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Chapter 12. BEHAVIOR OF GASES. BEHAVIOR OF GASES. Importance of Gases. Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3 . 2 NaN 3 2 Na + 3 N 2 if bag ruptures 2 Na + 2 H 2 O 2 NaOH + H 2. THREE STATES OF MATTER. - PowerPoint PPT Presentation
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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
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.
Dr. S. M. Condren
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)
Dr. S. M. Condren
PressurePressure of air is measured with a BAROMETER (developed by Torricelli in 1643)
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
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.
Dr. S. M. Condren
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
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.
Dr. S. M. Condren
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).
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
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)
Dr. S. M. Condren
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
Dr. S. M. Condren
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.
Dr. S. M. Condren
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 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
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
Dr. S. M. Condren
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?
Dr. S. M. Condren
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.
Dr. S. M. Condren
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.
Dr. S. M. Condren
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
Dr. S. M. Condren
Velocity of Gas MoleculesMolecules of a given gas have a Molecules of a given gas have a rangerange of speeds.of speeds.
Dr. S. M. Condren
Velocity of Gas MoleculesAverage velocity decreases with increasing mass.Average velocity decreases with increasing mass.
All gases at the same temperature
Dr. S. M. Condren
GAS DIFFUSION AND EFFUSION
DIFFUSIONDIFFUSION is the gradual mixing of is the gradual mixing of molecules of different gases.molecules of different gases.
Dr. S. M. Condren
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.
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
• 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
Dr. S. M. Condren
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.
Dr. S. M. Condren
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 + ----- )(
Dr. S. M. Condren
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
Dr. S. M. Condren
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
Dr. S. M. Condren
Some Oxides of Nitrogen
• N2O
• NO
• NO2
• N2O4
2 NO2 = N2O4
brown colorless
• NOx
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