1. Chapter 5 Gases and the Kinetic-Molecular Theory
Chp 5 Gas Laws
Chapter 5
Gases and the Kinetic-Molecular Theory
If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

2. Why do we need to learn about gases? To protect our earth!

3. Table 5.1 from 4th ed. Some Important Industrial Gases
Chp 5 Gas Laws
Table 5.1 from 4th ed. Some Important Industrial Gases
Name
Origin and Use
Methane (CH4)
Ammonia (NH3)
Chlorine (Cl2)
Oxygen (O2)
Ethylene (C2H4)
natural deposits; domestic fuel
from N2+H2; fertilizers, explosives
electrolysis of seawater; bleaching and
disinfecting
liquefied air; steelmaking
high-temperature decomposition of natural gas; plastics
Atmosphere-Biosphere Redox Interconnections

4. An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities under normal conditions.
5. Gases are miscible.

5. The three states of matter.
Chp 5 Gas Laws
Figure 5.1
The three states of matter.
Many pure substances, such as bromine (Br2), can exist under appropriate conditions of pressure and temperature as A, a gas; B, a liquid; or C, a solid. The atomic-scale views show that molecules are much farther apart in a gas than in a liquid or solid.
Bromine as a gas, a liquid and a solid.

6. Four measurementss define the state of a gas
1. Quantity in moles, n
2. Volume in Liters
3. Temperature in Kelvin
4. Pressure in Atm

7. Gas Pressure Pressure = Force/unit area
Force causes something to move a distance D in work
Gravity is a weak force, F = ma
On earth F = mg, where g is the acceleration due to earth’s gravity

8. A BAROMETER Measures pressure of atmosphere above
Essential principle of the barometer –
force of air on surface of water keeps water in the tube
If we had a long enough tube, about 34 ft., we'd reach the limit of air pressure to hold the water in

9. A mercury barometer. Figure 5.3 Chp 5 Gas Laws
Patm is due to weight of atmosphere above the earth
The barometer is still basically just a tube about 1 m long, closed at one end, filled with mercury, and inverted into a dish containing more mercury. When the tube is inverted, some of the mercury flows out into the dish, and a vacuum forms above the mercury remaining in the tube, as shown in Figure 5.3. At sea level under ordinary atmospheric conditions, the outward flow of mercury stops when the surface of the mercury in the tube
is about 760 mm above the surface of the mercury in the dish. It stops at 760 mm because at that point the column of mercury in the tube exerts the same pressure (weight/area) on the mercury surface in the dish as does the column of air that extends from the dish to the outer reaches of the atmosphere.

10. Barometers
Why do our barometers contain mercury? Because of its density. Hg is 13.6 g/mL vs. water at ~1.0 g/mL.
P = Force/Area
= mass of Hg column * grav. constant/area of column
= vol of Hg column * density of Hg/area of column
= height of Hg*area of column* mass/area of column
= height of Hg column * density of Hg
Normal atmospheric pressures is measured at sea level on a calm day: it is 760 mm Hg = one standard atmosphere pressure (both are exact quantities)

11. Effect of atmospheric pressure on objects
Chp 5 Gas Laws
Figure 5.2
Effect of atmospheric pressure on objects
at the Earth’s surface.
A, A metal can filled with air has equal pressure on the inside and outside.
B, When the air inside the can is removed, the atmospheric pressure crushes the can.
When the air inside the can is removed, atmospheric pressure crushes the can.
A metal can filled with air has equal pressure on the inside & outside.

12. Gas Pressure MEMORIZE:
Chp 5 Gas Laws
Gas Pressure
MEMORIZE:
1 atm = 760 mm Hg exactly = 760 torr exactly = kPa = 14.7 psi
Practice: convert 29.5 inches of Hg to mm Hg, torr, atm and psi. (Hint – do you remember how to convert inches to cm?)
749 mm Hg = 749 torr = atm = 14.4 psi

13. MANOMETERS
Definition: A manometer is a scientific instrument used to measure gas pressures. Open manometers measure gas pressure relative to atmospheric pressure. A mercury or oil manometer measures gas pressure as the height of a fluid column of mercury or oil that the gas sample supports.

14. Two types of manometers
Chp 5 Gas Laws
closed-end
Figure 5.4 from 4th ed.
Pgas = ht of Hg, Dh
Two types of manometers
open-end
A,B: In a closed-end manometer, there is a vacuum in the tube initially, on both sides of the mercury. The levels are equal. When a gas in introduced into the flask, the gas pressure is measured as the difference in height between the two sides.
C,D.E: In an open-ended manometer, the atmospheric pressure pushed down on one side and the gas pressure pushes on the other side. In C, the two sides are equal. In D, the gas pressure is less than atmospheric, therefore you subtract the difference in height from atm P. In E, the gas pressure is greater than atmospheric, therefore you add the difference in height from atm P.

15. The Simple Gas Laws
Memorize Boyle’s, Charles’, Avogadro’s, and Amonton’s (Gay-Lussac’s) laws

16. The relationship between the volume and pressure of a gas.
Chp 5 Gas Laws
Figure 5.4
The relationship between the volume and pressure of a gas.
Boyle’s Law
A. A small amount of air (the gas) is trapped in the short arm of a J tube; n and T are fixed. The total pressure on the gas (Ptotal) is the sum of the pressure due to the difference in heights of the mercury columns (h) plus the pressure of the atmosphere (Patm). If Patm 760 torr, Ptotal 780 torr. B. As mercury is added, the total pressure on
the gas increases and its volume (V) decreases. Note that if Ptotal is doubled (to 1560 torr), V is halved (not drawn to scale). C. Some typical pressure-volume data from the experiment. D. A plot of V vs. Ptotal shows that V is inversely proportional to P. E. A plot of V vs. 1/Ptotal is a straight line whose slope is a constant characteristic of any gas that behaves ideally.
See notes below or look in textbook.

17. Boyle’s Law
As shown in Fig 5.4, as Pressure  the Volume , which means V ~ 1/P (inversely proportional)
P*V is a constant
P1V1 = Cb and P2V2 = Cb
Therefore, P1V1 = P2V2
Example: gaseous CO2 at 55 torr occupies 125mL. It is transferred to new flask and P increases to 78 mm of Hg. What is the new volume?
First think about whether V inc or dec?
Dec. Why? P Inc.
V2 = P1V1/P2 = 55 mm*125 mL/78 mm = 88 mL

18. The relationship between the volume and temperature of a gas.
Chp 5 Gas Laws
Figure 5.5
The relationship between the volume and temperature of a gas.
Notice all lines extrapolate to oC! What is that?
At constant P, the volume of a given amount of gas is directly proportional to the absolute temperature. A fixed amount of gas (air) is trapped under a small plug of mercury at a fixed pressure. A, The sample is in an ice water bath. B, The sample is in a boiling water bath. As the temperature increases, the volume of the gas increases. C, The three lines show the effect of amount (n) of gas (compare red and green) and pressure (P) of gas (compare red and blue). The dashed lines extrapolate the data to lower temperatures. For any amount of an ideal gas at any pressure, the volume is theoretically zero at °C (0 K).
Charles’ Law
See notes below or in textbook.

19. Charles’ Law
As shown in Fig 5.5, as Temperature  the Volume , i.e., directly proportional. Therefore:
V/T = Cc
BUT Temperature has to be in positive increments, never negative.
This is why the absolute scale, Kelvin, was developed.
V1/T1 = Cc. V2/T2 = Cc. Therefore, V1/T1 = V2/T2
Example: A balloon is filled with He at 4.5 L and 25.0°C. Take it outside in winter in Chicago, -10.0°C. What is new volume?
First: is new V larger or smaller?
Smaller, because T dec.
V2 = V1*T2/T1 = 4.5 L*263 K/298 K = 4.0 L

20. Chp 5 Gas Laws
Figure 5.6
An experiment to study the relationship between the volume and amount of a gas comparing 0.1 mol and 0.2 mol.
Avogadro’s Law
At a given external P & T, a given amt of CO2 is placed in tube. When it changes from solid to gas, it pushes the piston up until Pgas = Patm. When twice as much CO2 is used, twice the volume is occupied, showing V is proportional to amt of gas.

21. Avogadro’s Law
As shown in Fig 5.6, as the quantity of a gas increases, the volume increases.
V1/n1 = Ca = V2/n2
therefore, V1/n1 = V2/n2
n = moles, which we know represents #'s of molecules

22. Amonton’s Law (also Gay-Lussac) P & T are directly related
P1/T1 = P2/T2 if n and V are constant
Does T have to be in Kelvin? Why?

23. Practice with the Simple Gas Laws
Work on problems 8a, 16a and 18. 8a: Convert 76.8 cm Hg to atm 16a: What is the effect on the volume of 1 mole of an ideal gas if the pressure is reduced by a factor of four at constant temperature? 18: A 93 L sample of dry air is cooled from 145oC to -22oC at constant pressure. What is its final volume?

24. Video Lecture on Ideal Gas Law
Caveat: you will use R = L-atm/mol-K, because we want 5 sig figs for many of our calcs. You will also use more sig figs for temperatures, so add to convert to Kelvin. (Also her final answer’s sig figs should be at least 3.)

25. IDEAL GAS LAW:
Combining the work of Boyle, Charles, Amonton, and Avogadro:
V=Cb/P V=Cc*T V=Ca*n
V=(Cb*Cc*Ca)*(n*T/P)
V = R*nT/P which rearranges to PV = nRT
R = L-atm/mol-K (memorize)

26. EXAMPLE of Ideal Gas Law
A big balloon is filled with moles H2, T is 20.0°C, P = 750. torr. What is Volume of balloon?
P = 750. torr*(1 atm/760 torr) = atm
V = nRT/P =
1300. mol* * K
atm
= 3.17 x 104 L

27. Standard molar volume. Figure 5.7
Chp 5 Gas Laws
Figure 5.7
Standard molar volume.
One mole of any ideal gas occupies L at STP. Notice these three gases behave ideally. Masses are different, therefore density will be different.

28. STP: standard temperature & pressure
Defined as K and atm
At STP, one mole of any gas occupies L, called the standard molar volume of a gas
Try V = nRT/P using this STP for one mole of gas

29. Simple combined gas law:
P1V1/T1 = P2V2/T2
Works if moles of a gas are constant.

30. Example
If you have 20.0 L of He at 150. atm and 30.0°C, then how many balloons can be filled with 5.00 L each at 755 mm and 22.0°C?
(Really asking for V2 at new T & P, note that n is constant.)
Convert P2 to atm: 755 mm (1 atm/760 mm)
= atm
V2 = P1V1T2/P2T1 = 150atm*20.0L*295.15K atm*303.15K
= 2940 L
2940L/5.00L = 588 balloons

31. MY SUPER-COMBINED GAS LAW:
Rearrange R = P1V1 = P2V2
n1T1 n2T2
Then "uncombine" these to get the simple gas laws:
@ const T & n : cross out all n & T to get Boyle's law
@ const P & n : cross out all n & P to get Charles' law
@ const P & T : cross out all P & T to get Avogadro's law
@ const V & n : cross out all V & n to get Amonton’s Law

32. IDEAL AND COMBINED GAS LAWS:
Chp 5 Gas Laws
IDEAL AND COMBINED GAS LAWS:
Practice: A sample of O2 gas occupies mL at 25.5oC and torr. (First determine the number of moles present.) If the temperature drops to 15.5oC and the pressure drops to torr, and you add moles of gas, what will the new volume read?
(You should get mL)
(275.3 mL)

33. Practice with Combined Gas Laws and Ideal Gas Law
Work on problems 20, 24 and : Calculate V of a sample of CO that is at -14oC and 367 torr if it first occupies 3.65 L at 298K and 745 torr. 24: A 75.0 g sample of N2O is held in a 3.1 L vessel at 115oC. What is its pressure? 83: (look in book)

34. The Density of a Gas Recall that Density = m/V
Chp 5 Gas Laws
The Density of a Gas
Recall that Density = m/V
Recall we can find moles: n = m/M
PV = nRT; substitute PV = (m/M)RT
Rearrange: m/V = M P/ RT
or D = M P/RT
The density of a gas is directly proportional to its molar mass.
The density of a gas is inversely proportional to the temperature.
Rearrange again to find molar mass: M = DiRT/P
M = DiRT/Pea (like gardening)

35. Example:
An unknown gas has a density of g/L at STP. Find its molar mass. (Hard way & easy way if at STP.)
MA = DRT/P
= g/L* *273.15K/1 atm
= g/mol
or: MA = g/L * L/mol

36. Mixtures of Gases and Dalton’s Law of Partial Pressures
Gases mix homogeneously in any proportions.
Each gas in a mixture behaves as if it were the only gas present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + …

37. Mixtures of Gases and Mole Fraction:
cA = nA/nT = the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture
PO2= nO2RT/V = nO2 = cO2
PT nTRT/V nT
Therefore, PO2 = cO2 * PT

38. Example:
We react N2 and H2 to make NH3. We start with 10.3 moles of H2 and 3.71 moles of N2. We use up the H2 to make 6.87 mol of NH3. Find the total pressure in the vessel and the partial pressure of each gas, assuming that the product is collected in a 125 L tank at 25.0°C.
N2(g) + 3 H2(g)  2 NH3(g)
Step 1: Determine moles NH3 produced and also find how much excess N2 remains:
10.3 mol H2 * 2NH3/3H2 = 6.87 mol NH3 produced
10.3 mol H2 * 1N2/3H2 = 3.43 moles N2 used
3.71moli molused = 0.28 moles N2 remaining
Add that to moles of ammonia produced for total moles:
nT = mol NH3 + mol N2 =
= moles total in tank

39. Example continued
Step 2: Determine total pressure, then pressure of each gas.
PT=nTRT/V = 7.15 mol * * 298K/125L
= atm
Partial pressure for each gas could be determined using moles of each in the ideal gas law, or by using mole fraction, as shown below.
cN2 = nN2/nT = 0.28/7.15 = ,
cNH3 = (from )
PN2 = cN2*PT = *1.399 = atm
PNH3 = cNH3*PT 0.961*1.399 = atm

40. Chp 5 Gas Laws
Figure 5.10
Collecting a water-insoluble gaseous reaction product and determining its pressure.
FIGURE 5.10 Collecting a water-insoluble gaseous product and determining its pressure.

41. Example:
We collected L of hydrogen gas over water at a room temperature of 26.0°C and atmospheric pressure of 745 torr. How many moles of hydrogen gas were collected?
Look up PH2O at 26.0oC and find it is 25.2 mm Hg.
Find PH2: Pbar = PH2 + PH2O
745 torr = PH torr
PH2 = 720. torr
Convert: torr (1 atm/760 torr) = atm
Rearrange Ideal Gas Law: n = PV/RT
n = ( atm) * L
L.atm/mol.K*299.15K
n = mol of H2

42. Practice Problems Work on problems 32, 34, 38, and 72ab.
32: What is the density of Freon gas (CFCl3) at 120.oC and 1.50 atm?
34: The density of a noble gas is 2.71 g/L at 3.00 atm and 0.00oC. Name the gas.
38: A 355 mL container holds g of Ne and some amt of Ar, at 35.0oC and total pressure of 626 torr. Calculate moles of Ar present.

43. STOICHIOMETRY OF GASES:
Stoichiometry problems that you will love to do!
At STP, one mole of any gas occupies L.
WE CAN USE THIS A LOT!
This means that equal volumes of gases at STP have the same # of moles!!!!
11.2 L of N2 = 0.5 moles = 11.2 L of O2 = 0.5 moles = 11.2 L of He, etc.
We can relate volumes through stoich coeff just like moles

44. Stoichiometry of Gases
Example:
N2(g) + 3 H2(g)  2 NH3(g) (over Fe catalyst at 500°C)
Given 355 L of hydrogen, how many L of nitrogen required?
355L H2 * 1 N2/3H2 = 118 L
How many L of ammonia produced?
355L H2 * 2NH3/3H2 = 237 L

45. Stoichiometry at STP: regular?
Example 2: g Zn react with excess sulfuric acid, how many liters of gas will you have at STP?
Zn(s) + H2SO4(aq)  ZnSO4(aq) + H2(g)
12.0 g/(65.4g/mol) = mol Zn * 1 H2/1 Zn
= mol H2(g)
0.183 mol gas * L/mol = 4.11 L
Note: if at other conditions, must use Ideal Gas Law to find volume, etc.

46. You Try “Regular” Stoichiometry:
Chp 5 Gas Laws
You Try “Regular” Stoichiometry:
During ammonia production, with 355 L of 25.0°C & 542 mm of Hg plus 105 L of 20.0°C & 645 torr.
Find volume and mass of ammonia produced at STP.
(154 L, 117 g)
Step 1: convert to moles with ideal gas law:
nH2 = (542/760)atm*355L = 10.3 mol
* 298K
nN2 = (645/760)atm*105L = 3.71 mol
* 293K
Step 2: Find which one is the limiting reagent:
10.3 mol * 2NH3/3H2 = 6.87 mol
3.71 mol * 2NH3/1N2 = 7.42mol
Step 3: mol * L/mol = 154 L
6.87 mol * g/mol = 117 g

47. Postulates of the Kinetic-Molecular Theory
Postulate 1: Particle Volume
The volume of an individual gas particle is so small compared to the volume of its container, that the gas particles are considered to have mass, but essentially no volume.
Postulate 2: Particle Motion
Gas particles are in constant, rapid, random, straight-line motion except when they collide with each other or with the container walls.
Postulate 3: Particle Collisions
Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant (although energy can be transferred).

48. Distribution of molecular speeds at three temperatures.
Chp 5 Gas Laws
Figure 5.12
Distribution of molecular speeds at three temperatures.
At a given temperature, a plot of the relative number of N2 molecules vs. molecular speed (u) results in a skewed bell-shaped curve, with the most probable speed at the peak. Note that the curves spread at higher temperatures and the most probable speed is directly proportional to the temperature.
See note below or in textbook.

49. Chp 5 Gas Laws
Figure at end of section 5.6
Diffusion of a gas particle through a space filled with other particles.
distribution of molecular speeds
mean free path
collision frequency
A bizarre and dangerous molecular highway!

50. GRAHAM'S LAWS OF DIFFUSION AND EFFUSION:
Diffusion is the gradual mixing of molecules of 2 or more gases in a container, owing to the continual molecular motion.
Effusion is the escape of molecules through a tiny hole in a pressurized container into a vacuum or lower pressure system.
Rate of effusion is related inversely to the molar mass of the gas, because of its relation to average speed of the gas molecules.
Rategas1/Rategas2 = (MAgas2/MAgas1)½

51. GRAHAM'S LAWS OF DIFFUSION AND EFFUSION:
Gas molecules are moving with various velocities; we calculate the average, called root-mean-square speed as shown, where R = J/mol.K and M is molar mass in kg/mol:
urms = (3RT/M)1/2
For example, calculate the different velocities of H2 and O2 at 25.0oC.
1921 m/s for H2 vs m/s for O2
Velocity depends inversely on molar mass! Lighter molecules move faster!

52. Relationship between molar mass and molecular speed.
Chp 5 Gas Laws
Figure 5.17
Relationship between molar mass and molecular speed.
Ek = 3/2 (R/NA) T
At a given temperature, gases with lower molar mass (in parenthesis) have higher most probable speeds (peak of each curve)

53. Example of Graham’s Law:
Methane and an unknown gas are observed to escape. When counted, 1.00x104 of molecules take 1.50 minutes for methane and the same number take 4.73 minutes for the unknown. Find MA for the unknown.
RateCH4/Rateunk = (1.00x104/1.50min)
(1.00x104/4.73min)
= (Munk/16g/mol)½
Square both sides: = Munk/16g/mol
Munk = 159.8g/mol
NOTE: these ratios are also related to urms1/urms2

54. REAL GASES DEVIATE FROM IDEAL GAS BEHAVIOR
When gases are at low temperature or high pressure, they approach a phase change to a liquid and their behavior deviates from ideal gas behavior.
When a gas molecule exhibits a force of attraction for another gas molecule, the behavior will deviate from ideal gas behavior. (Review KMT)

55. The behavior of several real gases with increasing external pressure.
Chp 5 Gas Laws
Figure 5.18
The behavior of several real gases with increasing external pressure.
The horizontal line shows the behavior of 1 mol of ideal gas: PV/RT 1 at all Pext. At very high pressures, all real
gases deviate significantly from such ideal behavior. Even at ordinary pressures, these deviations begin to appear
(expanded portion).

56. Practice Work on problems 57, 61 and 63a.
57: What is the ratio of effusion rates for O2 and Kr?
61: An unknown gas effuses in 11.1 min. An equal volume of H2 effuses in 2.42 min. What is the molar mass of the unknown gas?
63a: Calc urms for He at 0.00oC and at 30.0oC.