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- 1. Gases Jupiter Primarily made of airbag hydrogen, a quarter ofA chemical reaction its mass is from Helium.produces N2 gas to fill anairbag, this happenes in0.04 seconds, much fasterthan the blink of an eye(about 0.2 seconds)
- 2. Elemental GasesElements that exist as gases at 250C and 1 atmosphere www.corribaker.com 2
- 3. Some Substances Found as Gases at 1 atm and 25oC Elements CompoundsH2 (molecular hydrogen) HF (hydrogen fluoride)N2 (molecular nitrogen) HCl (hydrogen chloride)O2 (molecular oxygen) HBr (hydrogen bromide) O3 (ozone) HI (hydrogen iodide) F2 (molecular fluorine) CO (carbon monoxide)Cl2 (molecular chlorine) CO2 (carbon dioxide) He (helium) NH3 (ammonia) Ne (neon) NO (nitric oxide) Ar (argon) NO2 (nitrogen dioxide) Kr (krypton) N2O (nitrous oxide) Xe (xenon) SO2 (sulfur dioxide) Rn (radon) H2S (hydrogen sulfide) *The boiling point of HCN is 26oC, but it is HCN (hydrogen cyanide)* gas at close enough to qualify as a ordinary atmospheric conditions. www.corribaker.com 3
- 4. 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. NO2 gas www.corribaker.com 4
- 5. Pressure Force Pressure = Area(force = mass x acceleration)Gravitational acceleration on Earth = 9.8 ms-2 What is 1 atmosphere? (atm) Pascal (Pa) 1.01325 ×105 N/m2 1.01325 ×105 bar (bar) 1.01325 Torr (Torr) 760 mmHg 760 Pounds per 14.69595 square inch (psi) Barometer Making a barometer: http://www.youtube.com/watch?v=GgBE8_SyQCU www.corribaker.com 5
- 6. 1 cm 1 cmWhat is the massof a 1cm x 1cm aircolumn from sealevel to top ofatmosphere? Cruising Jet 0.2 atm (12 000m) Everest (8840m) 0.33 atmANSWER = 1 kg Sea level 1 atm www.corribaker.com 6
- 7. The Gas Laws• Relate the temperature, pressure and volume of an ideal gas. – Boyle’s law: pressure is inversely proportional to volume at constant temperature – Charles’ law: volume is directly proportional to temperature at constant pressure – Avogadro’s law: volume is directly proportional to number of moles of gas present at constant temperature and pressure• These three laws are summarised in the ideal gas equation: pV = nRT. R = 0.082057 L • atm / (mol • K) www.corribaker.com 7
- 8. Proportionality• Proportionality: if two variables have a ratio that is constant, they are said to be proportional.• Put another way, one variable is a constant multiple of the other.• Direct Proportionality – y = kx (where k is a non zero constant) – eg, the circumference of a circle is proportional to its diameter, constant is equal to π. (c = π d)• Inverse Proportionality – y = k x 1/x – eg, the volume of gas is inversely proportional to the pressure exerted, constant is equal to nRT. (PV = nRT) www.corribaker.com 8
- 9. Apparatus for Studying the Relationship Between Pressure and Volume of a Gas a) Pressure b) Add more c) Add more exerted on gas is mercury – an mercury again – same as increase in another increase atmospheric pressure in pressure pressure exerted on exerted on gas., gas, volume volume decreases decreases As P (h) increases V decreases www.corribaker.com 9
- 10. Boyle’s Law P α 1/V Conditions:P x V = constant Constant temperature Constant amount of gasP1 x V1 = P2 x V2 www.corribaker.com 10
- 11. Boyle’s LawA sample of chlorine gas occupies a volume of 946 mL at apressure of 726 mmHg. What is the pressure of the gas (inmmHg) if the volume is reduced at constant temperature to 154mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 726 mmHg x 946 mLP2 = = = 4460 mmHg V2 154 mL www.corribaker.com 11
- 12. Gas expanding and contracting As T increases V increases www.corribaker.com 12
- 13. Charles’ LawVariation of gas volume with temperature at constant pressure. VαT V = constant x T V1/T1 = V2 /T2 Temperature must be in Kelvin T (K) = t (0C) + 273.15 www.corribaker.com 13
- 14. Charles’ LawA 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 ifthe 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 1.54 L x 398.15 K T2 = = = 192 K V1 3.20 L www.corribaker.com 14
- 15. Avogadro’s LawV α number of moles (n) Conditions:V = constant x n Constant temperature Constant pressureV1 / n1 = V2 / n2 www.corribaker.com 15
- 16. Avogadro’s LawAmmonia burns in oxygen to form nitric oxide (NO) and watervapor. How many volumes of NO are obtained from one volumeof 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 www.corribaker.com 16
- 17. Ideal Gas EquationBoyle’s law: V α 1 (at constant n and T) PCharles’ law: V α T (at constant n and P)Avogadro’s law: V α n (at constant P and T) nTVα P nT nTV = constant x =R R is the gas constant P P PV = nRT www.corribaker.com 17
- 18. Gases at STP• The conditions 0 °C 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 PV (1 atm)(22.414L) R= = nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) www.corribaker.com 18
- 19. Ideal Gas EquationWhat is the volume (in litres) occupied by 49.8 g of HClat STP? PV = nRT T = 0 0C = 273.15 K nRT V= P = 1 atm P 1 mol HCl n = 49.8 g x = 1.37 mol 36.45 g HCl L•atm 1.37 mol x 0.0821 mol•K x 273.15 K V= 1 atm V = 30.6 L www.corribaker.com 19
- 20. Dimensional Analysis www.corribaker.com 20
- 21. Ideal Gas EquationArgon is an inert gas used in lightbulbs to retard the vaporizationof the filament. A certain lightbulb containing argon at 1.20 atmand 18 0C is heated to 85 0C at constant volume. What is thefinal pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR = P = constant P1 = 1.20 atm P2 = ? V T T1 = 291 K T2 = 358 K P1 P2 = T1 T2 T2 P2 = P1 x = 1.20 atm x 358 K = 1.48 atm T1 291 K www.corribaker.com 21
- 22. Density and Molar Mass calculations Density (d) Calculations m = PM m is the mass of the gas in g d= V RT M is the molar mass of the gasMolar Mass (M ) of a Gaseous Substance dRTM= d is the density of the gas in g/L P Practice the derivations of these equations from PV = nRT yourself! www.corribaker.com 22
- 23. Density and Molar Mass calculations 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 m = 4.65 g = 2.21 g M= d= P V 2.10 L L g L•atm 2.21 L x 0.0821 mol•K x 300.15 K M= 1 atm M = 54.6 g/mol www.corribaker.com 23
- 24. Gas StoichiometryWhat is the volume of CO2 produced at 37 0C and 1.00 atmwhen 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 5.60 g C6H12O6 = 0.031 mol C6H12O6 x 6 mol = 0.187 mol CO2 180 gmol -1 L•atm 0.187 mol x 0.0821 x 310.15 K nRT mol•K V= = = 4.76 L P 1.00 atm www.corribaker.com 24
- 25. Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2 P1 www.corribaker.com 25
- 26. Dalton’s Law of Partial Pressures• Daltons Law: the total pressure of a mixture of gases or vapours is equal to the sum of the partial pressures of its components.• ie, the sum of the pressures that each component would exert if it were present alone and occupied the same volume as the mixture of gases. www.corribaker.com 26
- 27. Dalton’s Law of Partial PressuresConsider a case in which two gases, A and B, are in acontainer of volume V. nART nA is the number of moles of A PA = V nBRT nB is the number of moles of BPB = V www.corribaker.com 27
- 28. Dalton’s Law of Partial Pressures nA nBPT = PA + PB XA = XB = nA + nB nA + nB number of moles of gas G nmole fraction (XG) = G nT TOTAL number of moles of all gases in the vessel PA = XA PT PB = XB PT PG = XG PT www.corribaker.com 28
- 29. Dalton’s Law of Partial Pressures 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 Xpropane = = 0.0132 8.24 + 0.421 + 0.116 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm www.corribaker.com 29
- 30. Kinetic Molecular Theory of Gases• Kinetic Molecular Theory of Gases explains what is happening to gases at the molecular level.• We know that gas will expand on heating (Charles’ Law) but WHY?• The physical properties of gases can be explained in terms of the motion of individual molecules. This molecular movement is a form of energy. www.corribaker.com 30
- 31. Kinetic Molecular Theory of Gases1. 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.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 m = mass of molecule KE = ½ mu2 u = molecule speed www.corribaker.com 31
- 32. Kinetic Molecular Theory of Gases• According to the kinetic molecular theory, gas pressure is the result of collisions between molecules and the walls of their container.• The absolute temperature of a gas is a measure of the average kinetic energy of the molecules.• The higher the temperature, the more energetic the molecules. www.corribaker.com 32
- 33. Kinetic theory of gases and …• Compressibility of Gases• Boyle’s Law *number density = P α collision rate with wall number of molecules per unit volume Collision rate α number density* Number density α 1/V P α 1/V• Charles’ Law P α collision rate with wall Collision rate α average kinetic energy of gas molecules Average kinetic energy α T PαT www.corribaker.com 33
- 34. Kinetic theory of gases and …• Avogadro’s Law P α collision rate with wall Collision rate α number density Number density α n Pαn• Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = ΣPG www.corribaker.com 34
- 35. Distribution of Molecular Speed The distribution of speeds for nitrogen gas molecules at three different temperatures The higher the temperature of the gas, the faster the molecules of the gas move. www.corribaker.com 35
- 36. Distribution of Molecular Speed The distribution of speeds of three different gases at the same temperature The heavier the gas, the more slowly its molecules move. www.corribaker.com 36
- 37. Root-Mean-Square Speed• How fast does a molecule move, on average at any temperature, T?• The root-mean-square (rms) speed (urms) is an average molecular speed urms = √ 3RT M M = molar mass www.corribaker.com 37
- 38. Gas Diffusion• Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. The path traveled by a single gas molecule. Each change in direction is a collision with another molecule• A lighter gas will diffuse through a certain space more quickly than will a heavier gas – (Since the rms of a light gas is greater than that of a heavier gas) www.corribaker.com 38
- 39. Gas Diffusion √r1 Graham’s Law of = M2 diffusionr2 M1 (r1 and r2 are the diffusion rates of the gases) NH4Cl Since NH3 is lighter than HCl, it diffuses faster, so solid NH4Cl first appears nearer to the HCl bottle on the right NH3 HCl 17 g/mol 36 g/mol www.corribaker.com 39
- 40. Gas Effusion• Gas effusion is the process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening.• Effusion is different from diffusion in nature, but the rate of effusion of a gas follows the same form as Graham’s law of diffusion. √ r1 t2 M2 = = r2 t1 M1 www.corribaker.com 40
- 41. Gas Effusion 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 2r1 = 3.3 x r2 M2 = (r ) 2 x M1 = (3.3)2 x 16 = 174.2M1 = 16 g/mol 58.7 + x • 28 = 174.2 x = 4.1 ~ 4 www.corribaker.com 41
- 42. Credits• McGraw Hill Publishing• Wikipedia Creative Commons Images www.corribaker.com 42

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