1. Chapter 12
BEHAVIOR OF GASES
Dr. S. M. Condren

2. BEHAVIOR OF GASES
Dr. S. M. Condren

3. Importance of Gases Airbags fill with N2 gas in an accident.
Gas is generated by the decomposition of sodium azide, NaN3.
2 NaN3  2 Na N2
if bag ruptures 2 Na H2O  2 NaOH + H2
Dr. S. M. Condren

4. THREE STATES OF MATTER
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5. THREE STATES OF MATTER
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6. General Properties of Gases
There is a lot of “free” space in a gas.
Gases can be expanded infinitely.
Gases occupy containers uniformly and completely.
Gases diffuse and mix rapidly.
Dr. S. M. Condren

7. Properties of Gases V = volume of the gas (L) T = temperature (K)
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

8. Pressure
Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)
Dr. S. M. Condren

9. Pressure
Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up).
P of Hg pushing down related to
Hg density
column height
Dr. S. M. Condren

10. Pressure Column height measures P of atmosphere
1 standard atm = 760 mm Hg
= 29.9 inches Hg
= about 34 feet of water
SI unit is PASCAL, Pa,
where 1 atm = kPa
Dr. S. M. Condren

11. Effect of Pressure Differential
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12. P V = n R T IDEAL GAS LAW Brings together gas properties.
Can be derived from experiment and theory.
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13. Boyle’s Law If n and T are constant, then PV = (nRT) = k
This means, for example, that P goes up as V goes down.
Robert Boyle ( ). Son of Earl of Cork, Ireland.
Dr. S. M. Condren

14. Boyle’s Law A bicycle pump is a good example of Boyle’s law.
As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.
Dr. S. M. Condren

15. Boyle’s Law
Dr. S. M. Condren

16. Charles’s Law If n and P are constant, then V = (nR/P)T = kT
V and T are directly related.
Jacques Charles ( ). Isolated boron and studied gases. Balloonist.
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17. Charles’s original balloon
Modern long-distance balloon
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18. Charles’s Law
Balloons immersed in liquid N2 (at -196 ˚C) will shrink as the air cools (and is liquefied).
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19. Charles’s Law
Dr. S. M. Condren

20. Avogadro’s Hypothesis
Equal volumes of gases at the same T and P have the same number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many molecules
Dr. S. M. Condren

21. Avogadro’s Hypothesis
The gases in this experiment are all measured at the same T and P.
2 H2(g) O2(g)
2 H2O(g)
Dr. S. M. Condren

22. Combining the Gas Laws V proportional to 1/P V prop. to T V prop. to n
Therefore, V prop. to nT/P
V = 22.4 L for 1.00 mol when Standard pressure and temperature (STP)
ST = 273 K
SP = 1.00 atm
Dr. S. M. Condren

23. Using PV = nRT
How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
R = L•atm/K•mol
memorize
Solution
1. Get all data into proper units
V = 27,000 L
T = 25 oC = 298 K
P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm
Dr. S. M. Condren

24. Using PV = nRT
How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
R = L•atm/K•mol
Solution
2. Now calc. n = PV / RT
n = 1.1 x 103 mol (or about 22 kg of gas)
Dr. S. M. Condren

25. R has other values for other sets of units.
Ideal Gas Constant
R = L*atm/mol*K
R has other values for other sets of units.
R = mL*atm/mol*K
= J/mol*K
= cal/mol*K
Dr. S. M. Condren

26. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O?
Solution
Strategy:
Calculate moles of H2O2 and then moles of O2 and H2O.
Finally, calc. P from n, R, T, and V.
Dr. S. M. Condren

27. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O?
Solution
#mol H2O2 = 1.1g H2O2 (1mol/ 34.0g H2O2) = 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

28. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
What is P of H2O? Could calculate as above. But recall Avogadro’s hypothesis.
V  n at same T and P
P  n at same T and V
There are 2 times as many moles of H2O as moles of O2. P is proportional to n. Therefore, P of H2O is twice that of O2.
P of H2O = 0.32 atm
Dr. S. M. Condren

29. Dalton’s Law
John Dalton
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30. Dalton’s Law of Partial Pressures
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
0.32 atm atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = P(H2O) + P(O2) = atm
Dalton’s Law: total P is sum of PARTIAL pressures.
Dr. S. M. Condren

31. Collecting Gases over Water
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32. Example A student generates oxygen gas and collects it over water
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 = ( ) torr = 734 torr
P1= PO2 = 734 torr; P2= SP = 760. torr
V1= 245mL; T1= 298K; T2= 273K; V2= ?
(V1P1/T1) = (V2P2/T2)
V2= (V1P1T2)/(T1P2)
= (245mL)(734torr)(273K)
(298K)(760.torr)
= 217mL
Dr. S. M. Condren

33. GAS DENSITY Low density helium PV = nRT n = P V RT m = P MV RT
Where m => mass
M => molar mass
and density (d) = m/V
Higher
Density air
d = m/V = PM/RT
d and M are proportional
Dr. S. M. Condren

34. USING GAS DENSITY
The density of air at 15 oC and 1.00 atm is 1.23 g/L. What is the molar mass of air?
What is air?
79% N2; M a 28g/mol
20% O2; M a 32g/mol
1. Calc. moles of air.
V = 1.00 L P = 1.00 atm T = 288 K
n = PV/RT = mol
2. Calc. molar mass
mass/mol = g/ mol = 29.1 g/mol
Reasonable?
Dr. S. M. Condren

35. KINETIC MOLECULAR THEORY (KMT)
Theory used to explain gas laws KMT assumptions are
Gases consist of atoms or molecules in constant, random motion.
P arises from collisions with container walls.
No attractive or repulsive forces between molecules. Collisions elastic.
Volume of molecules is negligible.
Dr. S. M. Condren

36. Kinetic Molecular Theory
Because we assume molecules are in motion, they have a kinetic energy.
KE = (1/2)(mass)(speed)2
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

37. Kinetic Molecular Theory
Maxwell’s equation
where u is the speed and M is the molar mass.
speed INCREASES with T
speed DECREASES with M
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38. Velocity of Gas Molecules
Molecules of a given gas have a range of speeds.
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39. Velocity of Gas Molecules
Average velocity decreases with increasing mass.
All gases at the same temperature
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40. GAS DIFFUSION AND EFFUSION
DIFFUSION is the gradual mixing of molecules of different gases.
Dr. S. M. Condren

41. GAS EFFUSION
EFFUSION is the movement of molecules through a small hole into an empty container.
Dr. S. M. Condren

42. GAS DIFFUSION AND EFFUSION
Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is
proportional to T
inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T. He
Dr. S. M. Condren

43. GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules.
Rate of effusion is inversely proportional to its molar mass.
Thomas Graham, Professor in Glasgow and London.
Dr. S. M. Condren

44. Gas Diffusion relation 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.
Dr. S. M. Condren

45. Deviations from Ideal Gas Law
Real molecules have volume.
There are intermolecular forces.
Otherwise a gas could not become a liquid.
Dr. S. M. Condren

46. Deviations from Ideal Gas Law
Account for volume of molecules and intermolecular forces with VAN DER WAALS’s EQUATION.
J. van der Waals, , Professor of Physics, Amsterdam. Nobel Prize 1910.
Measured V = V(ideal)
Measured P
nRT
V nb V 2 n a P ) (
vol. correction
intermol. forces
Dr. S. M. Condren

47. Deviations from Ideal Gas Law
Cl2 gas has a = 6.49, b =
For 8.0 mol Cl2 in a 4.0 L tank at 27 oC.
P (ideal) = nRT/V = 49.3 atm
P (van der Waals) = 29.5 atm
Dr. S. M. Condren

48. Dr. S. M. Condren

49. Dr. S. M. Condren

50. Carbon Dioxide and Greenhouse Effect
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51. Composition of Air at Sea Level
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52. Some Oxides of Nitrogen
N2O NO
NO2
N2O4
2 NO2 = N2O4
brown colorless
NOx
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53. Air Pollution in Los Angles
Dr. S. M. Condren