1. Gases and Liquids

2. Topics Properties of gases Gas laws Gas Stoichiometry
Diffusion / Effusion, defined
Liquids - properties

3. The Properties Of Gases
Gases exert pressure
Gases can be compressed
A gas takes up the volume of the container it is in.
Gases have no free surfaces
Gases are fluid

4. Free Surface...
A free surface is one that does not have to be contained by a wall of a vessel. A gas must be contained on all sides or it will flow out of the container.

5. Gases exert pressure
Pressure is defined as the force per unit area.
Gas particles are always colliding with each other and with the walls of the container in which they are held.

6. The collisions with the walls of the container produce pressure.
Gas also gets less dense when heated

7. Why does a gas get less dense when heated ?
When matter is heated, its particles begin to vibrate more. They have more kinetic energy.

8. The particles get farther apart when they possess more kinetic energy.
The more kinetic energy they possess, the more the particles will overcome the intermolecular forces that hold them close to each other.

9. Since density = mass per unit of volume
The density decreases because the particles move farther apart, thus decreasing the amount of mass in a given unit of volume.

10. Gases can be compressed
Gas is compressible because the particles are far apart.
Ex. A scuba tank may have 58 L of O2 and He compressed into a 5 L tank. The pressure in the tank would then be 11.7 atm.

11. Atmospheric pressure is measured with a barometer.
Some of the units associated with the measurement of pressure are:
mm Hg (millimeters of mercury)
torr
atm. (atmospheres)
bar (we will not use this one)

12. Converting from one unit to another
1 atm. = 760 mm Hg = 760 torr
Ex.1 Convert mm Hg to atm.
752.1 mmHg x 1 atm. =0.989= 0.99
760 mmHg atm.

13. Pressure is inversely proportional to volume.
Boyle’s Law -
This gas law states that as the pressure on a gas is increased, its volume decreases.
Pressure is inversely proportional to volume.

14. P 1 V 1 = P 2 V 2
If the pressure of a gas is changed while the temperature is held constant, the new volume can be calculated, using the above equation, (Boyle’s law). Or…
If the volume is changed the new pressure can be calculated.

15. Ex Boyle’s Law
A sample of Chlorine gas occupies 946 mL at a pressure of 726 mmHg. What is the pressure of the gas if the volume is reduced to 154 mL at constant temperature ?
946 mL x 726 mmHg = P2 x 154 mL
P2 = 4.46 x 10 3 mm Hg

16. Volume is directly proportional to temperature.
Charles’s Law -
This gas law states that as the temperature of a gas is increased, the volume of the gas increases when pressure is constant.
Volume is directly proportional to temperature.

17. V 1 / T 1 = V 2 / T 2
If the temperature of a certain volume of gas is changed while the pressure is held constant, the new volume can be calculated using the above equation, (Charles’s law).
If the volume is changed the new temperature can be calculated.

18. Ex Charles’s Law
452 mL of F2 is heated from 22.0o C to 187o C at constant pressure. What is the final volume ?
PROBLEM ! ALL GAS CALCULATIONS MUST BE DONE USING ABSOLUTE TEMPERATURE, KELVIN.

19. Convert the o C to kelvin...
K = 22.0 oC = 295 K
K = 187 oC = 460 K

20. V 1 / T 1 = V 2 / T 2
452 mL / 295 k = V2 / 460 k
V2 = 705 mL

21. The combined gas law - P1 V1 / T1 = P 2 V 2 / T 2
This is a combination of the last two laws and one other law.
P1 V1 / T1 = P 2 V 2 / T 2

22. Ex. 4 Combined gas law -
A small bubble with volume of 2.1 mL at the bottom of a lake where the pressure is 6.4 atm and the temperature is 8.0oC rises to the surface. The surface pressure is 1.0 atm. and the temperature is 25oC. Calculate the new volume.

23. It is helpful to write out the variables...
P1 = 6.4 atm. P2 = 1.0 atm.
V1 = 2.1 mL V2 = ?
T1 = T2 =
T1 = 281 k T2 = 298 k

24. Rearrange to solve for V2
V 2 = V 1 x P 1 x T2
P 2 x T 1
V2 = 2.1 atm. x 6.4 atm x 298 k
1.0 atm. x 281 k
V2 = 14 mL

25. THE IDEAL GAS LAW -
This equation explains how a gas would behave if it were an ideal or perfect gas. Real gases are not ideal but at high temperatures and low pressure they follow this equation fairly closely.

26. The Ideal Gas Law -
The ideal gas law accounts for moles, pressure, volume, temperature, and uses a proportionality constant called the gas constant, R.

27. P V = n R T The gas constant, R, has the value, 0.0821 L atm / mole k
These units will cancel with the undesired units in the calculation.

28. Ex. 5
Ideal Gas Law -
Calculate the pressure (in atm) exerted by 1.82 moles of SF6 gas in a steel vessel with a volume of 5.43 L at 69.5oC.
Note: When using the ideal gas law, you must use units of L, atm., and K to match the units in R.

29. Ex. continued...
Write out the variables...
P = ? n = moles = 1.82 moles
V = 5.43 L T = 69.5oC =342.5 k
R = L atm / mole k

30. Rearrange to solve for P.
P = nRT V
P=(1.82 mol)( L atm /mol k)(342 k)
5.43 L
P = 9.42 atm

31. Dalton’s Law Of Partial Pressures
This law states that the sum of the individual pressures of the gases in a mixture of gases is equal to the total pressure of the mixture.
PT = P1 + P Pn
and...
Pi = xi (PT)

32. P i = xi (P T )
In this equation the ‘ i ’ stands for the individual gas, the ‘ x ’ stands for the mole fraction of the individual gas in the mixture.
mole fraction is the desired gas moles, divided by the total moles of gas.

33. Xi = moles i Ex. 6 Dalton’s Law of Partial Pressures P i = x i (P T )
and
Xi = moles i
moles i + j + k +…

34. Ex. 6 continued...
A mixture of gases contains 4.46 moles of Ne, 0.74 moles Ar, and 2.15 moles Xe. Calculate the partial pressures of the gases if the total pressure is 2.00 atm.

35. First calculate the mole fraction
XNe = nNe = 0.607
nNe +n Ar + n Xe
The same procedure is followed to find the mole fractions of Ar and Xe. They are 0.10 and respectively.

36. Finally, Pi = Xi (PT)
PNe = x 2.00 atm = 1.21 atm
PAr = 0.10 x 2.00 atm = 0.20 atm
PXe = x 2.00atm = atm

37. PT = P1 + P Pn
Ex. 7
Calculate the partial pressure of oxygen gas collected through water if the total pressure is 762 mm Hg and the water vapor pressure is 22.4 mm Hg.
POxygen = PT - P water = 740 mm Hg

38. At STP, 1 mole of gas occupies 22.4 L
Gas Stoichiometry
An important piece of information;
At STP, 1 mole of gas occupies 22.4 L

39. STANDARD TEMPERATURE AND PRESSURE
STP
STP is defined as ...
STANDARD TEMPERATURE AND PRESSURE
STANDARD PRESSURE IS EQUAL TO 1 atm. STANDARD TEMPERATURE IS EQUAL TO 0oC OR 273 k.

40. 22.4 L x grams
mole liter
With this information, the molar mass of a gas at STP can be calculated, given the density. or
The density can be calculated, given the molar mass.

41. Ex. 8 Molar Mass of a Gas
The density of a gas at STP is equal to g/L. Calculate its molar mass.
0.761 g x L = g / mole
Liter mole
This gas would be ammonia.

42. Molar Mass and Density
If given the quantity, in grams, the Ideal gas law can be used to calculate the molar mass of a gas when not at STP.
Rearrange... n = PV / RT
Solve for n, then divide grams /n

43. n = PV / RT
Ex. 9 Calculate the molar mass of a g sample of a gas that fills L at 298 k and atm.
n = atm x L =
L atm k mol
mole k

44. Ex. 9 continued...
0.100 g = g/mol
mole

45. Gas Stoichiometry
At STP, the coefficients in a balanced chemical equation are related to the volumes of gas that react.

46. Ex Stoichiometry
Calculate the volume of O2 at STP, required for the complete combustion of 2.64 L of acetylene (C2H2) at STP.
Remember the first step in stoichiometry is to write the balanced equation.

47. volume of O2 = 2.64 L C2 H2 x 5 L O2 2 L C2 H2 = 6.60 L O2
2C2 H2 (g) + 5 O2 (g) ----->4CO2 (g) + 2H2O(l)
volume of O2 =
2.64 L C2 H2 x 5 L O2
2 L C2 H2
= L O2

48. Ex Stoichiometry
Sodium Azide (NaN3) is used in some automobile air bags. The NaN3 is decomposed when the collision impact triggers the reaction. Calculate the volume of N2 produced at 21oC and 1.08 atm for 60.0 g of NaN3.

49. Since the reaction is NOT at STP we must use the Ideal Gas Law.
2 NaN3 (s) ----> 2 Na (s) + 3 N2(g)
Since the reaction is NOT at STP we must use the Ideal Gas Law.
We were given T and P and grams, but not moles, n.
We must first find the moles in order to use the Ideal gas law to find volume.

50. moles N2 = = 1.38 mol N2 continued...
=60.0 g NaN3 x 1 mol NaN x 3 molN2
65.02g NaN3 2mol NaN3
= 1.38 mol N2
continued...

51. continued...
V = nRT / P
= (1.38 mol)(0.0821Latm)(294 k)
1.08 atm mol k
= 30.8 L N2

52. Diffusion and Effusion
Diffusion is the mixing of the molecules of one gas with the molecules of another gas.
Ex. An open bottle of ammonia in a room will soon diffuse through the air in the room. The vapors will be noticeable in a short time.

53. Rate of Diffusion
A lighter gas will diffuse more quickly than a heavier gas. For instance, NH3 gas will diffuse about twice as fast as HCl gas.
mass of NH3 = g / mol
mass of HCl = g / mol

54. Effusion
Effusion is defined as the process in by which a gas under pressure escapes out of a container through a small opening.

55. Rates of both diffusion and effusion can be determined by...
Graham’s Law of Diffusion -
The rates of diffusion for gases are inversely proportional to the square roots of their molar masses.
r 1 / r 2= M 2 / M 1

56. Ex Rate of Diffusion
Compare the effusion rates of helium and oxygen gas at the same temperature and pressure.
rHe / rO =
32.00 g / mol
4.003 g/mol 2

57. Ex continued...
= 2.827:1 is the rate of He compared to oxygen.
The Helium effuses times faster than the oxygen.

58. Liquids Some properties of liquids are; They are not very compressible
liquids have one free surface
liquids are fluid
liquids have a definite volume

59. Liquids are not compressible...
This is because though their particles are loosely held enough to be able to flow past one another, they are still very close together.

60. Liquids have one free surface
A liquid must be surrounded on all sides but the top or it will flow out of its container.

61. Surface Tension
Liquids have surface tension; some more so than others. Water, for instance, seems to almost have a skin on the surface. This is due to the very strong intermolecular forces that hold like molecules close together, (COHESION).

62. Comparing surface tensions...
Water has a relatively high surface tension. Methylene chloride, (CH2Cl 2) a non polar molecule has a relatively low surface tension. Why ?

63. It all has to do with intermolecular forces.
Water has hydrogen bonding, a very strong type of intermolecular force of attraction holding its molecules together. Methylene chloride is non polar and thus has only London Dispersion forces which are very weak forces.

64. Vapor Pressure
The pressure of a gas above a liquid is the vapor pressure.
The weaker the intermolecular forces of the liquid, the more molecules will be able to escape as vapor, thus, the higher the vapor pressure.

65. The Boiling point of a liquid
As temperature increases, the kinetic energy of the liquid increases. More and more vapor molecules escape, thus increasing the vapor pressure.

66. Boiling Point
When the vapor pressure above the liquid is equal to the atmospheric pressure, it is at the boiling point.

67. General Rules On Comparing Boiling Points
Particles with the stronger intermolecular forces have higher boiling points:
Ionic > Hydrogen bonded > Polar Covalent > Non polar covalent

68. Then the heavier particle will have the higher boiling point.
If the intermolecular forces are the same type...
Then the heavier particle will have the higher boiling point.

69. If they also have the same weight...
...as in two hydrocarbons in which one of them is straight and one is branched, the straight chain will have the higher boiling point.

70. Why ???
Because the straight, less bulky particles can get closer together to have more surface contact for the London forces to take effect.

71. Circle the one with the higher b.p. and explain why ?
HF or HCl
HCl or C 2 H 6
C 2 H6 or C 3 H 8

72. Answers...
HF - Hydrogen bonding vs. polar covalent (dipole - dipole)
HCl - dipole dipole vs. non polar covalent
C 3 H 8 - Heavier molecule wins

73. Remember, or else ! Properties of gases and liquids - explain
Gas laws - define and put to use
Gas stoichiometry
Diffusion and Effusion - define and use
Vapor pressure - define
Boiling point - define, compare substances for higher boiling point