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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.

5.1 Substances that exist as gases.

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

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).

5.1

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

5.2 Pressure of gas.

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.

(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=1.01325x105Pa 1 atm = 1.01325X102 kPa 5.2

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

Worked Example 5.1

1 atm = 760 mmHg X = 688 mmHg X= 68/760 = 0.905 atm

1.01325 × 105 Pa = 760 mmHg x Pa = 732 mmHg x Pa = (732 × 1.01325 × 105) / 760 mmHg x Pa = 9.76 × 104 Pa = 97.6 kPa

Manometers Used to Measure Gas Pressures 5.2

5.3 The gas laws.

The Pressure –Volume Relationship: Boyle`s Law

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas (Boyles’s law) As P (h) increases V decreases 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 P a 1/V Constant temperature Constant amount of gas P x V = constant P1 x V1 = P2 x V2 5.3

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

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

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

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) + 273.15 5.3

Absolute zero:theoretically,the lowest the attainable temperature=-273 Absolute zero:theoretically,the lowest the attainable temperature=-273.150C(kelven temperture scale).

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 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K 5.3

The Volume-Amount Relationship: Avogadro`s Law

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

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 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO 5.3

Ideal Gas Equation

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

5.3

5.3

5.3

General gas law: Constant moles

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 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) 5.4

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 = 273.15 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 0.0821 x 273.15 K L•atm mol•K V = 30.6 L 5.4

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

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

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

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

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 27.0 0C. 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 0.0821 x 300.15 K L•atm mol•K M = M = 54.6 g/mol 5.4

Worked Example 5.3

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

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

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

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

d PM RT = d 0.999 × 44.01 0.0821 × 328 = d = 1.62 g/L

M dRT P = M 7.71 × 0.0821× (36+273) 2.88 = M = 67.9 g/mol

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 = 33.0/28.09 = 1.17 mol Si nF = m/M = 67.0/10.0 = 3.53 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

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 Molar mass / empirical molar mass = 169/85.09 = 1.986 ≈ 2 Si (28.09 ×1) + F (19×3) = 85.09 SiF3 = (SiF3)2 = Si2F6

Gas Stoichiometry

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) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = 0.187 mol CO2 0.187 mol x 0.0821 x 310.15 K L•atm mol•K 1.00 atm = nRT P V = = 4.76 L 5.5

2 moles C2H2 5 moles O2 2 L C2H2 5 L O2 7.64 L C2H2 x L O2 x = 7.64×5/2 x = 19.1 L

أولاً يتم حساب عدد مولات N2 عن طريق إيجاد عدد مولات NaN3 n mol of NaN3 60/65.02 = 0.923 moles 2 moles of NaN3 3 moles of N2 0.923 moles of NaN3 x moles of N2 x moles of N2 = 0.923 × 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 = 36.93 L

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 0.250-L vessel contains 1.25 g of a gas at 715 mmHg and 28.0 0C. 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

Dalton’s Law of Partial Pressures

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

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

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, 0.421 moles of C2H6, and 0.116 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 8.24 + 0.421 + 0.116 Xpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

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 + 0.74 mol + 2.15 mol) XNe = 0.607 ثانياً يتم حساب الضغط الجزئي بالتعويض في المعادلة PNe = XNePT PNe = 0.607 × 2 atm PNe = 1.21 atm

بنفس الطريقة يتم حساب الضغط الجزئي لغازي الأرجون والزينون XAr = 0.74 mol/ (4.46 mol + 0.74 mol + 2.15 mol) = 0.1 PAr = 0.1 × 2 atm = 0.2 atm XXe = 2.15 mol/ (4.46 mol + 0.74 mol + 2.15 mol) = 0.293 PXe = 0.293 × 2 atm = 0.586 atm ملاحظة يجب أن يكون مجموع الضغوط الجزئية للغازات يساوي الضغط الكلي PT= 1.21 + 0.20 + 0.586 = 1.996 ≈ 2 atm

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, 0.421 moles of C2H6, and 0.116 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 8.24 + 0.421 + 0.116 Xpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

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

5.6

أولاً يتم حساب ضغط غاز الأوكسجين من العلاقة PT = PO2 + PH2O PO2 = PT - PH2O PO2 = 762 mmHg – 22.4 mmHg PO2 = 739.6 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 = 0.164 g 0.0821 L.atm/K.mol × (24+273) K

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.