Gases Composition of air: 78% Nitrogen, 21% of Oxygen, 1% others Diatomic Gases: H2, N2 , O2, F2, Cl2 Monoatomic Gases: All the Noble gases O3 (ozone) is an allotrope of Oxygen Ionic compounds do not exist as gases at 25 oC temperature and 1 atm pressure. Boiling point of NaCl is above 1000 oC Elemental Gases and Gaseous Compounds Elements H2 (Molecular hydrogen) N2 (Molecular Nitrogen O2 (Molecular Oxygen) O3 (Ozone) F2 (Molecular Fluorine) Cl2 (Molecular Chlorine He (Helium)
Ne (Neon) Ar (Argon) Kr (Krypton) Xe (Xenon) Ra (Radon) Compounds HF (Hydrogen fluoride) HCl (Hydrogen chloride CO (Carbon monoxide CO2 (Carbon dioxide) CH4 (Methane) NH3 (Ammonia) NO (Nitric oxide) NO2 (Nitrogen dioxide) N2O (Nitrous oxide) SO2 (Sulfur dioxide) H2S (Hydrogen sulfide)
HCN (Hydrogen cyanide) Behavior of Gases: Most gases are colorless. F2, Cl2 and NO2 have color. NO2 can be visible in the polluted air (dark brown) Nobles gases are inert, they do not react with any other substances. O2 is Essential. Ozone is hazardous Compounds: H2S and HCN are poisonous gases. CO, NO2 , SO2 are toxic Physical Properties of gases 1) Gases have volume and shape of the containers 2) Gases are the most compressible of the states of the matter 3) Gases will mix evenly and completely when confined to the same container 4) Gases have much lower densities than liquids and solids Pressure of Gases Gases exert pressure Pressure is related to velocity and time
Velocity: Distance moved Elapsed time Acceleration = Change of velocity Elapsed time SI units of Velocity: m/s 2 units : m/s2 or cm/s2 Newtons second law: Force = mass x Acceleration force 1N = 1 kg m/s2 Pressure = Force/area Si units (pascal) pa 1pa = 1N/m2 Atmospheric pressure Barometer: It is an instrument to measure the atmospheric pressure Standard atmospheric pressure (1 atm) is equal to the pressure that supports 760 mm (76 cm) column of mercury at 0 oC at sea level. 1 atm = 760 mmHg or torr 1 atm = 101.325 Pa or 1.01325 x 10 5 Pa 1000 Pa = 1 kPa
1 atm = 1.01325 x 10 2 kPa Manometer Pressure-Volume Relationship-Boyles Law P (mmHg) 724 869 951 998 1230 1891 2250 V (volume) 1.50 1.33 1.22 1.18 0.94 0.61 0.58 PV 1.09 x 10 1.16 x 10 1.16 x 10 1.18 x 10 1.2 x 10 1.2 x 10 1.3 x 10 3 3 3 3 3
3 3 The data shows the inverse relationship between pressure and volume Pa 1 V Mathematical expression P = K1 x 1 V or PV = K1 Boyles Gas Law PV = K1: Pressure and Volume of gas at constant temperature and amount of gas is a constant The variation in pressure or volume can be expressed P1V1 = K1 and P2V2 = K1
Therefore P1V1 = P2V2 The Temperature Volume relationship At constant pressure, the volume of gas expands with the increase of temperature Charless law: Volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas VaT V = K2T or V = K2 T Absolute zero 0 K (kelvin scale) -273.15 0C (~ 273 0C) Variable volume and temperature equation: V1 = K2 V2 = K2 Therefore T1
T2 V 1 = V2 T1 T2 Charless Law of Temperature-pressure Relationship PaT P = K3 T P = K3 T Variable temperature/ pressure relation P1 T1 P2
T2 Relationship of Volume, Pressure, Temperature P1V1 = P2V2 Boyles Law V1 T1 V2 T1 Charless Law P1 T1
P2 T2 Charless Law Avogadros Law: Volume-Amount Relationship At the same temperature and pre ssure, e qual volumes of different gases contain the same number of atoms (monoatomic gases) or molecules (diatomic or polyatomic gases) Volume is proportional to number of moles of molecules. V an V = k4n n = number of moles K4 = Proportional constant Re lation of reactants and products in a gase ous re action 3H2 (g) + N 2(g) 3 mol es 1 mole
2NH3 (g) 2 moles At the same temperature and pres sures vol ume is proportional to number of moles Therefore: 3H2 (g) + N 2(g) 3 volumes 1 volume 2NH3 (g) 2 volumes When 2 gases react, their volumes have asimpleratio:
The ratio of molecular hydrogen and molecular nitrogen is 3 :1 If theproduct is gas, thevolume of product and reactants are in simpleratio: The ratio of ammonia to the sum of both H2 and N2 is 2 :4 or 1 :2 Relationsof gaslaws Boyles Law:At constant Temperature Vol um e decreas es Volume Vol ume increas es Higher P ress ure P = K 11 Lower P ress ure
or P = (nRT) 1 V Charless Law: At constant pressure Volume De cre ase s Lowe r Te mpe rature nRT = Cons tant V Volume V = K2 T or V = (nR) T P Volum e incre as e s Hi ghe r Te mpe rature nR/ P cons tant
Relations of Gas Laws Avogadros Law: Relation between amount and volume at constant temperature and pressure Volume decreases Volume More molecules V = K 4n Volume increases Less molecules V = (RT) n P RT/P = Constant
Ideal Gas Equation Boyles Law: V a 1/P Charless Law: V a T Avogadro: V a n V a nT P or V = R nT or P PV = nRT (R = Propor tionality constant) An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for the ideal gas equation Gas Constant, R PV = nRT or R = PV
nT At 0 oC (273.5 K) and 1 atm pressure, many gases behave like ideal gas Standard Temperature and pressure (STP) : 0 oC and 1 atm. Volume of 1 mole of an ideal gas is 22.414 L (22.41L) Therefore the at STP R = 1 atm x 22.4L 1 mol x 273.5K = 0.082057 L.atm/K.mol or 0.0821 L.atm/K.mol Practice: What is the pressure (atm) of SF6 gas in a container having the volume 5.43 L at 69.5 oC Modified ideal gas equation R = PV nT R = P1V1 n1T1
R = P 2V2 n2T2 Modified equation = P1V1 = P2V2 n1T1 n2T2 n1 = n2 as the amount of gas does not change Therefore P 1V1 = P 2V2 T1 T2 Practice: At constant temperature, at 1 atm pressure the volume of helium is 0.55 L. What is the volume at 0.40 atm.? Density relation with Ideal Gas equation PV = nRT or n
= P V RT number of moles, n = m M m = mass in gram s M = Molar m ass Therefore Therefore m = MV d = M
P P RT or RT d = m V d = PM RT Calcul ate de ns ity of Carbon dioxide in g /L at 0 .9 9 0 atm and 5 5 oC Molar mass of a Gas d = PM RT
or M = dRT P 1) The density of a gas compound containing Chlorine and oxygen has 7.71g/L density at 36 oC and 2.88 atm. What is the molar mass of the gas.? 2) The density of a gas is 3.38g/L at 40oC and 1.97 atm. What is the molar mass.? Derive from both M = dRT P and n = PV RT Stoichiometry Calculate the volume of Oxygen gas (in liters) required for the combustion of 7.64 L of Acetylene at same temperature and pre ssure 2C2H 2 + 5O2 4CO2 + 2H2O
Avogadros Law: V a n at constant temperature and pressure As per e quation 2 moles of C 2H 2 = 5 moles of O2 Therefore 2L of C2H 2 = 5 L of O2 For 7.64 L of C2H2 = 7.64 L C2H2 x 5 L O2 2L C2H2 = 19.1 L of Oxyge n = Liters of O2 Stoichiometry Practice: Calculate the volume of N 2 Ge nerate d at 8 0 oC and 82 3 mmHg from the decomposition of 6 0g of NaN3 2NaN 3 (s) 2Na(s) + 3N 2(g)
Calculate the volume CO2 produced at 37 oC and 1.00 atm. when 5.60g of Glucose is decomposed C 6H12(s) + 6O2 6CO2 + 6H2O Daltons Law of Partial Pressures The total pressure of mixture of gases is the sum of the pressure of each gas in the mixture PT = P1 + P2 + P3 + . In a mixture of A and B gases the pressure of each gas can be expressed as PA = nART V Therefore PT If n = nA + nB and =
PB = nART V + PT = nRT nB RT V nBRT V V
= PT = PA + PB RT (nA + nB ) V Mole Fraction PA = nART/V Therefore PT = (nA + nB)RT/V PA =
nA PT (nA + nB = XA PA PT = nART/V (nA + nB)RT/V (mole fraction of A) The ratio of the number of moles of one to the number of moles of all gases in the mixture
Xi = ni nT Mole Fraction-Partial Pre ssure Xi = ni/n T Therefore the mole fraction is always less than 1 The relation of Partial pressure of A and B can be expressed in terms of mole fraction as XA + XB = nA nB
n A + nB PA /PT = XA or + nA + nB = 1 PA = XAPT General e quation for individual gas in a mixture: Pi = XiPT
Mole Fraction-Partial Pressure Practice 1: A mixture of gases contains 4.46 moles of Neon, 0.74 moles of Argon and 2.15 moles of Xenon. When the total pressure of the mixture is 2 atm., what is the partial pressure of each individual gases 1) calculate mole fraction of each gases with X i = ni/nT 2) Using mole fraction and total pressure, calculate the partial pressure of each gas (Pi = XiPT) Practice 2: A sample of natural gas contains 8.24 moles of methane, 0.421 mole of ethane, 0.116 mole of propane. The total pressure is 1.37 atm. Calculate partial pressure of the gases Kinetic Molecular Theory of Gases M olecular movement: Is a form of energy Energy: Capacity to do work or work done Work = Force x Distance or Energy = Force x Distance Units of Energy: J oule (J) or kilo joules (kJ ) 1 J = 1kg m 2/s2 = 1 N m 1 kJ = 1000 J Kinetic Molecular Theory of Gases 1. A gas is composed of molecules that are separated from each
other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic (transfer of energy occurs during collision). 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = mu2 m = mass and u = speed, bar means average value 5.7
Average Kinetic Ene rgy KE = 1 mu2 2 u2 = u12 + u 22 + + un2 N (u = speed, N = # of molecules) KE a T 1/2mu2 a T KE = 1 /2mu2 = CT (C = proportionality constant, T is absolute temperature) Kinetic Molecular theory: Application to the Gas laws Compressibility of Gases: In the gas phase, the molecules are separated by large distance. For this reason, gases can be compressed easily to occupy less volume Boyles Law: Molecular collisions a number density
(number density: number of molecules per unit volume) Numbe r density a 1/V Pressure a molecular collisions Therefore P a 1/V Kinetic Molecular theory: Application to the Gas laws Charless Law: KE a T Temperature results in the increase of Kinetic energy. The molecular collis ions on wall increase which in turn increases pressure KE a P PaT Kinetic Molecular theory: Application to the Gas laws Avogadros Law: P a d and P a T or P a dT d a n/V Therefore P a n/V x T
For two gases P1 a n1 T1 = V1 C n 1T1 V1 P2 a n2T2 = C n2 T2 V2 V2 If P, V, T are same n1 and n2 are e qual Kinetic Molecular theory: Application to the Gas laws
Daltons Law of Partial pressure: PT = P1 + P2 + .. If molecules do not attract or repel one another, then the pressure of one molecule is unaffected by another molecule. Consequently the total pressure is the sum of individual gas pressures Apparatus for studying molecular speed distribution 5.7 The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures urms =
3RT M 5.7 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r1 r2 = M2 M1
NH4Cl NH3 17 g/mol HCl 36 g/mol 5.7 Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. r1 r2 = t2
t1 = M2 M1 Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound? r1 2 r1 = 3.3 x r2 x M1 = (3.3)2 x 16 = 174.2 M2 = r2
x = 4.1 ~ 4 58.7 + x 28 = 174.2 M1 = 16 g/mol ( ) 5.7 Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT PV = 1.0 n= RT Repulsive Forces
Attractive Forces 5.8 Effect of intermolecular forces on the pressure exerted by a gas. 5.8 Van der Waals equation nonideal gas 2 an ( P + V2 ) (V nb) = nRT } }
P1/T1 = P2/T2 Charless Law: Pressure Temperature V = K4n Avogadros Law: Volume - # moles at Constant P and V PV = nRT P1V1/n1T1 = P2V2/n2T2 P1V1/T1 = P2V2/T2 Ideal Gas equation Relation of Pressure-Volume-Temperature-amount of gas Pressure-Volume-Temperature at constant n Density-Molar mass- Pressure-temperature d = PM/RT
Xi = ni/nT KE = 1/2mu2 = CT Mole Fraction Average Kinetic energy-Absolute temperature Key Equations Urms = 3RT/M Root-mean-square speed and Temperature r1/r2 = M1/M2 Grahams Law of diffusion and effusion (P + an2/V2)[V-nb] = nRT
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