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Gases Chapter 14 Sections 1. The Gas Laws 2. The Combined Gas Laws and Avogadros Principle 3. Ideal Gas Law 4. Gas Stoichiometry.

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Presentation on theme: "Gases Chapter 14 Sections 1. The Gas Laws 2. The Combined Gas Laws and Avogadros Principle 3. Ideal Gas Law 4. Gas Stoichiometry."— Presentation transcript:


2 Gases Chapter 14

3 Sections 1. The Gas Laws 2. The Combined Gas Laws and Avogadros Principle 3. Ideal Gas Law 4. Gas Stoichiometry

4 The Gas Laws Chapter 14.1

5 The Gas Laws Boyles Law Charles Law Gay-Lussacs Law Each law relates two variables to the behavior of gasses. –Pressure –Temperature –Volume –amount

6 Kinetic Theory Review K-M theory suggests gas particles behave differently than liquids and solids. K-M assumes the following are true about gasses: –Gas particles do not attract or repel each other –Gas particles are much smaller than the distances between them –Gas particles are in constant, random motion –No kinetic energy is lost when gas particles collide with each other or with the walls of the container –All gasses have the same average kinetic energy at a given temperature

7 The Nature of Gasses Actual gasses dont obey all the assumptions made by K-M theory –But their behavior approximates the theory Notice that all assumptions of K-M theory are based on the four factors mentioned about the gas laws: –Number of particles present –Temperature –Pressure –Volume of gas

8 Boyles Law Robert Boyle (1627-1691), Irish, studied relationship between pressure and volume. –An inverse relationship Boyles Law: –The volume of a given amount of gas at constant temperature varies inversely with the pressure –So at any two different times, for the same gas, the product of pressure and volume will be the same –P 1 V 1 = P 2 V 2

9 Boyles Law ojectfolder/flashfiles/gaslaw/boyles_law_graph.html ojectfolder/flashfiles/gaslaw/boyles_law_graph.html

10 Boyles Law Problem The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what is the new volume? –P 1 V 1 = P 2 V 2 –99.0kPa x 300.0mL = 188kPa x ? 99.0kPa x 300.0mL = ? 188kPa 158.0mL

11 Charless Law Jacque Charles (1746-1823) French –Studied relationship between volume and temperature –As temperature increases, so does volume of gas MUST BE IN KELVIN T k = T c + 273 –K-M Theory says as temperature increases, particles move faster, so must be further away. –Relationship between temperature and volume is a straight line with positive slope V1T1V1T1 V2T2V2T2 =

12 Charless Law Law: The volume of a gas is directly proportional to the temperature, expressed in Kelvin, at a constant pressure

13 Charless Law Problem A gas at 89C occupies a volume of 0.67L. At what Celsius temperature will the volume increase to 1.12L? A gas at 89°C occupies a volume of 0.67L. At what Celsius temperature will the volume increase to 1.12L? V1T1V1T1 V2T2V2T2 = 0.67L 89°C + 273 1.12L ? = ? x 0.67L 362 ° K 1.12L = ? 1.12L x 362°K 0.67L = = 605.1 K 605.1°K – 273K = 332 °C

14 Gay-Lussacs Law Joseph Gay-Lussac studied relationship between temperature and pressure of a constained gas at fixed volume: –Found direct proportion exists between temperature (KELVIN) and pressure –Mathematically: P1T1P1T1 P2T2P2T2 =

15 Gay-Lussacs Law Problem A gas in a sealed container has a pressure of 125kPa at a temperature of 30°C. If the pressure of the container is increased to 201kPa, what is the new temperature? Convert temp to K: 30+273 = 303K P1T1P1T1 P2T2P2T2 = 125kPa 303K 201kPa T 2 = T2T2 201kPa x 303K 125kPa == 487K °C 487K – 273 = 214 °C

16 Combined Gas Law and Avogadros Principle Chapter 14, Section 2

17 Combined Gas Law Combine Boyles Law, Charless Law, and Gay-Lussacs Law into one: –P 1 V 1 /T 1 = P 2 V 2 /T 2 –Lets you work out problems involving more variables Use known variables under one set of conditions to find a value of a missing variable.

18 Avogadros Principle 1811, Avogadro suggested that equal volumes of gasses at the same temperature and pressure contain the same number of particles: –K-M theory says the same thing Particles are so far apart that size doesnt matter –1 mole of particles is 6.02 x 10 23 particles –1 mole of gas particles is 22.4 L at 0.00° C and 1.00 atm pressure Standard Temperature and Pressure (STP)

19 Example Problems Determine the volume of a container that holds 2.4 mole of gas at STP: –2.4 mol x 22.4L/1mol = 53.76 L = 54L What volume will 1.02 mol of carbon monoxide gas occupy at STP? –1.02 mol x 22.4L/1 mol = 22.8 L A balloon will rise off the ground when it contains 0.0226 mol of He in a volume of 0.460L. How many moles of He are needed to make the balloon rise when volume is 0.885L, assuming temperature and pressure are constant? –0.0226 mol x 0.865L/0.460L = 0.0425 mol

20 Ideal Gas Law Chapter 14, Section 3

21 Ideal Gas Law P 1 V 1 /T 1 = P 2 V 2 /T 2 can be interpreted as saying that PV/T is a constant, k –k is based on the amount of gas present, n –Experiments have shown that k = nR, where n is number of moles and R is a constant. Ideal Gas Law is therefore: –PV = nRT Ideal Gas Constant (R), is: –0.0821 L*atm/mol-K (most often used) –8.314 L*kPa/mol K –62.4 L*mm Hg/mol K

22 Real vs. Ideal Ideal Gas Assumptions –Particles take up no space –No intermolecular forces –Follows ideal gas law under all temperatures and pressures Real Gas –All gas particles take up some space –All gas particles are subject to intermolecular forces –Most gasses act like ideal gasses at many/most temperatures and pressures –Deviations occur at extremely high pressures and extremely low temperatures because intermolecular forces start to show effects.

23 Applying Ideal Gas Law If you know any three variables, you can solve for the fourth: –You can solve for n (number of moles) –Combined gas law, you cannot. You can use ideal gas law allows you to solve for molar mass and density, if mass is known –n (number of moles) = m (mass) / M (molar mass) –Sustitute m/M for n PV = mRT/M

24 Solving for Density Density is m (mass) / V (volume) –Substitute D (Density) for m/V –M = mRT/PV = DRT/P –Or D = MP/RT

25 Example Problem Calculate the number of moles of gas contained in a 3.0L vessel at 3.00x10 2 K with a pressure of 1.50 atm –Find: moles –Known: V = 3.0L –Known: T = 3.00x10 2 –Known: P = 1.50 atm –Known: R = 0.0821 L*Atm/mol*K –PV=nRT –Solve for n –N = PV/RT

26 Example Problem n = PV/RT – substitute in known information: n = (1.50atm)(3.0L) (0.0821 L*atm/mol*K)(3.00 x 10 2 K) n = 0.18 mole Evaluation: 1 mole of gas occupies 22.4L at STP. The volume is much less than 1 mole. Temperature and pressures are not dramatically different than STP. Slightly higher pressure means more gas. Slightly higher temperature means a little less gas. Units of the answer is moles – every unit cancels except 1/ 1/mol.

27 Example Problem – Using Molar Mass What is the molar mass of a pure gas that has a density of 1.40g/L at STP? –Find: molar mass (g/mole) –Known: D = 1.40g/L –Known: T=0.00°C –Known: P = 1.00atm –Known: R = 0.0821 L*atm/mol*K –Step1: Convert T to Kelvin T k = T c +273 T k = 273

28 Molar Mass Problem Use density form of ideal gas law –M = DRT/P –Substitute known values: M = (1.40 g/L)(0.0821 L*atm/mol*K)(273K) 1 atm M = 31.4 g/mol

29 Gas Stoichiometry Chapter 14, Section 4

30 Gas Stoichiometry – Volume only Example: –2C 4 H 10(g) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O (g) –Remember: Avogadros principle states that 1 mole of gas is 22.4 L –For volume to volume calculations, you only need to know mole ratios –2 L of butane reacts, it involves 13 L of O 2 (2L x 13 O 2 /2 C 4 H 10 ) 8 L of CO 2 (2L x 8CO 2 /2 C 4 H 10 ) 10 L of H 2 O

31 Gas Stoichiometry – Volume & Mass Ammonia is synthesized from hydrogen and nitrogen –N 2(g) + 3 H 2(g) 2 NH 3(g) –If 5.00 L of nitrogen reacts completely by this reaction at 3.0 atm and 298K, how many grams of ammonia are produced? –Analyze: you are given L, pressure, temperature. You are asked to find grams of NH 3. –Known: –V N = 5.00L, P = 3.00 atm, T = 298K

32 Gas Stoichiometry (Cont) 1. Determine the mole ratio needed 2 volumes NH 3 /1 volume N 2 2. Determine volume of ammonia produced: 5.00L N 2 x 2 volumes NH3 1 volume N2 = 10L NH 3 3. Rearrange Gas Law to solve for n (moles): PV = nRT n = PV RT 4. Substitute values and solve for n

33 Gas Stoichiometry (Cont). 4. Substitute values and solve for n n = (3.00 atm)(10.00L) (0.0821 L*atm/mol*k)(298K) = 1.23 mol NH 3 5. Find the mass (M) of NH 3 by finding the molecular mass and converting the moles in grams Molecular mass = 1N(14.01amu) + 3 H(1.01amu) = 17.04amu 1.23 mol NH 3 x 17.04 g/mol = 21.0 g NH 3

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