1. Gases
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2. Chapter 5 (p170-196) 5.1 Substances that exist as gases.
5.2 Pressure of gas.
5.3 The gas laws.
5.4 The ideal gas equations.
5.5 Gas Stoichiometry.
5.6 Dalton’ s law of partial pressures.

3. 5.1 Substances that exist as gases.

4. Elements that exist as gases at 250C and 1 atmosphere
5.1

5. H2,N2,O2,F2 and Cl2 exist as gaseous diatomic molecules.
Allotrope of oxygen(O2) ,ozon(O3) is also agas at room temperature.
All the element in group 8A,the noble gases are monatomic
gases:He,Ne,Ar,Kr,Xe and Rn.
Ionic compounds do not exist as gases at 25oC and 1 atm(why).
Some molecular (example: CO,CO2,HCl,NH3 nd CH4 are gases,but the
majority of molecular compounds are liquid or solids at room temperature.
On heating they are converted to gases much more esily than ionic
compounds(why).

6. 5.1

7. Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
5.1

8. 5.2 Pressure of gas.

9. The density of air decreases very rapidly with increasing distance from earth.
Atmospheric pressure is the pressure exerted by Earth´s atmosphere.
The actual value of atmospheric pressure depends on
location,temperature and weather conditions.
Barometer:the most familiar instrument for measuring
atmospheric pressure
Standard atmospheric pressure(1 atm):is equal to the pressure that supports a columm of mercury exactly 760 mm(or 67 cm)high at 0oC at sea level.

10. (force = mass x acceleration)
Pressure of a gas
Force
Area
Barometer
Pressure =
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 76 cmHg
=760 mmHg = 760 torr
1 atm = 101,325 Pa= x105Pa
1 atm = X102 kPa
5.2

11. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm
Atmospheric pressure
10 miles
0.2 atm
4 miles
0.5 atm
Sea level
1 atm
5.2

12. Worked Example 5.1

13. 1 atm = 760 mmHg
X = 688 mmHg
X= 68/760
= atm

15. × 105 Pa = 760 mmHg
x Pa = 732 mmHg
x Pa = (732 × × 105) / 760 mmHg
x Pa = 9.76 × 104 Pa
= 97.6 kPa

16. Manometers Used to Measure Gas Pressures
5.2

17. 5.3 The gas laws.

18. The Pressure –Volume Relationship:
Boyle`s Law

19. Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas (Boyles’s law)
As P (h) increases
V decreases
5.3

20. Boyle’s Law: the pressure of a fixed amount of gas at
a constant temperature is inversely proportional to the volume of the gas
P a 1/V
Constant temperature
Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2
5.3

21. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg
5.3

22. The Temperature –Volume Relationship: Charles`s and Gay-Lussac`s Law

23. Charles’s & Gay-Lussac’s law
As T increases
V increases
5.3

24. The volume of a fixed amount of gas maintained at
constant pressure is directly proportional to the absolute temperature of the gas
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C)
5.3

25. Absolute zero:theoretically,the lowest the attainable temperature=-273
Absolute zero:theoretically,the lowest the attainable temperature= C(kelven temperture scale).

26. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K
5.3

27. The Volume-Amount Relationship:
Avogadro`s Law

28. V a number of moles (n) V = constant x n V1 / n1 = V2 / n2
Avogadro’s Law: at constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present
V a number of moles (n)
Constant temperature
Constant pressure
V = constant x n
V1 / n1 = V2 / n2
5.3

29. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO
5.3

30. Ideal Gas Equation

31. Boyle’s law: V a (at constant n and T) 1 P
Ideal Gas Equation An ideal gas is a hypothetical gas whose pressure – volume – temperature behavior can be completely accounted for by the ideal gas equation.
Boyle’s law: V a (at constant n and T) 1 P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT
5.4

32. 5.3

33. 5.3

34. 5.3

35. General gas law:
Constant moles

36. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)
5.4

37. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.6 L
5.4

38. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in light bulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the light bulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm
5.4

39. Summarizing of the gases laws
Conditions
Relationship
Law
Constant temperature and moles
Boyle’s Law
Constant pressure and moles OR
Charles’s Law
Constant volume and moles
Amonten’s Law
Constant pressure and temperature
Avogadro’s Law
Constant moles
General Law
No conditions
PV = nRT
Ideal Gases Law

40. d is the density of the gas in g/L
Density (d) Calculations m M m M
PV = nRT = RT n= m M
PV= RT PV M m = RT m V = PM RT
m is the mass of the gas in g PM RT
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L
5.4

41. d is the density of the gas in g/L
Density (d) Calculations m M
PV = nRT = RT = PM RT m V
m is the mass of the gas in g PM RT
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L
5.4

42. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.6 g/mol
5.4

43. Worked Example 5.3

44. PV = nRT
P = nRT/V
P = 1.82×0.0821×( )/5.43
P = 9.43 atm

46. 17.03 g L
7.40 g x L
x L = 7.40×22.41/17.03
x= 9.74 L

48. P1V1
n1T1
P2V2
n2T2 =
n1 = n2, T1 = T2
P1V1
P2V2 =
P1 = 1.0 atm P2 = 0.40 atm
V1 = L V2 = ? V2
P1V1/P2 = V2
1.0 × 0.55/0.4 = V2
1.375 L = V2
1.4 L ≈

50. P1 T1 P2 T2 =
P1 = 1.20 atm P2 = ? atm
T1 = (18+273) = 291 K T2 = (85+273) = 358 K P2
P1T2 T1 = P2
1.2×358
291 = P2
= 1.48 atm

52. d PM RT = d
0.999 × 44.01
× 328 = d
= 1.62 g/L

54. M
dRT P = M
7.71 × × (36+273)
2.88 = M
= 67.9 g/mol

56. nSi = m/M = 33.0/28.09 = 1.17 mol Si nF = m/M = 67.0/10.0 = 3.53 mol F
أولا يتم تعيين الصيغة الأوليةempirical formula للمركب بحساب عدد المولات للذرات المكونة للمركب
nSi = m/M = 33.0/ = mol Si
nF = m/M = 67.0/10.0 = mol F
بالقسمة على عدد المولات الأصغر
nSi = 1.17/1.17 = 1
nF = 3.53/1.17 = 3.02 ≈ 3
إذاً الصيغة الأولية SiF3

57. ثانياً حساب عدد مولات المركبn PV RT = n
1.7 × 0.21
× 308 = n
= mol
ثالثاً حساب الكتلة المولية M
2.38
mol = M
= 169 g/mol
رابعاً بقسمة الكتلة المولية على كتلة الصيغة الأولية نحصل على الرقم الذي يضرب في عدد ذرات العناصر المكونة للمركب
Molar mass / empirical molar mass = 169/ = ≈ 2
Si (28.09 ×1) + F (19×3) = 85.09
SiF3 = (SiF3)2 = Si2F6

58. Gas Stoichiometry

59. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L
5.5

61. 2 moles C2H moles O2
2 L C2H L O2
7.64 L C2H x L O2 x
= 7.64×5/2 x
= 19.1 L

63. أولاً يتم حساب عدد مولات N2 عن طريق إيجاد عدد مولات NaN3
n mol of NaN3 60/65.02 = moles
2 moles of NaN moles of N2
0.923 moles of NaN x moles of N2
x moles of N = × 3/2 = 1.38 mol
ثانيا يتم حساب حجم N2 باستخدام معادلة الغاز المثالي V
nRT P = V
1.38 ×0.0821× (80+273)
823/760 = V
= L

64. 3-What is the density of the Cl2 at 250C,1 atm? Ans=2.9 g/L
1-Calculat e the volume of CO2 generated at 1000C and 1 atm by the decomposition of 50 g of CaCO3
Ca CO3 (s) CaO(s)+CO2(g)
Ans=15.3L
2-A L vessel contains 1.25 g of a gas at 715 mmHg and C. What is the molar mass of the gas?
Ans=131 g / mol
3-What is the density of the Cl2 at 250C,1 atm?
Ans=2.9 g/L

65. Dalton’s Law of Partial Pressures

66. Dalton’s Law of Partial Pressures: the total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone.
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

67. PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
mole fraction (Xi) = ni nT
5.6

68. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
5.6

70. يحسب الضغط الجزيئي partial pressure من العلاقة
PNe = XNePT
أولاً يتم إيجاد قيمة الكسر المولي mole fraction
XNe = nNe/ (nNe + nAr + nXe)
XNe = 4.46 mol/ (4.46 mol mol mol)
XNe = 0.607
ثانياً يتم حساب الضغط الجزئي بالتعويض في المعادلة
PNe = XNePT
PNe = × 2 atm
PNe = 1.21 atm

71. بنفس الطريقة يتم حساب الضغط الجزئي لغازي الأرجون والزينون
XAr = 0.74 mol/ (4.46 mol mol mol) = 0.1
PAr = 0.1 × 2 atm = 0.2 atm
XXe = 2.15 mol/ (4.46 mol mol mol) = 0.293
PXe = × 2 atm = atm
ملاحظة يجب أن يكون مجموع الضغوط الجزئية للغازات يساوي الضغط الكلي
PT= = ≈ 2 atm

72. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
5.6

73. PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6
Bottle full of oxygen gas and water vapor
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2
5.6

74. 5.6

76. أولاً يتم حساب ضغط غاز الأوكسجين من العلاقة
PT = PO2 + PH2O
PO2 = PT - PH2O
PO2 = 762 mmHg – 22.4 mmHg
PO2 = mmHg
ثانياً يتم حساب كتلة الأوكسجين من قانون الغازات المثالية m PV
nRT = RT = M m
PVM = RT m
(740/760) atm ×0.128L×32 g/mol =
m = g
L.atm/K.mol × (24+273) K

77. Problems
5.1 – 5.2 – 5.3 – 5.8 – 5.14 – 5.22
5.24 – 5.32 – 5.36 – 5.38 – 5.40 – 5.44 – 5.48 –
5 52 – 5.54 – 5.64 – 5.66 – 5.68 – 5.70.