 1.00 atm T = 288 KT = 288 K

n = PV/RT = 0.0423 moln = PV/RT = 0.0423 mol2. Calc. molar mass2. Calc. molar mass

Reasonable?

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KINETIC MOLECULAR THEORY(KMT)

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

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

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Kinetic Molecular TheoryBecause 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 have the same average KE.As T goes up for a gas, KE also increases – and so does the speed.

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Kinetic Molecular Theory

where u is the speed and M is the where u is the speed and M is the molar mass.molar mass.

• speed INCREASES with Tspeed INCREASES with T• speed DECREASES with Mspeed DECREASES with M

Maxwell’s equationMaxwell’s equation

root mean square speed

2u M3RT

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Velocity of Gas MoleculesMolecules of a given gas have a Molecules of a given gas have a rangerange of speeds.of speeds.

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Velocity of Gas MoleculesAverage velocity decreases with increasing mass.Average velocity decreases with increasing mass.

All gases at the same temperature

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GAS DIFFUSION AND EFFUSION

DIFFUSIONDIFFUSION is the gradual mixing of is the gradual mixing of molecules of different gases.molecules of different gases.

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GAS EFFUSIONEFFUSION is the movement of is the movement of

molecules through a small hole molecules through a small hole into an empty container.into an empty container.

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GAS DIFFUSION AND EFFUSIONMolecules effuse thru holes in a Molecules effuse thru holes in a

rubber balloon, for example, at rubber balloon, for example, at a rate (= moles/time) that isa rate (= moles/time) that is

• proportional to Tproportional to T

• inversely proportional to M.inversely proportional to M.

Therefore, He effuses more Therefore, He effuses more rapidly than Orapidly than O22 at same T. at same T.

HeHe

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GAS DIFFUSION AND EFFUSIONGraham’s law Graham’s law

governs effusion governs effusion and diffusion of and diffusion of gas molecules.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

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• 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.

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

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Deviations from Ideal Gas Law

• Real molecules Real molecules

have have volume.

• There are There are

intermolecular forces.

– Otherwise a gas Otherwise a gas could not could not become a liquid.become a liquid.

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Deviations from Ideal Gas LawAccount for volume of molecules and Account for volume of molecules and

intermolecular forces with intermolecular forces with VAN DER WAALS’s EQUATION.

Measured V = V(ideal)Measured P

intermol. forcesvol. correction

J. van der Waals, J. van der Waals, 1837-1923, 1837-1923, Professor of Professor of Physics, Physics, Amsterdam. Amsterdam. Nobel Prize 1910.Nobel Prize 1910.

nRTV - nbV2

n2aP + ----- )(

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ClCl22 gas has gas has aa = 6.49, = 6.49, bb = 0.0562 = 0.0562

For 8.0 mol ClFor 8.0 mol Cl22 in a 4.0 L tank at 27 in a 4.0 L tank at 27 ooC.C.

P (ideal) = nRT/V = 49.3 atm

P (van der Waals) = 29.5 atm

Measured V = V(ideal)Measured P

intermol. forces

vol. correction

nRTV - nbV2

n2aP + -----

Deviations from Ideal Gas Law

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Real versus Ideal Gases

0

0.5

1

1.5

2

2.5

0 100 200 300 400 500 600 700 800 900

Pressure, atm

Vob

s/V

idea

l

ideal

H2

O2

N2

CH4

CO2

SO2

Cl2

H2O

Dr. S. M. Condren

Real versus Ideal Gases

0.9820.9840.9860.9880.99

0.992

0.9940.9960.998

11.0021.004

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Pressure, atm

Vob

s/V

idea

l

ideal

H2

O2

N2

CH4

CO2

SO2

Cl2

H2O

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Carbon Dioxide and Greenhouse Effect

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Composition of Air at Sea Level

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Some Oxides of Nitrogen

• N2O

• NO

• NO2

• N2O4

2 NO2 = N2O4

brown colorless

• NOx

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Air Pollution in Los Angles