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A Review

Gases. A Review. Elements that exist as gases at 25 0 C and 1 atmosphere. 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.

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A Review

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  1. Gases A Review

  2. Elements that exist as gases at 250C and 1 atmosphere

  3. 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.

  4. Properties of Gases Gas properties defined using only 4 varibles • V = volume of the gas (L) • T = Kelvin temperature (K) • n = number of moles (mol) • P = pressure in either (atm), (mm Hg) or (kPa)

  5. Force Area Barometer Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

  6. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

  7. Atmospheric pressure Manometer Barometer

  8. Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

  9. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

  10. Constant temperature Constant amount of gas Boyle’s Law Pa 1/V P x V = constant P1 x V1 = P2 x V2

  11. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg

  12. Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  13. Charles’s original balloon Modern long-distance balloon

  14. As T increases V increases

  15. Temperature must be in Kelvin Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s Law VaT V = constant x T T (K) = t (0C) + 273.15 V1/T1 = V2/T2

  16. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A 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 if the 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 T2 = = 192 K

  17. twice as many molecules Avogadro’s Law Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related.

  18. Constant temperature Constant pressure Avogadro’s Law Va number of moles (n) V = constant x n V1/n1 = V2/n2

  19. 4NH3 + 5O2 4NO + 6H2O 1 volume NH3 1 volume NO 1 mole NH3 1 mole NO Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? At constant T and P

  20. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT

  21. The conditions 0 0C 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. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) PV = nRT R = 0.082057 L • atm / (mol • K)

  22. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 273.15 K V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT V = 30.6 L

  23. What is the volume (in liters) occupied by 49.8 g of HCl at STP?

  24. P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 K nR P = = T V P1 P2 T2 358 K T1 T1 T2 291 K = 1.20 atm x P2 = P1 x Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? n, V and Rare constant PV = nRT = constant = 1.48 atm

  25. m V PM = RT dRT P Density (d) Calculations m is the mass of the gas in g d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance d is the density of the gas in g/L M =

  26. d = = m M = V 4.65 g 2.10 L g 2.21 L x 0.0821 x 300.15 K 54.6 g/mol M = 1 atm g = 2.21 L dRT L•atm mol•K P A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas? M =

  27. Gas Stoichiometry • You can apply the discoveries of Gay -Lussac and Avogadro to calculate the stoichiometry of chemical reaction. • Step 1  Write a balanced chemical equation. • Step 2  Convert all units to common units • Step 3  Start with given and Use factor-label skills (Via MOLES) to get to desired units. • Step 4 If gas laws are needed use appropriately at this point or prior to factor- label step.

  28. What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mol CO2 g C6H12O6 mol C6H12O6 mol CO2V CO2 x 1 mol C6H12O6 1 mol C6H12O6 x 180 g C6H12O6 L•atm mol•K nRT 0.187 mol x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.187 mol CO2 V = = 4.76 L

  29. Gas Stoichiometry problems • 1) Ammonia can react with Oxygen to produce nitrogen and water. If 1.78 L of O2 reacts, what volume of nitrogen will be produce? (Temp and pressure remain constant) • 2) Ethylene gas( C2H4) combust in Air to form H2O and CO2. If 13.8 L of C2H4 measured at 21°C and 1.038 atm burns completely with O2, calculate the volume of CO2 produced assuming the CO2 is measured at 44°C and 0.989 atm. • 3) When Arsenic (III) sulfide is roasted in air, it reacts with oxygen to produce arsenic (III) oxide and sulfur dioxide. When 89.5 g As2S3 is roasted with excess O2, what volume of SO2 is produced? The gaseous product is measured at 20°C and 98.0 kPa. Ans.= 1.19 L of N2 Ans.= 32.6 L of CO2 Ans.= 27.1 L of SO2

  30. Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2

  31. PA = nART nBRT V V PB = nB nA nA + nB nA + nB XB = XA = ni mole fraction (Xi) = nT Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B PT = PA + PB PA = XAPT PB = XBPT Pi = XiPT

  32. 0.116 8.24 + 0.421 + 0.116 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 = XiPT PT = 1.37 atm = 0.0132 Xpropane = Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

  33. P V Chemistry in Action: Scuba Diving and the Gas Laws PV = nRT

  34. Kinetic Molecular Theory of Ideal Gases • 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. • Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. • Gas molecules exert neither attractive nor repulsive forces on one another. • 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

  35. Kinetic theory of gases and … • Compressibility of Gases • Boyle’s Law • Pa collision rate with wall • Collision rate a number density • Number density a 1/V • Pa 1/V • “12m10an3.mov” • Charles’ Law • Pa collision rate with wall • Collision rate a average kinetic energy of gas molecules • Average kinetic energy aT • PaT • “12m10an2.mov”

  36. Kinetic theory of gases and … • Avogadro’s Law • Pa collision rate with wall • Collision rate a number density • Number density an • Pan • “12m10an1.mov” • 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 = SPi

  37. Deviations from Ideal Gas Law • Real molecules have volume. • There are intermolecular forces. • Otherwise a gas could not become a liquid.

  38. n = = 1.0 PV RT Deviations from Ideal Behavior 1 mole of ideal gas Repulsive Forces PV = nRT Attractive Forces

  39. Measured V = V(ideal) Measured P ( ) 2 n a nRT V - nb P + ----- J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. 2 V vol. correction intermol. forces Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with Van der Waal’s Equation.

  40. Deviations from Ideal Gas Law Cl2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl2 in a 4.0 L tank at 27 oC. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm

  41. ( ) P + (V – nb) = nRT } } corrected pressure corrected volume n2a V2 Van der Waals equation nonideal gas During Calculations plug in a and b values into the equation and solve algebraically

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