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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 6 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

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CHAPTER 6 THE GASEOUS STATE

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CHARACTERISTICS OF GASES - Gases are composed of nonmetals - Gases form homogeneous mixtures of each other Air mixture of 78% N 2 21% O 2 0.9% Ar other substances (CO 2, H 2, Ne, Kr, He)

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Gases at Ordinary Temperature and Pressure Noble gases (monatomic gases) He, Ne, Ar, Kr, Xe Diatomic Gases H 2, N 2, O 2, F 2, Cl 2 Other Common Gases propane, ammonia, carbon dioxide, hydrogen sulfide, methane, carbon monoxide, sulfur dioxide CHARACTERISTICS OF GASES

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PRESSURE - Gases exert pressure on any surface they come in contact with - Pressure is force applied per unit area A = area (m 2 ) F = force (newton, N = kg-m/s 2 ) P = pressure (N/m 2 = pascal, Pa)

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F = m x a m = mass (kg) a = acceleration (m/s 2 ) PRESSURE

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THE GAS LAWS Four Variables Define the Physical State of any Gas Amount (mole) Temperature (K) Volume (L) Pressure (bar, Pa, mm Hg, torr, atm, psi) 1 bar = 10 5 Pa 1 atm = 760 mmHg = 760 torr = x 10 5 Pa = kPa = 14.7 psi

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mm Hg: millimeters mercury atm: atmosphere (atmospheric pressure = 1atm) Pa: Pascal psi: pound per square inch (Ib/in 2 ) Pressure Instruments barometers, manometers, gauges 760, 700, 650 mm Hg - Considered to have 3 significant figures THE GAS LAWS

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BOYLE’S LAW - The volume of a fixed amount of a gas is inversely proportional to the pressure applied to the gas if the temperature is kept constant PV = constant P 1 V 1 = P 2 V 2 - P 1 and V 1 are the pressure and volume of a gas at an initial set of conditions - P 2 and V 2 are the pressure and volume of the same gas under a new set of conditions - The temperature and amount of gas remain constant

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A sample of N 2 gas occupies a volume of 3.0 L at 6.0 atm pressure. What is the new pressure if the gas is allowed to expand to 4.8 L at constant temperature? P 1 = 6.0 atmV 1 = 3.0 L P 2 = ?V 2 = 4.8 L P 1 V 1 = P 2 V 2 (6.0 atm)(3.0 L) = (P 2 )(4.8 L) P 2 = 3.8 atm BOYLE’S LAW

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CHARLES’S LAW - The volume of a fixed amount of gas is directly proportional to its absolute temperature if the pressure is kept constant - V 1 and T 1 are the volume and absolute temperature of a gas at an initial set of conditions - V 2 and T 2 are the volume and absolute temperature of the same gas under a new set of conditions - The pressure and amount of gas remain constant

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A sample of Ar gas occupies a volume of 1.2 L at 125 o C and a pressure of 1.0 atm. What is the new temperature, in Celsius, if the volume of the gas is decreased to 1.0 L at the same pressure? V 1 = 1.2 LT 1 = 125 o C = 398 K V 2 = 1.0 LT 2 = ? T 2 = 332 K = 59 o C CHARLES’S LAW

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AVOGADRO’S LAW - The volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas n = number of moles of a gas

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AVOGADRO’S LAW Avogadro’s Hypothesis - Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules At Standard Temperature and Pressure (STP) 1 mol of any gas (= x molecules) occupies a volume of L Conditions of STP Temperature = 0 o C = 273 K = 32 o F Pressure = 1.00 atm ( kPa or 100 kPa)

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THE IDEAL GAS LAW PV = nRT Considering all three gas laws V α 1/PV α TV α n

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R is the ideal gas constant = L-atm/mol.K = J/mol-K = m 3 -Pa/mol-K = cal/mol-K = L-torr/mol-K THE IDEAL GAS LAW

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RELATING THE GAS LAWS PV = nRT If n is constant

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A 1.00-L container is filled with mole of CO gas at 35.0 o C. Calculate the pressure, in atmospheres, exerted by the gas in the container PV = nRT P = ?V = 1.00 Ln = mol T = 35.0 o C = 308 KR = atm.L/mol.K (P)(1.00 L) = (0.500 mol)( atm.L/mol.K)(308 K) P = 12.6 atm RELATING THE GAS LAWS

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A balloon filled with helium initially has a volume of 1.00 x 10 6 L at 25 o C and a pressure of 752 mm Hg. Determine the volume of the balloon after a certain time when it encounters a temperature of -33 o C and a pressure of 75.0 mm Hg P 1 = 752 mm HgV 1 = 1.00 x 10 6 LT 1 = 25 o C = 298 K P 2 = 75.0 mm HgV 2 = ? T 2 = -33 o C = 240 K RELATING THE GAS LAWS

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GAS DENSITIES AND MOLAR MASS - Density (d) = mass/volume (m/V) - Rearrange the ideal gas equation and multiply both sides by molar mass (M) to give M x n = mass (m) and m/V = d

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Calculate the density of carbon dioxide gas at 0.45 atm and 252 K Density (d) = 0.96 g/L GAS DENSITIES AND MOLAR MASS

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Calculate the average molar mass of dry air if its density is 1.17 g/L at 21 o C and torr Molar mass = 29.0 g/mol GAS DENSITIES AND MOLAR MASS

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REACTION STOICHIOMETRY Consider the following reaction 4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) Calculate the mass of aluminum that would react completely with 2.00 L of pure oxygen gas at STP

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1 mol O 2 at STP = 22.4 L REACTION STOICHIOMETRY

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- At the same conditions volumes of gases combine in the same proportions as the coefficients of the chemical equation Example 3H 2 (g) + N 2 (g) → 2NH 3 (g) This implies that - 3 mol H 2 reacts with 1 mol N 2 to produce 2 mol NH L H 2 reacts with 1 L N 2 to produce 2 L NH 3 REACTION STOICHIOMETRY

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DALTON’S LAW OF PARTIAL PRESSURES - The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases present - The partial pressure is the pressure that a gas in a mixture of gases would exert if it were present alone under the same conditions

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- P total is the total pressure of a gaseous mixture - P 1, P 2, P 3,…. are the partial pressures of the individual gases - n total is the total number of moles of a gaseous mixture - n 1, n 2, n 3,….are the number of moles of the individual gases DALTON’S LAW OF PARTIAL PRESSURES

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The total pressure by a mixture of He, Ne, and Ar gases is 3.50 atm. Find the partial pressure of Ar if the partial pressures of He and Ne are 0.50 atm and 0.75 atm, respectively P total = 3.50 atm P 1 = 0.50 atm P 2 = 0.75 atm P 3 = ? P total = P 1 + P 2 + P 3 P 3 = P total - (P 1 + P 2 ) = 3.50 atm - (0.50 atm atm) = 2.25 atm DALTON’S LAW OF PARTIAL PRESSURES

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For gas 1 in a mixture of gases with n 1 moles Mole fraction (x 1 ) = n 1 /n total O 2 is 21% of air Mole fraction of O 2 (x 1 ) = 0.21 Total atmospheric pressure = 1 atm Partial pressure of O 2 = P 1 = (0.21)(1 atm) = 0.21 atm = (0.21)(760 mm Hg) = 160 mm Hg DALTON’S LAW OF PARTIAL PRESSURES

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EXPERIMENT TO COLLECT GAS OVER WATER - Magnesium reacts with HCl to produce hydrogen gas according to the following equation Mg(s) + 2HCl(aq) → MgCl 2 (aq) + H 2 (g) - The H 2 gas is collected in a gas collection tube initially filled with water and inverted in a water pan - The volume of H 2 gas is measured by raising or lowering the tube such that the water levels in the tube and pan are the same

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For this condition The pressure inside the tube is equal to the atmospheric pressure outside P total = P gas + P water P total = total atmospheric pressure P gas = pressure of gas in tube P water = vapor pressure of water at experimental temperature EXPERIMENT TO COLLECT GAS OVER WATER

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- Gases consist of large numbers of small molecules that are in continuous random motion - Attractive and repulsive forces between gas molecules are negligible - The combined volume of all the molecules of a gas is negligible relative to the total volume in which the gas is contained KINETIC MOLECULAR THEORY OF GASES

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- Collisions between molecules are perfectly elastic (the average kinetic energy of the colliding molecules does not change at constant temperature) - The average kinetic energy of the molecules is proportional to the absolute temperature (molecules of all gases have the same average kinetic energy at a given temperature) KINETIC MOLECULAR THEORY OF GASES

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AVERAGE KINETIC ENERGY ε = average kinetic energy m = mass of an individual molecule u = root-mean-square (rms) velocity at a given temperature

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AVERAGE KINETIC ENERGY Root-Mean-Square Velocity - Square-root of the average of the squares - The lower the molar mass of a gas the higher the u

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Diffusion - The spread of one substance throughout space or throughout a another substance Effusion - Escape of gas molecules through a tiny hole into an evacuated space GRAHAMS LAW OF EFFUSION

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- r 1 and r 2 are the effusion rates of gases 1 and 2 respectively - M 1 and M 2 are the molar masses of gases 1 and 2 respectively - The rate of effusion of a gas is inversely proportional to the square root of its molar mass

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- The rate of effusion of a gas is directly proportional to the rms velocity of the molecules GRAHAMS LAW OF EFFUSION - The time it takes for a gas to effuse is inversely proportional to the rate of effusion

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It took 4.5 minutes for 1.0 L helium to effuse through a porous barrier. How long will it take for 1.0 L Cl 2 gas to effuse under identical conditions? t 2 = 19 minutes GRAHAMS LAW OF EFFUSION

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REAL GASES Mean Free Path - The average distance traveled by a molecule between collisions - Real Gases deviate from ideal gas behavior at high pressures - Deviation is very small at low pressures (usually below 10 atm) - Deviation increases with decreasing temperature - Deviation is significant near the temperature at which a given gas changes to the liquid state

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- Molecules of real gases have finite volumes and attract one another - van der Waals equation REAL GASES

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a = constant (measure of how strongly gas molecules attract each other b = constant (measure of the small but finite volume occupied by gas molecules) n 2 a/V 2 accounts for the attractive forces and adjusts pressure upwards nb accounts for the finite volume occupied by molecules REAL GASES

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Calculate the pressure exerted by mol N 2 gas in a L container at 22.5 o C using both the ideal gas law and the van der Waals equation. Compare the results. PV = nRT P = nRT/V = ( mol)( L-atm/mol-K)(295.5 K)/(5.000 L) = atm REAL GASES

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