Gases Composition of air: 78% Nitrogen, 21% of Oxygen, 1% others Diatomic Gases: H 2, N 2, O 2, F 2, Cl 2 Monoatomic Gases: All the Noble gases O 3 (ozone) is an allotrope of Oxygen Ionic compounds do not exist as gases at 25 o C temperature and 1 atm pressure. Boiling point of NaCl is above 1000 o C

Elemental Gases and Gaseous Compounds ElementsCompounds H 2 (Molecular hydrogen) N 2 (Molecular Nitrogen O 2 (Molecular Oxygen) O 3 (Ozone) F 2 (Molecular Fluorine) Cl 2 (Molecular Chlorine He (Helium) Ne (Neon) Ar (Argon) Kr (Krypton) Xe (Xenon) Ra (Radon) HF (Hydrogen fluoride) HCl (Hydrogen chloride CO (Carbon monoxide CO 2 (Carbon dioxide) CH 4 (Methane) NH 3 (Ammonia) NO (Nitric oxide) NO 2 (Nitrogen dioxide) N 2 O (Nitrous oxide) SO 2 (Sulfur dioxide) H 2 S (Hydrogen sulfide) HCN (Hydrogen cyanide)

Behavior of Gases: Most gases are colorless. F 2, Cl 2 and NO 2 have color. NO 2 can be visible in the polluted air (dark brown) Nobles gases are inert, they do not react with any other substances. O 2 is Essential. Ozone is hazardous Compounds: H 2 S and HCN are poisonous gases. CO, NO 2, SO 2 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 SI units of Velocity: m/s 2 Elapsed time Acceleration = Change of velocity units : m/s 2 or cm/s 2 Elapsed time Newton’s second law: Force = mass x Acceleration force 1N = 1 kg m/s 2 Pressure = Force/area Si units (pascal) pa 1pa = 1N/m 2

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 o C at sea level. 1 atm = 760 mmHg or torr 1 atm = Pa or x 10 5 Pa 1000 Pa = 1 kPa 1 atm = x 10 2 kPa

Manometer

Pressure-Volume Relationship-Boyle’s Law P (mmHg) V (volume) PV 1.09 x x x x x x x 10 3 The data shows the inverse relationship between pressure and volume P  1 Mathematical expression P = K 1 x 1 V V or PV = K 1

Boyle’s Gas Law PV = K 1 : Pressure and Volume of gas at constant temperature and amount of gas is a constant The variation in pressure or volume can be expressed P 1 V 1 = K 1 and P 2 V 2 = K 1 Therefore P 1 V 1 = P 2 V 2

The Temperature – Volume relationship At constant pressure, the volume of gas expands with the increase of temperature Charles’s law: Volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas V  T V = K 2 T or V = K 2 T Absolute zero 0 K (kelvin scale) C (~ C) Variable volume and temperature equation: V 1 = K2 V 2 = K2 Therefore V 1 = V 2 T 1 T 2 T 1 T 2

Charles’s Law of Temperature-pressure Relationship P  T P = K 3 T P = K3 Variable temperature/ pressure relation T P 1 P 2 T 1 T 2

Relationship of Volume, Pressure, Temperature P 1 V 1 = P 2 V 2 Boyle’s Law V 1 V 2 Charles’s Law T 1 T 1 P 1 P 2 Charles’s Law T 1 T 2

Avogadro’s Law: Volume-Amount Relationship At the same temperature and pressure, equal 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  n V = k 4 n n = number of moles K 4 = Proportional constant

Relation of reactants and products in a gaseous reaction 3H 2 (g) + N 2 (g) 2NH 3 (g) 3 moles 1 mole 2 moles At the same temperature and pressures volume is proportional to number of moles Therefore: 3H 2 (g) + N 2 (g) 2NH 3 (g) 3 volumes 1 volume 2 volumes When 2 gases react, their volumes have a simple ratio: The ratio of molecular hydrogen and molecular nitrogen is 3 : 1 If the product is gas, the volume of product and reactants are in simple ratio: The ratio of ammonia to the sum of both H 2 and N 2 is 2 : 4 or 1 : 2

Relations of gas laws Boyle’s Law: At constant Temperature Volume decreases Volume Volume increases Higher Pressure Lower Pressure P = K 1 1 or P = (nRT) 1 nRT = Constant V V Charles’s Law: At constant pressure Volume Decreases Volume Volume increase s Lower Temperature Higher Temperature V = K 2 T or V = (nR) T nR/P constant P

Relations of Gas Laws Avogadro’s Law: Relation between amount and volume at constant temperature and pressure Volume decreases Volume Volume increases More molecules Less molecules V = K 4 n V = (RT) n RT/P = Constant P

Ideal Gas Equation Boyle’s Law: V  1/P Charles’s Law: V  T Avogadro: V  n V  nT or V = R nT or PV = nRT P P (R = Proportionality 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 o C (273.5K) and 1 atm pressure, many gases behave like ideal gas Standard Temperature and pressure (STP) : 0 o C and 1 atm. Volume of 1 mole of an ideal gas is L (22.41L) Therefore the at STP R = 1 atm x 22.4L 1 mol x 273.5K = L.atm/K.mol or 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 o C

Modified ideal gas equation R = PV R = P 1 V 1 R = P 2 V 2 nT n 1 T 1 n 2 T 2 Modified equation = P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 n 1 = n 2 as the amount of gas does not change Therefore P 1 V 1 = P 2 V 2 T 1 T 2 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 grams M = Molar mass Therefore m = P d = m M V RT V Therefore d = P or d = P M M RT RT Calculate density of Carbon dioxide in g/L at atm and 55 o C

Molar mass of a Gas d = P M or M = dRT RT P 1) The density of a gas compound containing Chlorine and oxygen has 7.71g/L density at 36 o C and 2.88 atm. What is the molar mass of the gas.? 2) The density of a gas is 3.38g/L at 40 o C and 1.97 atm. What is the molar mass.? Derive from both M = dRT and n = PV P RT

Stoichiometry Calculate the volume of Oxygen gas (in liters) required for the combustion of 7.64 L of Acetylene at same temperature and pressure 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O Avogadro’s Law: V  n at constant temperature and pressure As per equation 2 moles of C 2 H 2 = 5 moles of O 2 Therefore 2L of C 2 H 2 = 5L of O 2 For 7.64 L of C 2 H 2 = 7.64 L C 2 H 2 x 5 L O2 = Liters of O 2 2L C2H2 = 19.1 L of Oxygen

Dalton’s Law of Partial Pressures The total pressure of mixture of gases is the sum of the pressure of each gas in the mixture P T = P 1 + P 2 + P 3 + …. In a mixture of A and B gases the pressure of each gas can be expressed as P A = n A RT and P B = n B RT P T = P A + P B V V Therefore P T = n A RT + n B RT = RT (n A + n B ) V V V If n = n A + n B P T = nRT V

Mole Fraction P A = n A RT/V P T = (n A + n B )RT/V P A = n A RT/V P T (n A + n B )RT/V Therefore P A = n A P T (n A + n B = X A (mole fraction of A) The ratio of the number of moles of one to the number of moles of all gases in the mixture X i = n i n T X i = n i n T

Mole Fraction-Partial Pressure X i = n i /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 X A + X B = n A n B n A + n B + n A + n B = 1 P A /P T = X A or P A = X A P T General equation for individual gas in a mixture: P i = X i P T

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 = n i /n T 2) Using mole fraction and total pressure, calculate the partial pressure of each gas (P i = X i P T ) Practice 2: A sample of natural gas contains 8.24 moles of methane, mole of ethane, mole of propane. The total pressure is 1.37 atm. Calculate partial pressure of the gases

Kinetic Molecular Theory of Gases Molecular 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: Joule (J) or kilo joules (kJ) 1 J = 1kg m 2 /s 2 = 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 5.7 KE = ½ mu 2 m = mass and u = speed, bar means average value

Average Kinetic Energy KE = 1 mu 2 2 u 2 = u u …………+ u n 2 (u = speed, N = # of molecules) N KE  T 1/2mu 2  T KE = 1/2mu 2 = 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 Boyle’s Law: Molecular collisions  number density (number density: number of molecules per unit volume) Number density  1/V Pressure  molecular collisions Therefore P  1/V

Kinetic Molecular theory: Application to the Gas laws Charles’s Law: KE  T Temperature results in the increase of Kinetic energy. The molecular collisions on wall increase which in turn increases pressure KE  P P  T

Kinetic Molecular theory: Application to the Gas laws Avogadro’s Law: P  d and P  T or P  dT d  n/V Therefore P  n/V x T For two gases P 1  n 1 T 1 = C n 1 T 1 V 1 V 1 P 2  n 2 T 2 = C n 2 T 2 V2 V 2 If P, V, T are same n 1 and n 2 are equal

Kinetic Molecular theory: Application to the Gas laws Dalton’s Law of Partial pressure: P T = 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 for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 u rms = 3RT M 

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl r1r1 r2r2 M2M2 M1M1  =

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. 5.7 r1r1 r2r2 t2t2 t1t1 M2M2 M1M1  == Nickel forms a gaseous compound of the formula Ni(CO) x What is the value of x given that under the same conditions methane (CH 4 ) effuses 3.3 times faster than the compound? r 1 = 3.3 x r 2 M 1 = 16 g/mol M 2 = r1r1 r2r2 ( ) 2 x M 1 = (3.3) 2 x 16 = x 28 = x = 4.1 ~ 4

Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = Repulsive Forces Attractive Forces

Effect of intermolecular forces on the pressure exerted by a gas. 5.8

Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume

Key Equations P 1 V 1 = P 2 V 2 V 1 /T 1 = V 2 /T 2 P 1 /T 1 = P 2 /T 2 V = K 4 n PV = nRT P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 P1V1/T1 = P2V2/T2 d = P M /RT X i = n i /n T KE = 1/2mu 2 = CT Boyle’s Law: volume - pressure Charles’s Law: Volume – Temperature Charles’s Law: Pressure – Temperature Avogadro’s Law: Volume - # moles at Constant P and V Ideal Gas equation Relation of Pressure-Volume-Temperature-amount of gas Pressure-Volume-Temperature at constant n Density-Molar mass- Pressure-temperature Mole Fraction Average Kinetic energy-Absolute temperature

Key Equations U rms = 3RT/ M r1/r2 = M 1/ M 2 (P + an 2 /V 2 )[V-nb] = nRT Root-mean-square speed and Temperature Graham’s Law of diffusion and effusion Van der Waals equation-Deviation from ideal gas