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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% N2 21% O2 0.9% Ar other substances (CO2, H2, Ne, Kr, He)

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**CHARACTERISTICS OF GASES**

Gases at Ordinary Temperature and Pressure Noble gases (monatomic gases) He, Ne, Ar, Kr, Xe Diatomic Gases H2, N2, O2, F2, Cl2 Other Common Gases propane, ammonia, carbon dioxide, hydrogen sulfide, methane, carbon monoxide, sulfur dioxide

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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 (m2) F = force (newton, N = kg-m/s2) P = pressure (N/m2 = pascal, Pa)

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PRESSURE F = m x a m = mass (kg) a = acceleration (m/s2)

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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 = 105 Pa 1 atm = 760 mmHg = 760 torr = x 105 Pa = kPa = 14.7 psi

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**THE GAS LAWS mm Hg: millimeters mercury**

atm: atmosphere (atmospheric pressure = 1atm) Pa: Pascal psi: pound per square inch (Ib/in2) Pressure Instruments barometers, manometers, gauges 760, 700, 650 mm Hg - Considered to have 3 significant figures

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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 P1V1 = P2V2 - P1 and V1 are the pressure and volume of a gas at an initial set of conditions - P2 and V2 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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**BOYLE’S LAW A sample of N2 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? P1 = 6.0 atm V1 = 3.0 L P2 = ? V2 = 4.8 L P1V1 = P2V2 (6.0 atm)(3.0 L) = (P2)(4.8 L) P2 = 3.8 atm

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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 - V1 and T1 are the volume and absolute temperature of a gas at an initial set of conditions - V2 and T2 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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**CHARLES’S LAW A sample of Ar gas occupies a volume of 1.2 L at 125 oC**

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? V1 = 1.2 L T1 = 125 oC = 398 K V2 = 1.0 L T2 = ? T2 = 332 K = 59 oC

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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 1023 molecules) occupies a volume of L Conditions of STP Temperature = 0 oC = 273 K = 32 oF Pressure = 1.00 atm ( kPa or 100 kPa)

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**Considering all three gas laws**

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

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**R is the ideal gas constant**

THE IDEAL GAS LAW R is the ideal gas constant = L-atm/mol.K = J/mol-K = m3-Pa/mol-K = cal/mol-K = L-torr/mol-K

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RELATING THE GAS LAWS PV = nRT If n is constant

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

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RELATING THE GAS LAWS A balloon filled with helium initially has a volume of 1.00 x 106 L at 25 oC and a pressure of 752 mm Hg. Determine the volume of the balloon after a certain time when it encounters a temperature of -33 oC and a pressure of 75.0 mm Hg P1 = 752 mm Hg V1 = 1.00 x 106 L T1 = 25 oC = 298 K P2 = 75.0 mm Hg V2 = ? T2 = -33 oC = 240 K

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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**

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

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**Molar mass = 29.0 g/mol GAS DENSITIES AND MOLAR MASS**

Calculate the average molar mass of dry air if its density is 1.17 g/L at 21 oC and torr Molar mass = 29.0 g/mol

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**Consider the following reaction**

REACTION STOICHIOMETRY Consider the following reaction 4Al(s) + 3O2(g) → 2Al2O3(s) Calculate the mass of aluminum that would react completely with 2.00 L of pure oxygen gas at STP

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**REACTION STOICHIOMETRY**

1 mol O2 at STP = 22.4 L

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**REACTION STOICHIOMETRY**

- At the same conditions volumes of gases combine in the same proportions as the coefficients of the chemical equation Example 3H2(g) + N2(g) → 2NH3(g) This implies that - 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3 - 3 L H2 reacts with 1 L N2 to produce 2 L NH3

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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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**DALTON’S LAW OF PARTIAL PRESSURES**

- Ptotal is the total pressure of a gaseous mixture - P1, P2, P3,…. are the partial pressures of the individual gases - ntotal is the total number of moles of a gaseous mixture - n1, n2, n3,….are the number of moles of the individual gases

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**The total pressure by a mixture of He, Ne, and Ar gases is**

DALTON’S LAW OF PARTIAL PRESSURES 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 Ptotal = 3.50 atm P1 = 0.50 atm P2 = 0.75 atm P3 = ? Ptotal = P1 + P2 + P3 P3 = Ptotal - (P1 + P2) = 3.50 atm - (0.50 atm atm) = 2.25 atm

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**DALTON’S LAW OF PARTIAL PRESSURES**

For gas 1 in a mixture of gases with n1 moles Mole fraction (x1) = n1/ntotal O2 is 21% of air Mole fraction of O2 (x1) = 0.21 Total atmospheric pressure = 1 atm Partial pressure of O2 = P1 = (0.21)(1 atm) = 0.21 atm = (0.21)(760 mm Hg) = 160 mm Hg

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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) → MgCl2(aq) + H2(g) - The H2 gas is collected in a gas collection tube initially filled with water and inverted in a water pan - The volume of H2 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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**EXPERIMENT TO COLLECT GAS OVER WATER**

For this condition The pressure inside the tube is equal to the atmospheric pressure outside Ptotal = Pgas + Pwater Ptotal = total atmospheric pressure Pgas = pressure of gas in tube Pwater = vapor pressure of water at experimental temperature

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**KINETIC MOLECULAR THEORY OF GASES**

- 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

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**KINETIC MOLECULAR THEORY OF GASES**

- 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)

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**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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**GRAHAMS LAW OF EFFUSION**

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

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**GRAHAMS LAW OF EFFUSION**

- The rate of effusion of a gas is inversely proportional to the square root of its molar mass - r1 and r2 are the effusion rates of gases 1 and 2 respectively - M1 and M2 are the molar masses of gases 1 and 2 respectively

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**GRAHAMS LAW OF EFFUSION**

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

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**GRAHAMS LAW OF EFFUSION**

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 Cl2 gas to effuse under identical conditions? t2 = 19 minutes

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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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**REAL GASES - Molecules of real gases have finite volumes**

and attract one another - van der Waals equation

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REAL GASES a = constant (measure of how strongly gas molecules attract each other b = constant (measure of the small but finite volume occupied by gas molecules) n2a/V2 accounts for the attractive forces and adjusts pressure upwards nb accounts for the finite volume occupied by molecules

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**REAL GASES Calculate the pressure exerted by 0.2500 mol N2 gas in a**

5.000 L container at 22.5 oC 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

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REAL GASES

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