1. CHEM 230: Principles of Physical Chemistry

2. CHEM 230: Principles of Physical Chemistry
                                                   
Molecular kinetic theory of gases, first law of thermodynamics, thermo chemistry, second and third laws of thermodynamics, free energies, adsorption and heterogeneous catalysis.

3. Content of the Course: Molecular kinetic theory of gases
Fundamental concepts, the zeroth law, heat, internal energy, and work
First law of thermodynamics
Thermochemistry
Entropy and the second and third laws of thermodynamics
Free Energy and chemical equilibria
Adsorption and heterogeneous catalysis

4. Course Evaluation: Homework: 10 points
Midterm exams: 2 x 25 points 50 points
Final Exam: 1 x 40 points 40 points
Total 100 points
الإختبار الفصلي الأول يوم الأثنين 13 صفر
الإختبار الفصلي الثاني يوم الأثنين 18 ربيع الأول

5. Molecular kinetic theory of gases
Part 1
Molecular kinetic theory of gases

6. Characteristics of Gases
Unlike liquids and solids, gases
expand to fill their containers;
are highly compressible;
have extremely low densities.

7. Pressure F P = A Pressure is the amount of force applied to an area.
Atmospheric pressure is the weight of air per unit of area.

8. Units of Pressure Pascals (SI units) 1 Pa = 1 N/m2
𝑝= 𝐹 𝐴 = 𝑁 𝑚 2 = 𝑘𝑔. 𝑚 𝑠 𝑚 2 =𝑘𝑔. 𝑚 −1 . 𝑠 −2 = 𝑘𝑔 𝑚. 𝑠 2 =1𝑃𝑎
𝐹=𝑚.𝑎
1 𝑁=1𝑘𝑔. 𝑚 𝑠 2

9. Units of Pressure mm Hg or torr
These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.
Atmosphere
1.00 atm = 760 torr
A mercury barometer

10. Bar
1 bar = 105 Pa = 100 kPa
Do you Know another units ?

11. Standard Pressure
Normal atmospheric pressure at sea level is referred to as standard pressure.
It is equal to
1.00 atm
760 torr (760 mm Hg)
kPa

12. Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

13. Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

14. As P and V are inversely proportional
A plot of V versus P results in a curve.
V α (1/P)
Since
V = k (1/P)
This means a plot of V versus 1/P will be a straight line.
PV = k

15. Charles’s Law V α T V = kT V = k T
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.
V α T
V = kT
i.e., V T
= k
A plot of V versus T will be a straight line.

16. Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

17. Another form of Charles’s law shows that at constant amount of gas and volume, the pressure of a gas is proportional to temperature
P α T
P = kT
P/T=k
𝑃 1 𝑇 1 = 𝑃 2 𝑇 2

18. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
Mathematically, this means
V = kn

19. Ideal-Gas Equation V  nT P So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
V  n (Avogadro’s law)
Combining these, we get
V  nT P

20. Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship
then becomes or
PV = nRT

21. Ideal-Gas Equation
The constant of proportionality is known as R, the gas constant.

22. At Standard Conditions for Temperature and Pressure (STP) for any gas
T = 0 C˚ = 273 K
P = 1 atm
R = atm.L.mol-1.K-1
PV = nRT
𝑉= 𝑛𝑅𝑇 𝑃 = 1×0.0821×273 1 =22.4 𝐿 or
P = Pa so R = Pa.m3. mol-1.K-1
𝑉= 𝑛𝑅𝑇 𝑃 = 1×8.314× = 𝑚 3
V = dm3 or L

23. Relating the Ideal-Gas Equation and the Gas Laws
PV = nRT
When the quantity of gas and the temperature are held constant, n & T have fixed values.
PV = nRT = constant
P1V 1= P2V2

24. A metal cylinder holds 50 L of O2 gas at 18
A metal cylinder holds 50 L of O2 gas at 18.5 atm and 21oC, what volume will the gas occupy if the temperature is maintained at 21 oC while the pressure is reduced to 1.00 atm?
18.5 atm x 50 L = 1.00 atm x V2
V2 = 925 L
P1V 1= P2V2

25. Relating the Ideal-Gas Equation and the Gas Laws
PV = nRT
When the quantity of gas and the volume are held constant, n & V have fixed values. P T nR V =
= constant P1 T1 P2 T2 =

26. The gas pressure in an aerosol can is 1. 5 atm at 25 oC
The gas pressure in an aerosol can is 1.5 atm at 25 oC. Assuming that the gas inside obeys the ideal-gas equation, what would the pressure be if the can were heated to 450 oC?
P2 = 3.6 atm P1 T1 P2 T2 =
1.5 atm
298 K P2
723 K =

27. Relating the Ideal-Gas Equation and the Gas Laws
PV = nRT
When the quantity of gas is held constant, n has fixed values.
= nR = constant PV T
P1V1 T1
P2V2 T2 =

28. V2 = 11 L P1V1 T1 P2V2 T2 1.0 atm x 6.0 L 295 K 0.45 atm x V2 252 K
An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.45 atm. During ascent the temperature of the gas falls from 22 oC to -21 oC. Calculate the volume of the balloon at its final altitude.
V2 = 11 L
P1V1 T1
P2V2 T2 =
1.0 atm x 6.0 L
295 K
0.45 atm x V2
252 K =

29. 48.7 K
3.05 x 10-3 mol
11.2 L
10.3 atm

30. A scuba diver’s tank contains 0
A scuba diver’s tank contains 0.29 kg of O2 compressed into a volume of 2.3 L. (a) Calculate the gas pressure inside the tank at 9 oC. (b) What volume would this oxygen occupy at 26 oC and 0.95 atm?
PV = nRT
P x 2.3 L = ( 290 g/32 gmol-1) x x 282 K
P = 91 atm
V2 = 233 L
P1V1 T1
P2V2 T2 =

31. What is the molar mass of a gas if 0
What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C?
n = PV = (0.813 atm) (0.215 L) = mol
RT ( L-atm/molK) (303K)
Molar mass = g = g = g/mol
mol mol

32. How many L of O2 are need to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) NO(g) H2O(g)
Find mole of O2
28.0 g NH3 x 1 mol NH3 x 5 mol O2
17.0 g NH mol NH3
= 2.06 mol O2
V = nRT = (2.06 mol)(0.0821)(297K) = L
P atm

33. Densities of Gases n P n   = m V = RT m PM V = RT PV = nRT
moles  molecular mass = mass n V P RT =
n   = m m V PM RT =

34. Densities of Gases
Density = mass  volume
So, m V
d = PM RT m V =
d =

35. What is the density of carbon tetrachloride (CCl4) vapor at 714 torr and 125 oC?PM RT
d =
714 torr x 154 g mol -1
62.36 L torr mol -1K-1 x 398 K
d =
d = 4.43 g L-1

36. Which gas is most dense at 1.00 atm and 298 K? CO2, N2O, or Cl2.PM RT
d =
1 x 44
0.082 x 298
d =
= 1.8 g L-1
CO2
d = 1.8 g L-1
N2O
d = 2.91 g L-1 Cl

37. Molecular Mass P d = RT dRT P  =
We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT
d =
Becomes
dRT P
 =

38. Calculate the density of NO2 gas at 0. 970 atm and 35 oC
Calculate the density of NO2 gas at atm and 35 oC. (b) Calculate the molar mass of a gas if 2.50 g occupies L at 685 torr and 35 oC. PM RT
d =
0.97 x 46
0.082 x 308
d =
d = g L-1

39. d = d = = 2.86 g L-1 dRT  = P mass volume 2.5 g 0.875 L
2.86 x x 308
685
 =
 = g mol-1

40. A gas has a % composition by mass of 85. 7% carbon and 14. 3% hydrogen
A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas?
Calculate Empirical formula
85.7 g C x 1 mol C = mol C/7.14 = 1 C
12.0 g C
14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H
1.0 g H
Empirical formula = CH2
EF mass = (1.0) = 14.0 g/EF

41. Using STP and density ( 1 L = 2.50 g)
2.50 g x L = g/mol
1 L mol
56.0 g/mol = 4
14.0 g/EF
molecular formula
CH2 x = C4H8

42. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.
In other words,
Ptotal = P1 + P2 + P3 + …

43. Partial Pressures
When one collects a gas over water, there is water vapor mixed in with the gas.
To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

44. Partial Pressures Ptotal = P1 + P2 + P3 + …
P1 = n1 ( ); P2 = n2 ( ); and so on
Pt = (n1+n2+n3+..) = nt( ) RT V RT V
(1) RT V RT V
(2)
Dividing equation (1) by equation (2) P1 Pt
n1 RT/V
nt RT/V n1 nt = = n1 nt
P1 = Pt
P1 = X1 Pt

45. A 5. 00 L scuba tank contains 1. 05 mole of O2 and 0
A 5.00 L scuba tank contains 1.05 mole of O2 and mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?
P = nRT PT = PO + PHe V
PT = mol x L-atm x 298 K
5.00 L (K mol)
= 7.19 atm

46. Kinetic-Molecular Theory
This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

47. Main Tenets of Kinetic-Molecular Theory
Gases consist of large numbers of molecules that are in continuous, random motion.
The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

48. Main Tenets of Kinetic-Molecular Theory
Attractive and repulsive forces between gas molecules are negligible.
Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

49. Main Tenets of Kinetic-Molecular Theory
The average kinetic energy (u) of the molecules is proportional to the absolute temperature.

50. Kinetic-Molecular Theory (model & fundamental laws)
N number of molecules of ideal gas
m mass of each molecules present in a cube has length l cm
Each molecule move in deferent speed u1, u2, u3, uN (cm s-1)
Each molecule move in three dimensions (u1x, u1y, u1z) and each molecule make one collision when it cut distance l cm

52. Thus, the number of collisions that one molecule make in this direction in one second is 𝑢 1𝑥 𝑙
The impulse before collision is +mu1xand after collision is –mu1x, thus, the change in impulse due to collision with the wall of cube is 2mu1x.
The change of the impulse in one second equal 2mu1x× 𝑢 1𝑥 𝑙 = 2𝑚 𝑢 1𝑥 2 𝑙

53. Finally, the sum of change of molecules impulse in all dimensions in one second is:
2𝑚 𝑙 𝑢 2 ×𝑁
Where: 𝑢 2 = 𝑢 𝑢 𝑢 3 2 +…+ 𝑢 𝑁 2 𝑁
𝑢 2 called mean square speed of one molecule, and it’s the exerted force by all molecules in all dimensions :
𝐹= 2𝑚𝑁 𝑢 2 𝑙

54. The pressure is the exerted force on the area, so the sum of area in the cube is 6l2
𝑃= 𝐹 6 𝑙 2
𝑃= 2𝑚𝑁 𝑢 𝑙 3
l 3is the volume that the molecules move in it, so:
𝑃= 1 3 𝑚𝑁 𝑢 2 𝑉
𝑃𝑉= 1 3 𝑚𝑁 𝑢 2
This equation known as The Fundamental Equation of The Kinetic Theory

55. On the other hand, the average of kinetic energy 𝑘𝑒 of one molecule related to speed of it as shown:
𝑘𝑒 = 1 2 𝑚 𝑢 2
For all molecules: 𝑘𝑒 ×𝑁= 1 2 𝑚𝑁 𝑢 2
Or:
𝑘𝑒 𝑁 = 1 2 𝑚𝑁 𝑢 2

56. Based on the hypotheses of the theory:
𝑘𝑒 𝛼 𝑇
𝑘𝑒 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ×𝑇
1 2 𝑚 𝑢 2 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ×𝑇 ………(a)
From the Fundamental Equation of the Kinetic Theory:
𝑃𝑉= 1 3 𝑚𝑁 𝑢 2
3𝑃𝑉 𝑁 = 𝑚 𝑢 2
3𝑃𝑉 𝑛 𝑁 𝐴 = 𝑚 𝑢 2

57. Then the equation become: PV=nRT
3𝑃𝑉 2𝑛 𝑁 𝐴 = 1 2 𝑚 𝑢 2
𝑃𝑉 𝑛 3 2 𝑁 𝐴 = 1 2 𝑚 𝑢 2
From equation (a):
𝑃𝑉 𝑛 3 2 𝑁 𝐴 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ×𝑇
𝑃𝑉 𝑛 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ×𝑇
The constant is R,
Then the equation become: PV=nRT

58. Again, from the Fundamental Equation of The Kinetic Theory:
𝑃𝑉= 1 3 𝑚𝑁 𝑢 2
And from the ideal gas equation:
PV = nRT
1 3 𝑚𝑁 𝑢 2 =𝑛𝑅𝑇
𝑚 𝑁 𝑛 𝑢 2 =3𝑅𝑇
𝑚 𝑁 𝐴 𝑢 2 =3𝑅𝑇……… (b)

59. 1 2 𝑚 𝑁 𝐴 𝑢 2 = 3 2 𝑅𝑇
1 2 𝑚 𝑁 𝐴 𝑢 2 = 𝑘𝑒 𝑁 𝐴 =𝐾𝐸
KE is the kinetic energy of one mole, while 𝑘𝑒 for one molecule.
𝐾𝐸= 3 2 𝑅𝑇
𝐾 𝐸 𝑡𝑜𝑡𝑎𝑙 = 3 2 𝑛𝑅𝑇
𝑘𝑒 = 3 2 𝑅 𝑁 𝐴 𝑇
𝑘𝑒 = 3 2 𝑘𝑇
Where 𝑘= 𝑅 𝑁 𝐴 called Boltzmann’s constant and has the value 1.37×10-23 J.K-1

60. Example:
A tank used for filling helium balloons has a volume of 0.3 m3 and contains 2.00 mol of helium behaves like an ideal gas.
1- What is the total translational kinetic energy of the gas molecules?
If n= 2.00 mol and T = 293 K, use the equation:
𝐾 𝐸 𝑡𝑜𝑡𝑎𝑙 = 3 2 𝑛𝑅𝑇= 𝑚𝑜𝑙 𝐽 𝑚𝑜𝑙 −1 𝐾 − 𝐾 =7.30 × 𝐽

61. 2- what is the average kinetic energy per molecule?
1 2 𝑚 𝑣 2 = 3 2 𝑘𝑇= × 10 −23 𝐽 𝐾 − 𝐾 =6.07× 10 −21 𝐽

62. Distribution of Molecular Speeds in an Ideal Gas
Root mean square speed is assumed that all molecules move at the same speed.
The motions of gas molecules should have distribution of molecular speeds in equilibrium.
Statistical mechanics help us to understand gas molecules behaviour.

63. Boltzmann Distribution:
𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝐸 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 0 = 𝑒 −𝐸 𝑘𝑇
𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝐸 2 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝐸 1 = 𝑒 𝐸 1 − 𝐸 2 𝑘𝑇
Kinetic energy E= 𝑘𝑒 = 1 2 𝑚 𝑢 2
𝑁 𝑖 𝑁 = 𝑒 −𝐸 𝑘𝑇
Ni : number of molecules between E and E+dE
N: total number of molecules

65. Maxwell-Boltzmann Distribution
The M-B distribution function can be written in terms of energy as:
𝑓 𝐸 = 𝐴 𝑒 − 𝐸 𝑘𝑇
The distribution function can be written in terms of velocity:
𝑓 𝑢 =𝐴 𝑒 − 𝑚 𝑢 2 2𝑘𝑡
But the velocity is in three dimensional:
u = ux , uy , uz and du = dux .duy .duz = d3u
The constant A is determined by normalization. We will treat this as a probability distribution function normalized so that:
𝑓 𝑢 𝑑 3 𝑢= 𝐴 𝑒 − 𝑚 𝑢 2 2𝑘𝑡 𝑑 3 𝑢=1

66. 𝐴= 𝑒 − 𝑚 𝑢 2 2𝑘𝑡 𝑑 3 𝑢 −1
So:
From calculus :
𝑒 − 𝑥 2 = 𝜋
𝐴= 2𝜋𝑘𝑡 𝑚
Therefore:
𝑓 𝑢 = 2𝜋𝑘𝑡 𝑚 𝑒 − 𝑚 𝑢 2 2𝑘𝑡
In spherical coordinates. The velocity will be:
𝑢 𝑥 =𝑢 sin 𝜃 cos 𝜑
𝑢 𝑦 =𝑢 sin 𝜃 sin 𝜑
𝑢 𝑧 =𝑢 cos 𝜃

67. And in spherical coordinates the 3D differential volume element is:
𝑑𝑢= 𝑢 2 𝑑𝑢 sin 𝜃𝑑𝜃𝑑𝜑
𝑓 𝑢 𝑑 𝑢 =𝑓(𝑢,𝜃,𝜑) 𝑢 2 𝑑𝑢 sin 𝜃𝑑𝜃𝑑𝜑
Where: <𝑢<∞, <𝜑<𝜋, <𝜑<2𝜋

68. 𝑓 𝑢 = 𝑑 𝑁 𝑖 𝑁 =4𝜋 𝑚 2𝜋𝑘𝑡 3 2 𝑒 − 𝑚 𝑢 2 2𝑘𝑡 𝑢 2 𝑑𝑢
dNi : number of gas molecules that have speed between u and u+du
N: total number of molecules
k: boltzmann constant which equal to R/NA
1 𝑁 𝑑𝑁 𝑑𝑢 =4𝜋 𝑚 2𝜋𝑘𝑡 𝑒 − 𝑚 𝑢 2 2𝑘𝑡 𝑢 2 𝑑𝑢

69. To find the average speed: 𝑢 or 𝑢
𝑢 = 0 ∞ 𝑢 𝑑 𝑁 𝑖 𝑁 =4𝜋 𝑚 2𝜋𝑘𝑡 ∞ 𝑒 − 𝑚 𝑢 2 2𝑘𝑡 𝑢 3 𝑑𝑢
𝑎= 𝑚 2𝑘𝑡
Let
𝑢 = 0 ∞ 𝑢 𝑑 𝑁 𝑖 𝑁 =4𝜋 𝑎 𝜋 ∞ 𝑒 −𝑎 𝑢 2 𝑢 3 𝑑𝑢
𝑢 =4𝜋 𝑎 𝜋 𝑎 2
mNA= M

70. To find the most probable speed uP :
𝑢 = 8𝑘𝑡 𝜋𝑚 = 8𝑅𝑇 𝜋 𝑁 𝐴 𝑚 = 8𝑅𝑇 𝜋𝑀
To find the most probable speed uP :
By differentiating f with respect to u and looking for the value of u at which the derivative is zero
𝑑𝑓(𝑢 𝑑𝑢 =0
i.e.:
𝑢 𝑃 = 2𝑅𝑇 𝑀
Therefore:

71. To find the root mean square (rms) speed urms:
𝑘𝑒 = 1 2 𝑚 𝑢 2 = 3 2 𝑘𝑇
𝑚 𝑢 2 =3𝑘𝑇
𝑢 2 = 3𝑘𝑡 𝑚
𝑢 2 = 3𝑘𝑡 𝑚 = 3𝑅𝑇 𝑁 𝐴 𝑚 = 3𝑅𝑇 𝑀
𝑢 𝑟𝑚𝑠 = 3𝑅𝑡 𝑀

72. Maxwell distribution of speeds

74. The molecular speed (Diffusion)
Diffusion is the spread of one substance throughout a space or throughout a second substance.

75. The molecular speed (Diffusion)
From the equation (b): 𝑚 𝑁 𝐴 𝑢 2 =3𝑅𝑇
𝑚 ×𝑁 𝐴 =𝑀 the molar mass (kg / mol)
𝑀 𝑢 2 =3𝑅𝑇
𝑢 2 = 3𝑅𝑇 𝑀
𝑢 2 = 3𝑅𝑇 𝑀

76. 𝑢 𝑟𝑚𝑠 = 𝑢 2
This molecular speed called root-mean-square speed urmsof the molecules, which is the square root of the mean of the squares of the speeds of the molecules.

77. Note!: 𝑢 2 ≠ 𝑢 Because: 𝑢 2 = 𝑢 1 2 + 𝑢 2 2 + 𝑢 3 2 +…+ 𝑢 𝑁 2 𝑁 While: 𝑢 = 𝑢 1 + 𝑢 2 + 𝑢 3 +…+ 𝑢 𝑁 𝑁

78. Also Note!!
𝑢 𝑟𝑚𝑠 = 3𝑅𝑇 𝑀
The R must be:
8.314 J.mol-1.K-1or m3.Pa. mol-1.K-1
The M must be in kg/mol units

79. R = m3.(kg.m-1.s-2).mol-1.K-1 = kg.m2. s-2.mol-1.K-1
Because: J.mol-1.K-1 = kg.m2.s-2. mol-1.K-1
From Einstein equation: E=m.c2, the units are:
J = kg.m2.s-2
Also Pa = kg.m-1.s-2 so
R = m3.(kg.m-1.s-2).mol-1.K-1 = kg.m2. s-2.mol-1.K-1
𝑢= 3𝑅𝑇 𝑀 = 𝑘𝑔. 𝑚 2 . 𝑠 −2 . 𝑚𝑜𝑙 −1 . 𝐾 −1 ×𝐾 𝑘𝑔. 𝑚𝑜𝑙 −1 = 𝑚 2 . 𝑠 −2 = 𝑚 . 𝑠 −1

80. Calculate the rms speed, u, of an N2 molecule at
Is this correct? If yes How?
𝑢 𝑟𝑚𝑠 = 3𝑃 𝑑

81. Effusion
Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

82. Effusion
The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.

83. Graham's Law From root-mean-sqaure speed urms equation:
𝑢 𝑟𝑚𝑠 = 3𝑅𝑇 𝑀
𝑟 1 𝑟 2 = 𝑢 𝑟𝑚𝑠1 𝑢 𝑟𝑚𝑠2 = 3𝑅𝑇 𝑀 𝑅𝑇 𝑀 = 𝑀 2 𝑀 1
𝑟 1 𝑟 2 = 𝑀 2 𝑀 1
Where r1 & r2 are rates of effusion of the two substances.

84. From kinetic energy law:
𝑘𝑒 = 1 2 𝑚 𝑢 2
For two gases:
𝑘𝑒 𝐴 = 1 2 𝑚 𝐴 𝑢 2 𝐴 And 𝑘𝑒 𝐵 = 1 2 𝑚 𝐵 𝑢 2 𝐵
When the temperature constant, then their kinetic are equal:
1 2 𝑚 𝐴 𝑢 2 𝐴 = 1 2 𝑚 𝐵 𝑢 2 𝐵

85. 𝑢 2 𝐴 𝑢 2 𝐵 = 𝑚 𝐵 𝑚 𝐴
𝑢 2 𝐴 𝑢 2 𝐵 = 𝑚 𝐵 𝑚 𝐴 = 𝑢 𝑟𝑚𝑠 𝑢 𝑟𝑚𝑠 = 𝑟 𝐴 𝑟 𝐵
Knowing that:
𝑑= 𝑚 𝑉 𝑎𝑛𝑑 𝑚=𝑉×𝑑
𝑟 𝐴 𝑟 𝐵 = 𝑉 𝐵 × 𝑑 𝐵 𝑉 𝐴 × 𝑑 𝐴

86. In the same volume:
𝑟 𝐴 𝑟 𝐵 = 𝑑 𝐵 𝑑 𝐴

87. Graham's Law
An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only times that of O2 at the same temperature. Calculate the molar mass of the unknown, and identify it.

88. The collision frequency and the mean free path
The kinetic model enables us to make that picture more quantitative. In particular, it enables us to calculate the frequency with which molecular collisions occur and the distance a molecule travels on average between collisions

89. To know how many collision that happened by molecules let suppose a vessel its volume is V and contains gas molecules A. The diameter of A molecule is d and move in average speed u in a cylindrical path its diameter is twice of A (i.e. 2d). Also supposed all other molecules which collide with A molecule is solid.

90. Note: 𝑛 𝐴 =𝑛× 𝑁 𝐴 therefore 𝑛 𝐴 = 𝑁 𝐴 𝑃 𝑅𝑇 = 𝑃 𝑘𝑇
Note that A molecule collide with only that molecules which center inside the cylindrical path.
If we suppose that length of cylindrical path equal to average speed 𝑢 , therefore the volume of the path is 𝜋 𝑑 2 𝑢 .
And if we suppose that number of molecules is 𝑛 𝐴 then the number of A molecules per volume unit is 𝑛 𝐴 = 𝑁 𝑉 .
Note: 𝑛 𝐴 =𝑛× 𝑁 𝐴 therefore 𝑛 𝐴 = 𝑁 𝐴 𝑃 𝑅𝑇 = 𝑃 𝑘𝑇
𝑢 𝐴 average speed of A molecules in unit (m.s-1)
𝑑 𝐴 the diameter of A molecule, also call the diameter of collision in unit (m)
nA number of A molecules
V the volume in unit (m3)

91. The relative mean speed 𝑢 𝑟𝑒𝑙 , is the mean speed with which one molecule approaches another, can also be calculated:
𝑢 𝑟𝑒𝑙 = 𝑢 𝐴 𝑢 𝐴 = 2 𝑢 𝐴
For different molecules A and B
𝑢 𝑟𝑒𝑙 = 𝑢 𝐴 𝑢 𝐵 = 8𝑅𝑇 𝜋 𝑀 𝐴 + 𝑀 𝐵 𝑀 𝐴 𝑀 𝐵
The term 𝑀 𝐴 𝑀 𝐵 𝑀 𝐴 + 𝑀 𝐵 is known as the reduced molar mass of the two molecules A and B

92. 𝑍 𝐴 = 𝜋 𝑑 𝐴 2 𝑢 𝑟𝑒𝑙 𝑁 𝑉 =𝜋 𝑑 𝐴 2 𝑢 𝑟𝑒𝑙 𝑛 𝐴 collision s-1
The collision number (frequency) that happened by one A molecule per one second ZA is:
𝑍 𝐴 = 𝜋 𝑑 𝐴 2 𝑢 𝑟𝑒𝑙 𝑁 𝑉 =𝜋 𝑑 𝐴 2 𝑢 𝑟𝑒𝑙 𝑛 𝐴 collision s-1
Also:
𝑍 𝐴 = 2 𝑢 𝐴 𝜋 𝑑 𝐴 2 𝑛 𝐴 = 𝑢 𝐴 𝜋 𝑑 𝐴 2 𝑃 𝑘𝑇 just for similar molecules A
But for different molecules A and B, we have to find the diameter of collision 𝑑 𝐴𝐵 :
𝑑 𝐴𝐵 = 𝑑 𝐴 + 𝑑 𝐵 2 =( 𝑟 𝐴 + 𝑟 𝐵 )

93. Therefore the number of collision that happed by one A molecules with B molecules:
𝑍 𝐴 =𝜋 𝑑 𝐴𝐵 2 𝑢 𝑟𝑒𝑙 𝑛 𝐵 =𝜋 𝑑 𝐴𝐵 𝑢 𝐴 𝑢 𝐵 𝑛 𝐵
If we want to know the total number of collision that happened by all A molecules, ZAA, we have to multiply ZA by the total number of molecules nA as follows:
𝑍 𝐴𝐴 = 1 2 𝑍 𝐴 × 𝑛 𝐴 = 1 2 𝜋 𝑢 𝑟𝑒𝑙 𝑑 𝐴 2 𝑛 𝐴 2 = 𝜋 𝑑 𝐴 𝑅𝑇 𝜋 𝑀 𝐴 𝑛 𝐴 2 = 𝑑 𝐴 𝜋8𝑅𝑇 𝑀 𝐴 𝑛 𝐴 2

94. Note that we divided on 2 to avoid calculating each collision twice because we deal with similar molecules.
But if the collided molecules different, such as A molecules collide with B molecules. Therefore the total number of collision (collision frequency) (this time will be denoted as ZAB) will be find through the equation:
𝑍 𝐴𝐵 =𝜋 𝑑 𝐴𝐵 𝑢 𝐴 𝑢 𝐵 𝑛 𝐴 𝑛 𝐵
Again, dAB is the diameter of collision and given by equation:
𝑑 𝐴𝐵 = 𝑑 𝐴 + 𝑑 𝐵 2 = 𝑟 𝐴 + 𝑟 𝐵

95. If A molecule collide with another molecule, it will take a specific distance before collide with another molecule. This distance called mean free path λ
𝜆= 𝑢 𝐴 𝑍 𝐴 = 𝑢 𝐴 𝜋 𝑑 𝐴 2 𝑢 𝑟𝑒𝑙 𝑛 𝐴 = 𝑢 𝐴 2 𝜋 𝑑 𝐴 2 𝑢 𝐴 𝑛 𝐴 = 𝜋 𝑑 𝐴 2 𝑛 𝐴
𝜆= 𝜋 𝑑 𝐴 2 𝑛 𝐴 = 𝑅𝑇 2 𝜋 𝑑 𝐴 2 𝑁 𝐴 𝑃 𝐴 = 𝑘𝑇 2 𝜋 𝑑 𝐴 2 𝑃 𝐴
This equation prove that reducing the pressure in a system will decrease the number of collision and increase length of the mean free path

96. Real Gases
In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

97. Real Gases
Even the same gas will show wildly different behavior under high pressure at different temperatures.

98. Deviations from Ideal Behavior
The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

99. Corrections for Non-ideal Behavior
The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.
The corrected ideal-gas equation is known as the van der Waals equation.

100. The van der Waals Equation
) (V − nb) = nRT
n2a V2
(P +

101. 𝑃 𝑖𝑑𝑒𝑎𝑙 = 𝑃 𝑟𝑒𝑎𝑙 + 𝑎𝑛 2 𝑉 2 𝑉 𝑖𝑑𝑒𝑎𝑙 = 𝑉 𝑟𝑒𝑎𝑙 −𝑛𝑏 Observed Correction
𝑃 𝑖𝑑𝑒𝑎𝑙 = 𝑃 𝑟𝑒𝑎𝑙 + 𝑎𝑛 2 𝑉 2
Observed
pressure
Correction
term
𝑉 𝑖𝑑𝑒𝑎𝑙 = 𝑉 𝑟𝑒𝑎𝑙 −𝑛𝑏
Observed
volume
Correction
term

102. 𝑃+ 𝑎𝑛 2 𝑉 2 𝑉−𝑛𝑏 =𝑛𝑅𝑇 Corrected volume Corrected pressure
𝑃+ 𝑎𝑛 2 𝑉 2 𝑉−𝑛𝑏 =𝑛𝑅𝑇
Corrected volume
Corrected pressure
The constant “a” is a measure of how strongly the gas molecules attract one another.
The constant “b” is a measure of the finite volume occupied by the molecules.

103. The term 𝑎𝑛 2 𝑉 2 accounts for the attractive forces.
The equation adjusts the pressure upward by adding 𝑎𝑛 2 𝑉 2 because attractive forces between molecules tend to reduce the pressure.
The term nb accounts for the small but finite volume occupied by the gas molecules.
The equation adjusts the volume downward by subtracting nbto give the volume that would be available to the molecules in the ideal case.
WATCH THE UNITS OF “a” and “b” constants!

104. If 1. 000 mol of an ideal gas were confined to 22. 41 L at 0
If mol of an ideal gas were confined to L at 0.0 °C, it would exert a pressure of atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by mol of Cl2(g) in L at 0.0 °C.