1. Chapter 5 The Gaseous State

2. G AS P ROPERTIES خصائص الغازات 5 | 2 Gases differ from liquids and solids: الغاز يختلف عن السوائل والصلب بما يلي : They are compressible. القابلية للضغط او الانضغاط The variables pressure, volume, temperature, and amount are related. التنوع من ناحية الضغط، الحجم، الحرارة، والكميات هي جميعها ذات علاقة

3. P RESSURE لنبدأ بالضغط Pressure, P The force exerted per unit area الضغط ويرمز له بالرمز P هو القوة المؤثرة عل وحدة من المساحة It can be given by two equations: يمكن توضيحها بمعادلتين The SI unit for pressure is the pascal, Pa. 5 | 3

4. P RESSURE Other Pressure Units atmosphere, atm mmHg torr Bar الضغط الجوي العادي 760 ملم زئبق حجم الغاز عل هذا الضغط هو 100 مل 5 | 4

5. P RESSURE A barometer is a device for measuring the pressure of the atmosphere. جهاز لقياس الضغط الجوي A manometer is a device for measuring the pressure of a gas or liquid in a vessel. جهاز لقياس ضغط الغاز او السائل في الانابيب 5 | 5

6. E XAMPLE 1 The water column would be higher because its density is less by a factor equal to the density of mercury to the density of water. طبعا كثافة الماء اقل وسترتفع اكثر بعامل مساوي لكثافة الزئبق مضروب بنسبو كثافة الزئبق الى كثافة الماء حسب معادلة الضغط 5 | 6 لنفترض ان لديك جهازين بارميتر لقياس الضغط الجوي واحد مليء بالزئبق وواحد بالماء اي العمودين سيرتفع اكثر

7. G AS L AWS قانون الغاز 5 | 7 Empirical Gas Laws قانون الغازات التجريبي All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties. تتصرف الغازات بحسب تعرضها لدرجة الحرارة الضغط والحجم الموضوعة فيه و كمية المولات. اذا تم تثبيت اثنان من هذه الخواص الفيزيائية للغازات فبالامكان اظهار العلاقة بين الخاصيتين المتبقيتين The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century. الدراسات الموجودة على نظرية الغازات التجريبية موجودة من منتصف القرن السابع عشر وحتى القرن التاسع عشر.

8. B OYLE ’ S L AW 5 | 8 Boyle’s Law قانون بويل The volume of a sample of gas at constant temperature varies inversely with the applied pressure. حجم عينة الغاز في درجات حرارة ثابتة تختلف عكسيا مع الضغط الذي تتعرض له The mathematical relationship: In equation form: اذن الحجم والضغط دائما مترافقين فلو كان هناك حجم غاز معين وعليه ضغط معين تبقى محصلة الحجم متناسبة مع اي زيادة او اختلاف في الضغط

9. B OYLE ’ S L AW 5 | 9 Figure A shows the plot of V versus P for 1.000 g O 2 at 0°C. This plot is nonlinear. Figure B shows the plot of ( 1 / V ) versus P for 1.000 g O 2 at 0°C. This plot is linear, illustrating the inverse relationship.

10. B OYLE ’ S L AW 5 | 10 At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to one- third, 33 mL.

11. B OYLE ’ S L AW 5 | 11 When a 1.00-g sample of O 2 gas at 0C is placed in a containerat a pressure of 0.50 atm, it occupies a volume of 1.40 L. When the pressure on the O 2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

12. E XAMPLE 2 5 | 12 A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant? V i = 38.7 mL P i = 751 mmHg T i = 21°C V f = ? P f = 359 mmHg T f = 21°C

13. 5 | 13 V i = 38.7 mL P i = 751 mmHg T i = 21°C V f = ? P f = 359 mmHg T f = 21°C = 81.0 mL (3 significant figures) Example 2 (Cont)

14. V OLUME - T EMPERATURE 5 | 14 A graph of V versus T is linear. Note that all lines cross zero volume at the same temperature, -273.15°C.

15. A BSOLUTE Z ERO ظارة الصفر المطلق 5 | 15 The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero. تسمى درجة حرارة -273.15 بدرجة الصفر المطلق وهي درجة الحرارة التي عندها تكون احجام الغازات نظريا صفر. This is the basis of the absolute temperature scale, the Kelvin scale (K). وهي بالمناسبة الدرجة الاساسية لتدريج كالفن الحراري

16. C HARLES ’ L AW قانون تشارل 5 | 16 Charles’ Law The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K). حجم عينة من الغاز على درجة ضغط ثابت يتناسب مباشرة مع قيمة الحرارة المطلقة بالكالفن. The mathematical relationship: In equation form:

17. C HARLES ’ L AW 5 | 17 A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume. As the air inside warms, the balloon expands to its orginial size.

18. C HARLES ’ L AW 5 | 18 A 1.0-g sample of O 2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L. When the absolute temperature of the sample is raised to 200 K, the volume of the O 2 is doubled to 0.52 L.

19. E XAMPLE 3 5 | 19 You prepared carbon dioxide by adding HCl( aq ) to marble chips, CaCO 3. According to your calculations, you should obtain 79.4 mL of CO 2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C? V i = 79.4 mL P i = 760 mmHg T i = 0°C = 273 K V f = ? P f = 760 mmHg T f = 27°C = 300. K

20. 5 | 20 V i = 79.4 mL P i = 760 mmHg T i = 0°C = 273 K V f = ? P f = 760 mmHg T f = 27°C = 300. K = 87.3 mL (3 significant figures) Example 3 (Cont)

21. G AY -L USSAC ’ S L AW قانون جاي لوساك 5 | 21 Gay-Lussac’s Law There is an analogous relationship to Charles’ Law that relates pressure and temperature. This relationship is called Gay-Lussac’s Law, and is described as follows: هناك علاقة متناظرة لقانون شارل والتي ترتبط منخلالها قيم الضعط بالحرارة The mathematical relationship: In equation form:

22. G AY -L USSAC ’ S L AW 5 | 22 Interestingly, a plot of pressure versus temperature can also be extrapolated to estimate absolute zero, and values from Charles’ Law and Gay- Lussac’s Law give equivalent estimates. رسم العلاقة بين الضغط والحرارة من الممكن تقدير نقطة الصفر المطلق فيها وقيم تشارلز بين الحرارة والحجم ومنها ايضا يمكن اشراك قانون غاي Problems using Gas-Lussac’s Law are done in the same fashion as Charles’ Law, and though we will not do any here, you will be accountable for solving them as well.

23. C OMBINED G AS L AW هذا ينقلنا الى قانون الغازات المشترك من هذه القوانين 5 | 23 Combined Gas Law The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature. وينص هذا القانون بان حجم عينة الغاز عند قيمة ضغط ثابتة تتناسب عكسيا مع الضغط وطرديا مع الحرارة المطلقة بالكالفن The mathematical relationship: In equation form:

24. E XAMPLE 4 5 | 24 Divers working from a North Sea drilling platform experience pressure of 5.0 × 10 1 atm at a depth of 5.0 × 10 2 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C? V i = 5.0 L P i = 5.0 × 10 1 atm T i = 4°C = 277 K V f = ? P f = 1.0 atm T f = 11°C = 284 K

25. E XAMPLE 4 (C ONT ) 5 | 25 V i = 5.0 L P i = 5.0 × 10 1 atm T i = 4°C = 277 K V f = ? P f = 1.0 atm T f = 11°C = 284. K = 2.6 x 10 2 L (2 significant figures)

26. E XAMPLE 5 5 | 26

27. E XAMPLE 5 (C ONT ) a.Decreasing the temperature at a constant pressure results in a decrease in volume. Subsequently increasing the volume at a constant temperature results in a decrease in pressure. b.Increasing the temperature at a constant pressure results in an increase in volume. Subsequently decreasing the volume at a constant temperature results in an increase in pressure. 5 | 27

28. A VOGADRO ’ S L AW قانون افوجادرو 5 | 28 Avogadro’s Law Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules. We could also say, الاحجام المتسوية من غازين مختلفين على نفس درجة الحرارة والضغط تحتوي على نفس العدد من الجزيئات، ونستطيع ان نقول ايضا :

29. S TANDARD T EMPERATURE AND P RESSURE ظروف الحرارة والضغط المثالية 5 | 29 Standard Temperature and Pressure (STP) The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure. هي الظروف المرجعية للغازات اتفق على ان تكون بحرارة صفر مئوي وواحد اتموسفير The molar volume, V m, of a gas at STP is 22.4 L/mol. الجم المولي للغاز في الظروف المثالية هو 22.4 لتر لكل مول The volume of the yellow box is 22.4 L. To its left is a basketball.

30. I DEAL G AS C ONSTANT 5 | 30 Ideal Gas Law قانون الغاز المثالي The ideal gas law is given by the equation PV=nRT The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T / P. ثابت الغاز المولي هو ثابت التناسب الذي يربط بين الحجم المولي للغاز مع الحرارة والضغط T/P

31. E XAMPLE 6 5 | 31 You put varying amounts of a gas into a given container at a given temperature. Use the ideal gas law to show that the amount (moles) of gas is proportional to the pressure at constant temperature and volume. وضعت كميات مختلفة من الغاز في وعاء بدرجة حرارة معينة. اتخدم قانون الغاز المثالي لاظهار بأن كمية الغاز بالمولات تتناب مع الضغط في درجات حرارة وحجم ثابتين

32. E XAMPLE 7 5 | 32 A 50.0-L cylinder of nitrogen, N 2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder? 50 لتر اسطوانة تحتوي على النيترجين ضغطها 17.1 اتموسفير على درجة حرارة 23 مئوية. ماهي كتلة النيتروجين؟ V = 50.0 L P = 17.1 atm T = 23°C = 296 K mass = 986 g (3 significant figures)

33. M OLAR M ASS AND D ENSITY الكتلة المولية والكثافة 5 | 33 Gas Density and Molar Mass Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter). باستخدام قانون الغاز المثالي من الممكن حساب المولات في لتر على درجة حرارة وضغط معينين، وبعدها نحول المولات الى غرامات To find molar mass, find the moles of gas, and then find the ratio of mass to moles. In equation form:

34. E XAMPLE 8 5 | 34 What is the density of methane gas (natural gas), CH 4, at 125°C and 3.50 atm? M m = 16.04 g/mol P = 3.50 atm T = 125°C = 398 K

35. E XAMPLE 9 5 | 35 A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? ( Note: The empirical formula of octane is C 4 H 9.) What is the molecular formula of octane?

36. E XAMPLE 9 (C ONT ) 5 | 36 P = 634 mmHg = 0.8342 atm d = 1.57 g/0.5000 L = 3.140 g/L T = 95.0°C = 368.2 K

37. E XAMPLE 9 (C ONT ) 5 | 37 Molecular formula: C 8 H 18 Molar mass = 114 g/mol Empirical formula: C 4 H 9 Empirical formula molar mass = 57 g/mol

38. E XAMPLE 10 5 | 38

39. E XAMPLE 10 (C ONT ) Assume the flasks are closed. a.All flasks contain the same number of atoms. b.The gas with the highest molar mass, Xe, has the greatest density. c.The flask at the highest temperature (the one containing He) has the highest pressure. d.The number of atoms is unchanged. 5 | 39

40. S TOICHIOMETRY AND THE G AS L AWS 5 | 40 Stoichiometry and Gas Volumes Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.

41. E XAMPLE 11 5 | 41 When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO 3 was required to neutralize the spill. What volume of CO 2 was released by the neutralization at 735 mmHg and 20.°C?

42. E XAMPLE 11 (C ONT ) 5 | 42 First, write the balanced chemical equation: CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + H 2 O(l) + CO 2 (g) Moles of CO 2 produced = 11.99 mol Second, calculate the moles of CO 2 produced: Molar mass of CaCO 3 = 100.09 g/mol

43. E XAMPLE 11 (C ONT ) 5 | 43 n = 12.0 mol P = 735 mmHg = 0.967 atm T = 20°C = 293 K = 3.0 × 10 2 L (2 significant figures)

44. D ALTON ’ S L AW 5 | 44 Gas Mixtures Dalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.

45. D ALTON ’ S L AW 5 | 45 Originally (left), flask A contains He at 152 mmHg and flask B contains O 2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg

46. D ALTON ’ S L AW Partial Pressure The pressure exerted by a particular gas in a mixture Dalton’s Law of Partial Pressures The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture: P = P A + P B + P C +... 5 | 46

47. E XAMPLE 12 5 | 47 A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N 2, 0.0194 g O 2, 0.00640 g CO 2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?

48. E XAMPLE 12 (C ONT ) 5 | 48

49. E XAMPLE 12 (C ONT ) 5 | 49 P = 1.00 atm

50. E XAMPLE 13 5 | 50 The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

51. E XAMPLE 13 (C ONT ) 5 | 51 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg

52. E XAMPLE 13 (C ONT ) 5 | 52 Mole fraction of N 2 Mole fraction of H 2 OMole fraction of CO 2 Mole fraction of O 2 Mole fraction N 2 = 0.7500 Mole fraction O 2 = 0.1355 Mole fraction CO 2 = 0.0526 Mole fraction O 2 = 0.0618

53. E XAMPLE 14 5 | 53 a.Nothing happens to the pressure of H 2. b.The pressures are equal because the moles are equal. c.The total pressure is the sum of the pressures of the two gases. Because the pressures are equal, the total pressure is double the individual pressures.

54. D ALTON ’ S L AW AND W ATER V APOR 5 | 54 Collecting Gas Over Water Gases are often collected over water. The result is a mixture of the gas and water vapor. The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. The partial pressure of water depends only on temperature and is known (Table 5.6). The pressure of the gas can then be found using Dalton’s law of partial pressures

55. C OLLECTING G AS O VER W ATER 5 | 55 The reaction of Zn( s ) with HCl( aq ) produces hydrogen gas according to the following reaction: Zn( s ) + 2HCl( aq )  ZnCl 2 ( aq ) + H 2 ( g ) The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.

56. A PPARATUS FOR T RAPPING G AS 5 | 56

57. W ATER V APOR 5 | 57

58. E XAMPLE 15 5 | 58 You prepare nitrogen gas by heating ammonium nitrite: NH 4 NO 2 ( s )  N 2 ( g ) + 2H 2 O( l ) If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH 4 NO 2 ? Molar mass NH 4 NO 2 = 64.05 g/mol P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K

59. 5 | 59 P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K Molar mass NH 4 NO 2 = 64.04 g/mol = 0.886 9 mol CO 2 gas Example 15 (Cont)

60. 5 | 60 P = 727 mmHg P vapor = 21.1 mmHg P gas = 706 mmHg T = 23°C = 296 K n = 0.8869 mol = 2.32 L of CO 2 (3 significant figures) Example 15 (Cont)