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.
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?
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
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 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
2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)
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.
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
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.
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
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
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
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