the gaseous state of matter

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  • <number>
  • A mole of water occupies 18 mL as a liquid but would fill this box (22.4 L) as a gas at the same temperature.
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  • Figure 12.3 Preparation of a mercury barometer. The full tube of mercury at the
    left is inverted and placed in a dish of mercury.
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  • Figure 12.4 The pressure exerted by a gas is directly proportional to the number of molecules present. In each case shown, the volume is 22.4 L and the temperature is 0°C.
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  • Figure 12.5
    The pressure of a gas in a fixed volume increases with increasing temperature. The increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature.
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  • Figure 12.6 Graph of pressure versus volume showing the inverse PV relationship of an ideal gas.
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  • Figure 12.8 Volume-temperature relationship of methane (CH4). Extrapolated portion of the graph is shown by the broken line.
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  • Figure 12.12 Oxygen collected over water.
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  • Figure 12.14 Avogadro’s Law proved the concept of diatomic molecules for hydrogen and chlorine
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  • Figure 12.16 Summary of the primary conversions involved in stoichiometry. The conversion for volumes of gases is included.
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  • the gaseous state of matter

    1. 1. The gases The Gaseous State of Matter The air in a hot air balloon expands When it is heated. Some of the air escapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.
    2. 2. The atmosphere The atmosphere protects the planet and provides chemicals necessary for life.
    3. 3. The Atmosphere Each of the gases, N2, O2, CO2 and H2O among others in the atmosphere, serve a purpose O2 supports metabolism N2 dilutes the O2 so explosive combustion does not take place and is also part of proteins CO2 and H2O trap heat
    4. 4. General Properties Gases • Have an indefinite volume Expand to fill a container • Have an indefinite shape Take the shape of a container • Have low densities d air = 1.2 g / L at 25°C d H2O = 1.0 g / mL • Have high kinetic energies
    5. 5. Kinetic Molecular Theory (KMT) Assumptions of the KMT and ideal gases include: 1. Gases consist of tiny particles 2. The distance between particles is large compared with the size of the particles. 3. Gas particles have no attraction for each other 4. Gas particles move in straight lines in all directions, colliding frequently with each other and with the walls of the container.
    6. 6. Kinetic Molecular Theory Assumptions of the KMT (continued): 5. Collisions are perfectly elastic (no energy is lost in the collision). 6. The average kinetic energy for particles is the same for all gases at the same temperature. 1 2 KE = mv where m is mass and v is velocity 2 7. The average kinetic energy is directly proportional to the Kelvin temperature.
    7. 7. Diffusion
    8. 8. Effusion Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure. Graham’s law of effusion: rate of effusion of gas A density B molar mass B = = rate of effusion of gas B density A molar mass A
    9. 9. Graham’s law of effusion Rates of effusion of gases are inversely proportional to the square roots of their molar masses © 2013 John Wiley & Sons, Inc. All rights reserved.
    10. 10. Measurement of Pressure Pressure = Force Area Pressure depends on the • Number of gas molecules • Temperature of the gas • Volume the gas occupies
    11. 11. Atmospheric Pressure Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface.
    12. 12. Barometer
    13. 13. Pressure Conversions Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm 1 atm 675 mm Hg × = 0.888 atm 760 mm Hg Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr. 760 torr 675 mm Hg × = 675 torr 760 mm Hg
    14. 14. Dependence of Pressure on Number of Molecules P is proportional to n (number of molecules) at Tc (constant T) and Vc (constant V). The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.
    15. 15. Dependence of Pressure on Temperature P is proportional to T at nc (constant number of moles) and Vc. The increased pressure is due to • more frequent collisions • higher energy collisions
    16. 16. Boyle’s Law 1 At Tc and nc : V α P or PV1 = PV2 1 2 What happens to V if you double P? • V decreases by half! What happens to P if you double V? • P decreases by half!
    17. 17. Boyle’s Law PV1 = PV2 1 2 A sample of argon gas occupies 500.0 mL at 920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature. Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL Set-Up PV1 P2 = 1 V2 Calculate 920. torr × 500. mL P2 = = 491 torr 937 mL
    18. 18. Boyle’s Law Another approach to the same problem: Since volume increased from 500. mL to 937 ml, the pressure of 920. torr must decrease. Multiply the pressure by a volume ratio that decreases the pressure:  500. mL P2 = 920. torr   937 mL  ÷= 491 torr 
    19. 19. Charles’ Law • The volume of an ideal gas at absolute zero (-273°C) is zero. • Real gases condense at their boiling point so it is not possible to have a gas with zero volume. • The gas laws are based on Kelvin temperature. • All gas law problems must be worked in Kelvin! At Pc and nc : V α T or V1 V2 = T1 T2
    20. 20. Charles’ Law V1 V2 = T1 T2 A 2.0 L He balloon at 25°C is taken outside on a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant? Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K Set-Up Calculate V1T2 V1 V2 rearranged gives V2 = = T1 T1 T2 (2.0 L)(258 K) V2 = = 1.7 L 298 K
    21. 21. Charles’ Law Another approach to the same problem: Since T decreased from 25°C to -15°C, the volume of the 2.0L balloon must decrease. Multiply the volume by a Kelvin temperature ratio that decreases the volume:  258K P2 = 2.0L   298 K  ÷= 1.7L 
    22. 22. Gay-Lussac’s Law At Vc and nc : P α T or P P2 1 = T1 T2
    23. 23. Combined Gas Laws Used for calculating the results of changes in gas conditions. • Boyle’s Law where Tc • Charles’ Law where Pc PV1 1 T1 = P2V2 T2 PV1 = PV2 1 2 V1 V2 = T1 T2 P P2 1 = T1 T2 • Gay Lussacs’ Law where Vc P1 and P2 , V1 and V2 can be any units as long as they are the same. T1 and T2 must be in Kelvin.
    24. 24. Combined Gas Law PV1 1 T1 = PV2 2 T2 If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr? STP: Standard Temperature 273K or 0°C Standard Pressure 1 atm or 760 torr Knowns Set-Up Calculate V1 = 500. mL T1 =273K P1= 760 torr T2 = 85°C = 358K P2= 560 torr PV1T2 V2 = 1 T1 P2 (760 torr)(500. mL)(358K) V2 = = 890. ml (273K)(560 torr)
    25. 25. Combined Gas Law PV1 1 T1 = PV2 2 T2 A sample of oxygen gas occupies 500.0 mL at 722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg. Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr V2 = 2.53 L = 2530 mL P2= 560 torr T1 PV2 T2 = 2 Set-Up PV1 1 Calculate ( 491 torr ) ( 2530 ml ) ( 248K ) =853K = T2 = ( 722 torr ) ( 500.0 ml ) 580°C
    26. 26. Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. PTotal = PA + PB + PC + …. Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air. PAir = PN2 + PO2 + PAr + PCO2 + PH 2O + ….
    27. 27. Collecting Gas Over Water • Gases collected over water contain both the gas and water vapor. • The vapor pressure of water is constant at a given temperature • Pressure in the bottle is equalized so that the Pinside = Patm Patm = Pgas + PH 2 O
    28. 28. Avogadro’s Law Equal volumes of different gases at the same T and P contain the same number of molecules. The ratio is the same: 1 volume 1 molecule 1 mol 1 volume 1 molecule 1 mol 2 volumes 2 molecules 2 mol
    29. 29. Mole-Mass-Volume Relationships Molar Volume: One mole of any gas occupies 22.4 L at STP. Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP. Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm Set-Up Calculate 22.4 L 1 mol = 22.4 L so the conversion factor is 1mol  3.94 g   22.4 L   ÷ ÷ = 58.1g/mol  1.52 L   1 mol 
    30. 30. Density of Gases mass g d= = volume L Calculate the density of nitrogen gas at STP. dSTP  1 mol  = molar mass  ÷  22.4 L   28.02 g  1 mol  dSTP =  = 1.25g/L ÷ ÷particular temperature, Note that densities are mol  22.4 La  1 always cited for  since gas densities decrease as temperature increases.
    31. 31. Ideal Gas Law L× atm PV = nRT where R = 0.0821 mol ×K Calculate the volume of 1 mole of any gas at STP. Knowns Set-Up n = 1 mole T = 273K P = 1 atm nRT V = P Calculate Molar volume! L ×atm )(273 K) mol ×K = 22.4 L (1 atm) (1 mol)(0.0821 V =
    32. 32. Ideal Gas Law L× atm PV = nRT where R = 0.0821 mol ×K How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg? Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm Set-Up Calculate PV n = RT (0.980 atm)(1.3 L) n= =0.052 mol L ×atm (0.0821 )(297 K) mol ×K
    33. 33. Ideal Gas Law Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr. Knowns m =4.12 g V = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm Set-Up g n= M so g PV = RT M gRT M= PV L ×atm (4.12 g)(0.0821 )(296 K) mol ×K Calculate M = =107 g/mol (0.988 atm)(0.943 L)
    34. 34. Gas Stoichiometry • Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L. • Use the Ideal Gas Law if the conditions are not at STP.
    35. 35. Gas Stoichiometry Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm. P4(s) + 6H2(g)  4PH3(g) Knowns V = 4.0 L T = 273 K Solution Map Calculate P = 1 atm L H2  mol H2  mol P4  1 mol H2   1 mol P4  = 0.030 mol P4 4.0 L H2  ÷  ÷ 22.4L   6 mol H2  
    36. 36. Gas Stoichiometry What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C2H6? 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) Knowns m = 3.2 g C2H6 Solution Map Calculate T = 25°C = 298K P = 1 atm m C2H6  mol C2H6  mol O2  volume O2  1 mol C 2 H 6   7 mol O2  3.2g C2 H 6  ÷ ÷= 0.37mol O2  30.08g C2 H 6   2 mol C2 H 6  L ×atm (0.37 mol)(0.0821 )(298 K) mol ×K V = = 9.1 L (1 atm)
    37. 37. Volume-Volume Calculations Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm. N2(g) + 3H2(g)  2NH3(g) Knowns Solution Map Calculate V = 9.0 L T = 450K P = 5.00 atm Assume T and P for both gases are the same. Use volume ratio instead of mole ratio! L H2  L N2 9.0 L H2  1 L N2   ÷ 3 L H2   = 3.0 L N2
    38. 38. Real Gases Most real gases behave like ideal gases under ordinary temperature and pressure conditions. Conditions where real gases don’t behave ideally: • At high P because the distance between particles is too small and the molecules are too crowded together. • At low T because gas molecules begin to attract each other. High P and low T are used to condense gases.
    39. 39. The law of effusion and diffusion The postulate of the kinetic theory states that the temperature of a system is proportional to the average kinetic energy of its particles and nothing else. In other words, two gases at the same temperature, such as and must have the same average kinetic energy, hence multiplying by 2 gives:
    40. 40. Law of Effusion and diffusion Then rearranging gives: Taking out the square root both sides we get

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