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

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Presentation on theme: "Gases. Elements that exist as gases at 25 0 C and 1 atmosphere."— Presentation transcript:

1 Gases

2 Elements that exist as gases at 25 0 C and 1 atmosphere

3 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. Physical Characteristics of Gases

4 Kinetic Theory of Gases KMT Based on 5 Assumptions: 1.G. have large number of particles that are very far apart from each other relative to their size. 2.Collisions between g. particles and between g. particles and container walls are elastic. (No net loss of kinetic energy)

5 Kinetic Theory of Gases cont. 3.G. particles are in constant motion 4.There are no forces of attraction between G. particles. 5.The temperature of a gas depends on the average kinetic energy of the particles of the gas. KE= ½ (mv 2 ). All gases have the same KE at the same temperature.

6 Ideal vs. Real Gases Ideal gas: hypothetical gas that perfectly fits all of the assumptions of the KMT. Real gas: one that does not behave exactly according to the assumptions of the KMT. Real gas behavior more likely at high pressures and low temperatures.

7 Questions 1.What happens to gas particles when a gas is compressed? 2.What happens to gas particles when a gas is heated? 3.Describe the conditions under which a real gas is most likely to behave ideally.

8 Gases and Pressure Pressure: Force per unit area. Pressure = Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

9 4/16/2012 Get out your notebooks and your books if you have them.

10 Gases Gases and Pressure Calculate the pressure of the following: 1.F=500N, A= 325cm 2 2.F=500N, A= 13cm 2 3.F=500N, A= 6.5cm 2

11 Gases Gases and Pressure Convert the following pressures: 1.P = 0.830atm to mmHg, and kPa 2.P = 1.75atm to mmHg and kPa 3.The critical pressure of carbon dioxide is 72.7atm, what is this value in Pa?

12 Gases Gases and Pressure Standard Temperature and Pressure (STP) = 1atm and 0°C

13 Gases Dalton’s Law of Partial Pressures The pressure of each gas in a mixture is the partial pressure of that gas. Dalton’s Law of partial pressures = the total pressure of a gas mixture is the sum of all the partial pressures of the component gases. P T = P 1 + P 2 + P 3 + ….

14 Gases Dalton’s Law of Partial Pressures Water Displacement, gas collected over water. Always mixed with water vapor, which has specific water pressure. P atm = P gas + P H2O

15 Gases Dalton’s Law of Partial Pressures Three of the primary components of air are CO 2, O 2, and N 2. In the sample containing a mixture of only these gases at exactly 1atm, the partial pressures of P CO2 = 0.285 torr and P N2 = 593.525 torr. What is the partial pressure of oxygen?

16 Gases Dalton’s Law of Partial Pressures Oxygen gas from decomposition of KClO 3, was collected by water displacement. The barometric pressure and temperature was 731.0 torr and 20.0°C. What was the partial pressure of the oxygen collected?

17 4/19/2012 Get out your notebooks

18 Boyle’s Law Volume of a fixed mass of gas varies inversely with pressure at a constant temperature. PV = k, k is a constant Product always remains the same, so can be written: P 1 V 1 = P 2 V 2

19 Boyle’s Law Constant temperature Constant amount of gas

20 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? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

21 Charle’s Law Volume of a fixed mass of gas at constant pressure varies directly with the Kelvin Temperature. V = kT, k is a constant or

22 Charle’s Law Kelvin Scale: -273.15ºC, or 0 K. Absolute Zero at which all movement stops. K = 273.15 + ºC Convert 50ºC to K K = 273.15 +50 K = 223.15K

23 Charle’s Law

24 A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2

25 Guy-Lussac’s Law Pressure of a fixed mass of gas at a constant temperature varies directly with the Kelvin Temperature. P = kT, k is a constant or

26 The gas in a container is at a pressure of 3.00atm at 25ºC. Directions on the container warn the user not to keep it in a place where the temp. exceeds 52ºC. What would the pressure in the container be at that temp? P 1 = 3.00 atm T 1 = 298.15 K P 2 = ? T 2 = 325.15K P 2 = T 2 x P 1 T1T1 325.15 K x 3.00atm 298.15 K = = 3.27atm P 1 /T 1 = P 2 /T 2

27 Combined Gas Law Expression of the relationship between pressure, volume and temperature of a fixed amount of gas.

28 Combined Gas Law A helium filled balloon has a volume of 50.0L at 25ºC and 1.08atm. What volume will it have at 0.855atm and 10.0 ºC. P 1 = 1.08 atm T 1 = 298.15 K P 2 = 0.855atm T 2 = 283.15 K V 2 =T 2 x P 1 x V 1 P 2 x T 1 283.15 K x 1.08atm x 50.0L 298.15 K x 0.855atm = = 60.0L V 1 = 50.0 LV 2 = ?

29 Avogadro’s Law V  number of moles (n) V = k x n Constant temperature Constant pressure

30 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? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO

31 Avogadro’s Law Standard Molar Volume: Volume occupied by 1 mole at STP = 22.4L 1mol gas = 22.4L gas What Volume does 0.0685mol of gas occupy at STP? Multiply the amount in moles by the conversion factor, 1mol gas = 22.4L gas

32 What quantity of gas, in moles, is contained in 2.21L at STP? Multiply the volume in liters by the conversion factor, 1mol gas = 22.4L gas

33 Ideal Gas Equation Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

34 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

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

36 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

37 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L

38 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm

39 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 3 17 g/mol HCl 36 g/mol NH 4 Cl Gas effusion is the process by which gas particles pass through a tiny opening.


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