1. Gases Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. C HAPTER 5 ( P 170-196) 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 S UBSTANCES THAT EXIST AS GASES.

4. Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

5.  H 2,N 2,O 2,F 2 and Cl 2 exist as gases diatomic molecules.  Allotrope of oxygen(O 2 ),ozon(O 3 ) is also a gas 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 25 o C and 1 atm(why).  Some molecular compounds (example: CO,CO 2,HCl,NH 3 and CH 4 are gases, but the majority of molecular compounds are liquid or solids at room temperature.  On heating they are converted to gases much more easily than ionic compounds (why).  H 2,N 2,O 2,F 2 and Cl 2 exist as gases diatomic molecules.  Allotrope of oxygen(O 2 ),ozon(O 3 ) is also a gas 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 25 o C and 1 atm(why).  Some molecular compounds (example: CO,CO 2,HCl,NH 3 and CH 4 are gases, but the majority of molecular compounds are liquid or solids at room temperature.  On heating they are converted to gases much more easily than ionic compounds (why). Strong electrostatic forces large amount of energy (heat)

6. 5.1 Deadly poisons

7. 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 Physical Characteristics of Gases

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. Sea level1 atm 4 miles0.5 atm 10 miles 0.2 atm Atmospheric pressure الضغط الجوي : وزن عمود الهواء المؤثر على وحدة المساحة ويمتد رأسيا من السطح إلى نهاية الغلاف الجوي. الضغط الجوي يكون أكبر ما يمكن بالقرب من سطح الأرض في أي مكان ويقل مع الارتفاع رأسيا إلى أعلى

10. 5.2 Barometer: the most familiar instrument for measuring atmospheric pressure Standard atmospheric pressure(1 atm):is equal to the pressure that supports a column of mercury exactly 760 mm(or 76 cm) high at 0 o C at sea level. How is atmospheric pressure measured Barometer Standard atmospheric pressure= 760 mm Hg 1atm= 760 mm Hg

11. SI Units of pressure m/s m/s 2 Force= mass ×acceleration Kg m/s 2 = N (Newton) N/m 2 =1 Pa SI Unit of pressure is the Pascal (Pa)

12. The relation between atmospheres and Pascals Units of pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 76 cm Hg =760 mmHg = 760 torr 1 atm = 101,325 Pa =1.01325x10 5 Pa 5.2 1 atm = 1.01325X10 2 kPa pressure can be converted from mmHg to atm and kPa

13. W ORKED E XAMPLE 5.1

14. 1 atm = 760 mmHg X = 688 mmHg X= 688/760 = 0.905 atm

16. 1 atm = 1.01325 × 10 5 Pa = 760 mmHg X Pa = 732 mmHg X Pa = (732 × 1.01325 × 10 5 ) / 760 mmHg X Pa = 9.76 × 10 4 Pa = 97.6 kPa

17. Manometers Used to Measure Gas Pressures Manometer is a device used to measure the pressure of gases other than atmosphere Closed-tube manometeropen-tube manometer Used to measure pressures below atmospheric pressure Used to measure pressures equal or greater than atmospheric pressure

18. 5.3 The gas laws.

19. Ideal Gas Equation

20. 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. Gases can be characterized by four variables: (1)Pressure (P) (2)Volume (V) (3)Temperature (T) (4)Amount of gas (mole No. (n)) V = constant x = R nT P P R is the gas constant PV = nRT

21. The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. Evaluation of gas constant (R)

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

23. PV = nRT Initial conditionFinal condition Modified form of ideal gas equation that take into account the initial and final conditions No condition

24. The Pressure –Volume Relationship: Boyle`s Law Boyle’s law: V  (at constant n and T) 1 P P 1 x V 1 = P 2 x V 2 As P increasesV decrease

25. P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 5.3 Boyle’s Law: the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas Constant temperature (T) Constant amount of gas (n)

26. 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 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3 P x V = constant

27. The Temperature –Volume Relationship: Charles`s and Gay-Lussac`s Law Charles’ law: V  T  (at constant n and P) V 1 /T 1 = V 2 /T 2 As T increases V increases

28. The volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas 5.3 V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin Absolute zero: theoretically, the lowest the attainable temperature=-273.15 0 C (kelvin temperature scale). 0 K

29. A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K 5.3 V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K

30. The Volume-Amount Relationship: Avogadro`s Law Avogadro’s law: V  n  (at constant P and T) V 1 / n 1 = V 2 / n 2

31. A VOGADRO ’ S L AW : AT CONSTANT PRESSURE AND TEMPERATURE, THE VOLUME OF A GAS IS DIRECTLY PROPORTIONAL TO THE NUMBER OF MOLES OF THE GAS PRESENT V  number of moles (n) V = constant x n V 1 / n 1 = V 2 / n 2 5.3 Constant temperature Constant pressure

32. 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? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 5.3

33. 5.4 The Pressure-Temperature Relationship: Amonten’s Law P 1 / T 1 = P 2 / T 2 Amonten’s law: P  T  (at constant V and n)

34. 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the light bulb (in atm)? P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4 18 o C + 273 85 o C + 273

35. General gas law (constant moles) Ideal Gases Law When n 1 =n 2 General gas law

36. ConditionsRelationshipLaw Constant temperature and moles Boyle’s Law Constant pressure and moles OR Charles’s Law Constant volume and moles OR Amonten’s Law Constant pressure and temperature OR Avogadro’s Law Constant moles General Law No conditionsPV = nRT Ideal Gases Law Summarizing of the gases laws Ideal Gases Law

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

38. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? 5.4 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K Latm molK M = 54.6 g/mol

39. W ORKED E XAMPLE 5.3

40. PV = nRT P = nRT/V P = 1.82 ×0.0821×(69.5+273.15) /5.43 P = 9.43 atm

42. 17.03 g 22.41 L 7.40 g x L x L = 7.40×22.41/17.03 x= 9.74 L at STP, 1 mole of an ideal gas occupies 22.414 L. Molar mass of ammonia

44. P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = n 1 = n 2, T 1 = T 2 P1V1P1V1 P2V2P2V2 = P 1 = 1.0 atm P 2 = 0.40 atm V 1 = 0.55 L V 2 = ? V2V2 P 1 V 1 /P 2 = V2V2 1.0 × 0.55/0.4= V2V2 1.375 L = V2V2 1.4 L ≈ Boyle’s Law

46. P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm P 2 = ? atm T 1 = (18+273) = 291 K T 2 = (85+273) = 358 K P2P2 P1T2P1T2 T1T1 =P2P2 1.2×358 291 = P2P2 = 1.48 atm P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = n 1 = n 2, V 1 = V 2 Amonten’s law

48. d PM RT = d 0.999 × 44.01 0.0821 × 328 = d= 1.62 g/L P=0.990 atmT= 55+273= 328 K R = 0.082057 L atm / (mol K) M (CO 2 )= (1 ×12.01)+ (2×16.0)=44.01 g/mol

50. M dRT P = M 7.71 × 0.0821× 309 2.88 = M= 67.9 g/mol d= 7.71 g/L P=2.88 atm T=36 +273= 309 K R = 0.082057 L atm / (mol K)

52. أولا يتم تعيين الصيغة الأوليةempirical formula للمركب بحساب عدد المولات للذرات المكونة للمركب n Si = mass percent/M = 33.0/28.09 = 1.17 mol Si n F = mass percent/M = 67.0/10.0 = 3.53 mol F بالقسمة على عدد المولات الأصغر n Si = 1.17/1.17 = 1 n F = 3.53/1.17 = 3.02 ≈ 3 إذاً الصيغة الأولية SiF 3

53. ثانياً حساب عدد مولات المركب n PV RT = n 1.7 × 0.21 0.0821 × 308 = n = 0.0141 mol ثالثاً حساب الكتلة المولية M 2.38 0.0141 mol = M = 169 g/mol رابعاً بقسمة الكتلة المولية على كتلة الصيغة الأولية نحصل على الرقم الذي يضرب في عدد ذرات العناصر المكونة للمركب g/mol Empirical molar mass= Si (28.09 ×1) + F (19×3) = 85.09 n= Molar mass of compound / empirical molar mass = 169/85.09 = 1.986 ≈ 2 Molecular formula= (SiF 3 ) n =(SiF 3 ) 2 = Si 2 F 6

54. Gas Stoichiometry

55. What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm == 4.76 L 5.5 from balanced equation nC 6 H 12 O 6 5.60 g 180 g/mol = = 0.0311 mol 1 mol C 6 H 12 O 6 6 mol CO 2 0.031 mol C 6 H 12 O 6 x = 0.187 mol CO 2 n CO 2

57. x = 7.64 ×5/2 2 moles C 2 H 2 5 moles O 2 2 L C 2 H 2 5 L O 2 7.64 L C 2 H 2 x L O 2 x= 19.1 L According to balanced equation: Avogadro's law

59. أولاً: يتم حساب عدد مولات N 2 عن طريق إيجاد عدد مولات NaN 3 n mol of NaN 3 60/65.02 = 0.923 moles 2 moles of NaN 3 3 moles of N 2 0.923 moles of NaN 3 x moles of N 2 x moles of N 2 = 0.923 × 3/2 = 1.38 mol ثانيا: يتم حساب حجم N 2 باستخدام معادلة الغاز المثالي V nRT P = V 1.38 ×0.0821× (80+273) 823/760 = V = 36.93 L 1 atm 760 mmHg X atm 823 mmHg

60. 1-Calculat e the volume of CO 2 generated at 100 0 C and 1 atm by the decomposition of 50 g of CaCO 3 Ca CO 3 (s) CaO (s) +CO 2(g ) Ans=15.3L 2-A 0.250-L vessel contains 1.25 g of a gas at 715 mmHg and 28.0 0 C. What is the molar mass of the gas? Ans=131 g / mol 3-What is the density of the Cl 2 at 25 0 C,1 atm? Ans=2.9 g/L

61. Dalton’s Law of Partial Pressures

62. 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 P1P1 P2P2 P total = P 1 + P 2 5.6 قانون دالتون للضغوط الجزئية : الضغط الكلي لخليط من غازات مختلفة يساوي مجموع الضغوط الجزئية لكل غاز. PT=Pa+Pb+……..

63. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT الكسر المولي: هو تعبير كيميائي يصف نسبة تواجد مادة معينة في مخلوط من المواد. ويعرف االكسر المولي لمادة i في المخلوط بأنها كمية المادة n i مقسومة على الكمية الكلية للمخلوط n T.

64. A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6

66. P Ne = X Ne P T X Ne = n Ne / (n Ne + n Ar + n Xe ) X Ne = 4.46 mol/ (4.46 mol + 0.74 mol + 2.15 mol) X Ne = 0.607 P Ne = X Ne P T P Ne = 0.607 × 2 atm P Ne = 1.21 atm يحسب الضغط الجزيئي partial pressure من العلاقة أولاً يتم إيجاد قيمة الكسر المولي mole fraction ثانياً يتم حساب الضغط الجزئي بالتعويض في المعادلة

67. بنفس الطريقة يتم حساب الضغط الجزئي لغازي الأرجون والزينون X Ar = 0.74 mol/ (4.46 mol + 0.74 mol + 2.15 mol) = 0.1 P Ar = 0.1 × 2 atm = 0.2 atm X Xe = 2.15 mol/ (4.46 mol + 0.74 mol + 2.15 mol) = 0.293 P Xe = 0.293 × 2 atm = 0.586 atm ملاحظة يجب أن يكون مجموع الضغوط الجزئية للغازات يساوي الضغط الكلي P T = 1.21 + 0.20 + 0.586 = 1.996 ≈ 2 atm

68. 2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O 22 5.6 Dalton's law of partial pressures is useful for calculating volumes of gases collected over water

69. 5.6

71. P T = P O 2 + P H 2 O P O 2 = P T - P H 2 O P O 2 = 762 mmHg – 22.4 mmHg P O 2 = 739.6 mmHg =740 mmHg أولاً يتم حساب ضغط غاز الأوكسجين من العلاقة PV nRT= = RT m M m PVM = RT m (740/760) atm ×0.128L×32 g/mol = ثانياً يتم حساب كتلة الأوكسجين من قانون الغازات المثالية 0.0821 L.atm/K.mol × (24+273) K m = 0.164 g

72. 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 – 5.50 - 5.94. 5 52 – 5.54 – 5.64 – 5.66 – 5.68 – 5.70.