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Chapter 5 Gases. The Nobel Gases Exist as Monatomic Gases He, Ne, Ar, Kr, Xe, Rn Diatomic Gases are H 2, N 2, O 2, F 2, Cl 2. Oxygen also exists as the.

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Presentation on theme: "Chapter 5 Gases. The Nobel Gases Exist as Monatomic Gases He, Ne, Ar, Kr, Xe, Rn Diatomic Gases are H 2, N 2, O 2, F 2, Cl 2. Oxygen also exists as the."— Presentation transcript:

1 Chapter 5 Gases

2 The Nobel Gases Exist as Monatomic Gases He, Ne, Ar, Kr, Xe, Rn Diatomic Gases are H 2, N 2, O 2, F 2, Cl 2. Oxygen also exists as the gas Ozone = O 3

3 Compound Gases Ionic Compounds do not exist as gases at standard temperature and pressure (STP = 0 o C and 1 atm). This is because the cations and anions are held together by very strong electrostatic forces. To overcome these forces, we would have to raise the temperature well above 1000 o C. Molecular Compounds vary but their boiling points are much lower. Most still exist as solids or liquids at STP. Some common molecular compounds that are gases at STP are the following: HF, HCl, HBr, HI, CO, CO 2, NH 3, NO, NO 2, N 2 O, SO 2, H 2 S, HCN Most of these gases are colorless, except F 2 (greenish yellow), Cl 2 (greenish yellow), and NO 2 (brown)

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5 Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible of the states of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

6 The SI Unit of Pressure We need to derive pressure from the 7 SI base units. First, we derive force: Force = mass x acceleration Force = kilogram x m/s 2 = 1 newton (N) = the SI unit for force Next, we derive pressure: Pressure = force / area = N / m 2 = (kg x m/s 2 )/m 2 = 1 pascal (Pa) Pascal is the SI unit for pressure

7 = 760 mm Hg Instrument used to measure the pressure of the air in the atmosphere.

8 Used to measure pressures below atmospheric pressure. The closed-tube manometerThe open-tube manometer Used for measuring pressures equal to or greater than atmospheric pressure.

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10 Boyle’s Law : P  1/V P = k 1 / V and PV = k 1 (k is the proportionality constant k 1 = nRT) At constant temperature

11 Using Boyle’s Law In Boyle’s Law: PV=k 1 and k 1 is the proportionality constant therefore it will be equal to nRT. For this reason, at constant temperature, a given sample of gas (with the same number of moles = n) at two different sets of conditions (different volumes and pressures) will follow the following equation: P 1 V 1 = P 2 V 2 Example: An inflated balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? P 1 V 1 = P 2 V 2 => V 2 = (P 1 / P 2 ) x V 1 (1.0 atm / 0.40 atm) x 0.55 L = 1.4 L

12 At Constant Pressure V  T

13 V = k 2 T V/T = k 2 k 2 = nR/P P  T P = k 3 T P/T = k 3 k 3 = nR/V

14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Absolute Zero = o C = 0 Kelvin At Absolute Zero, there is no volume for the gases. Volume is directly proportional to the Temperature in Kelvin. Meaning if you doubled the temp, the volume doubles!

15 Using Charles’s Law In Charles’s Law: V 1 / T 1 =k 2 and k 2 is the proportionality constant therefore it will be equal to nR / P. For this reason, at constant pressure, a given sample of gas (with the same number of moles = n) at two different sets of conditions (different volumes and temperatures) will follow the following equation: V 1 / T 1 = k 2 = V 2 / T 2 Example: A 452-mL sample of fluorine gas is heated from 22 degrees Celsius to 187 degrees Celsius at constant pressure. What is its final volume? V 2 = V 1 x (T 2 / T 1 ) = 452 mL x (460 K / 295 K) = 705 mL

16 V  n V = k 4 n k 4 = RT/P

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18 Gas Law Summary So Far: Boyle’s Law: V  1/P (at constant n and T) Charles’s Law: V  T (at constant n and P) Avogadro’s Law: V  n (at constant P and T) We can combine all three expressions to form a single master equation for the behavior of gases: V  nT / P = R (nT / P) or PV = nRT where R is the proportionality constant called the gas constant. The equation PV = nRT is called the ideal gas equation or ideal gas law. It describes the relationship among the four variables P, V, T and n.

19 The Ideal Gas Law PV = nRT An ideal gas is a hypothetical gas whose pressure- volume-temperature behavior can be completely accounted for by the ideal gas equation. The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container. Although there is no such thing as an ideal gas, discrepancies in the behavior of real gases over reasonable temperature and pressure ranges do not significantly affect calculations. Therefore, we can safely use the ideal gas law equation to solve many gas problems.

20 The Ideal Gas Law Constant, R At Standard Temperature and Pressure (STP) – O o C ( K) and 1 atm pressure, many real gases behave like an ideal gas. Experiments show that under these conditions, 1 mole of gas occupies L, which is somewhat greater than the volume of a basketball. From PV=nRT we can write: R = PV / nT R = (1 atm)( L) (1 mol)( K) R = L x atm K x mol

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22 Using the Ideal Gas Law Example: Sulfur hexafluoride (SF 6 ) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5 o C. P = nRT / V P = (1.82 mol) ( L x atm/K x mol) ( )K 5.43 L P = 9.42 atm Try this problem on your own: Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54 atm and 76 o C.

23 Molar Volume at STP Gas Calculations Example: Calculate the volume (in liters) occupied by 7.40 g of NH 3 at STP. Remember that 1 mole of an ideal gas at STP occupies 22.4 Liters. V = 7.40 g NH 3 x 1 mol NH 3 x 22.4 L g NH 3 1mol NH 3 V = 9.74 L

24 When conditions change for P, V, T and n we must use a modified form of the ideal gas equation that includes the final and initial conditions. R = P 1 V 1 (before change) n 1 T 1 R = P 2 V 2 (after change) n 2 T 2 So that…. P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 If n 1 =n 2 then, P 1 V 1 = P 2 V 2 T 1 T 2 Modified Ideal Gas Law

25 Using the Modified Ideal Gas Law Example: A small bubble rises from the bottom of a lake, where the temperature and pressure are 8 o C and 6.4 atm, to the water’s surface, where the temperature and pressure is 25 o C and 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL. Note that the moles do not change in this question. Therefore we can use the modified ideal gas law. P 1 V 1 = P 2 V 2 rearranged => V 1 P 1 T 2 = V 2 T 1 T 2 P 2 T 1 Therefore… 2.1 mL x 6.4 atm x 298 K 1.0 atm x 281 K V 2 = 14 mL

26 Electric Lightbulbs are usually filled with argon gas. Using the Modified Ideal Gas Law Example: Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.2 atm and 18 o C is heated to 85 o C at constant volume. Calculate its final pressure (in atm). P 1 V 1 = P 2 V 2 rearranged => P 1 = P 2 T 1 T 2 rearranged => P 2 = P 1 x T 2 T 1 P 2 = 1.20 atm x 358 K 291 K P 2 = 1.48 atm

27 Density Calculations If we rearrange the ideal gas equation, we can calculate the density of the gas: n = P V RT We can find the number of moles (n) by: n = mass (g = m) molar mass (g/mol = M ) Therefore, plugging the above in for n in the ideal gas law, we get: m = P M x V RT Which rearranges into: d = m = P x M V R x T

28 Finding the Molar Mass, M, of a Gas Many times we do not know the molar mass, M, of a gas. From our previous slide, we know that we can find the density of a gas from: density = m = P x M V R x T We can rearrange this equation in order to solve for the molar mass, M, of a gas: M = d x R x T P Example: A chemist has synthesized a greenish- yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36 o C and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula.

29 A Combined Gas Sample Problem Chemical analysis of a gaseous compound showed that it contained 33.0 % silicon and 67.0 % fluorine by mass. At 35 o C, L of the compound exerted a pressure of 1.70 atm. If the mass of L of the compound was 2.38 g, calculate the molecular formula of the compound. (R = L x atm/K x mol) Calculate the empirical formula: SiF 3 Calculate the molecular molar mass: M = d x R x T = 169 g/mol P The empirical molar mass of SiF 3 = g/mol. 169 / 85 = 2 Therefore the molecular formula is Si 2 F 6

30 Gas Stoichiometry When you notice that you have a gas involved in a chemical reaction, the rules for stoichiometry change. You will still convert the amount of reactants to the amount of products using moles, but remember that moles (n) vary depending on PV = nRT for gases. To solve Gas Stoichiometry problems you must read the question and: 1) Identify whether the problem states that the gas is at STP = 0 o C and 1 atm. If this is the case remember that 1 mole (n) of any gas will occupy 22.4 Liters. 2) If the problem states that each gas is not at STP, then you will have to determine the moles for each gas given based on PV = nRT for each gas.

31 Gas Stoichiometry Sample Problem Calculate the volume of O 2 (in liters) at STP required for the complete combustion of 2.64 L of acetylene (C 2 H 2 ) at STP: 2C 2 H 2(g) + 5O 2(g) 4CO 2(g) + 2H 2 O (l) 2.64 L C 2 H 2 ? L O L C 2 H 2 1 mol C 2 H 2 =.12 mol C 2 H L C 2 H 2.12 mol C 2 H 2 5 mol O L O 2 = 6.6 L O 2 2 mol C 2 H 2 1 mol O 2

32 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

33 Sodium azide (NaN 3 ) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN 3 as follows: 2 NaN 3(s) 2Na (s) + 3N 2(g) The nitrogen gas produced quickly inflates the bag between the driver and the windshield. Calculate the volume of N 2 generated at 80 o C and 823 mmHg by the decomposition of 60.0 g of NaN 3. 2 NaN 3(s) 2Na (s) + 3N 2(g) 60.0g NaN 3 ? Volume N g NaN 3 1 mol NaN 3 3 mol N 2 = 1.38 mol N g NaN 3 2 mol NaN 3 V = nRT = (1.38 mol) x ( L x atm) x ( )K = 36.9 L N 2 P (K x mol) x (1.08 atm) (Note : 760 mmHg = 1 atm and R = L x atm / K x mol)

34 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

35 Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. Therefore: The pressure exerted by gas A and B is: P A = n A RT and P B = n B RT V V P T = P A + P B = n A RT + n B RT = RT (n A + n B ) V V V X i is the mole fraction of gas i, within a mixture of gases: X i = n i n total The partial pressure, P i, of any substance, i, is equal to the mole fraction of the substance multiplied by the total pressure of a mixture of gases: P i = (X i )(P T )

36 Whenever Gas is collected over Water Use Dalton’s Law of Partial Pressure to Figure out the Pressure of the Gas Being Collected. Always look for this in problems!!!! 2KClO 3(s) 2KCl (s) + 3O 2(g) P T = P O2 + P H20 P O2 = P T - P H20

37 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. P  1 V

38 Scuba Diving and Partial Pressures The composition of the air that we breathe is 20% oxygen and 80% nitrogen gas by volume. To much oxygen can actually be harmful to us. At sea level, the pressure of the air around us is 1 atm. This means that the pressure of the oxygen we breathe is : P O =.20 moles x 1 atm = 0.2 atm = P O ( ) moles (Remember that moles (n)  V therefore we can replace the volume with moles.) Our bodies are therefore used to breathing in 0.2 atm of pressure of oxygen. But, as a diver swims deeper into the ocean, at 100 feet the pressure increases to 4 atm and the oxygen pressure now becomes: P O =.20 moles x 4 atm = 0.8 atm = P O ( ) moles This pressure of oxygen is much too high, therefore the oxygen needs to be diluted in order to keep it at a level of 0.2 atm. For this to happen, the oxygen must be diluted to a level of 5%. Nitrogen gas is not used to dilute the oxygen down because at a pressure of 1 atm, nitrogen gas in the lungs will start to dissolve into the blood. This causes nitrogen narcosis. This is also responsible for the bends. As a diver surfaces, the nitrogen dissolved in the blood transforms into bubbles of nitrogen gas and can stop blood flow and impair the nervous system. Helium is used instead to dilute oxygen down and a special valve automatically adjusts for the pressure of oxygen. Helium is used because it is an inert gas and it has a low solubility in blood.

39 The Kinetic Molecular Theory of an Ideal Gas 1.The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). They can be considered as “points”, where they posses mass but have zero volume. 2.The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. Collisions among molecules are perfectly elastic, in other words, energy can not be transferred from one molecule to another as a result of a collision, the total energy of all the molecules in a system remains the same. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Any two gases at the same temperature will have the same average kinetic energy.

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42 The Kinetic Molecular Theory and It’s Applications to the Gas Laws Boyle’s Law: V  1/P (at constant n and T) Charles’s Law: V  T (at constant n and P) P  T (at constant n and V) Avogadro’s Law: V  n (at constant P and T)

43 The Average Kinetic Energy of a Molecule Remember the term Kinetic Energy (KE)? It means molecules in motion. And do you remember that temperature is the average kinetic energy? The average KE for one molecule in a gas is represented by: KE = ½ mu 2 Where m = mass of one molecule and u = is speed. The term u 2 = mean square speed. It is the average of the square of the speeds of all of the molecules: u 2 = u u u …. + u 2 n N N = the number of molecules.

44 Maxwell Speed Distribution Curves The kinetic theory of gases allows us to investigate molecular motion in more detail. As long as we hold the temperature of a gas constant, the average KE and the means square speed will stay constant for the same length of time that the temperature is held constant. The motion of the molecules are totally random and unpredictable. At a given instant, different molecules within a sample are moving at different speeds. How many molecules are moving at a particular speed? Maxwell Boltzman analyzed the behavior of gas molecules at different temperatures to answer this question. Maxwell speed distribution curves show gas at different temperatures and graph the number of molecules that move at a particular molecular speed.

45 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

46 Root-Mean-Square Speed How fast does a molecule move, on the average, at any temperature T? One way to estimate molecular speed is to calculate the root- mean-square (rms) speed (u rms ), which is an average molecular speed. The Total KE of a mole of any gas = 3/2 RT Because Average KE is only for one molecule of a gas: N A (1/2mu 2 ) = 3/2 RT Because N A m = M, we can write: u 2 = 3RT / M Taking the square root of both sides gives us: u 2 = u rms = 3RT / M

47 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

48 Diffusion A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Diffusion always proceeds from areas of higher concentrations to areas of lower concentrations. The diffusion process takes a long time to complete even if the molecular speeds are great. This is because molecules experience numerous collisions while moving from one end of the bench to the other. Also, because the root mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Because NH 3 is lighter and therefore diffuses heavier than HCl, the solid NH 4 Cl first appears nearer the HCl bottle on the right. Calculating the ratio of the effusion rates of two gases or the distance traveled for two gases: (Distance traveled/Effusion rate) for gas 1: u rms = 3RT / M 1 = M 1 (Distance traveled/Effusion rate) for gas 2: u rms = 3RT / M 2 M 2

49 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

50 Deviation from Ideal Behavior Ideal gases are said to: 1)Not exert any attractive or repulsive forces upon one another. 2)Have negligibly small volumes compared to that of the container. To study real gases, the van der Waals equation is used: (P + an 2 /V 2 ) (V – nb) = nRT Where a and b are van der Waals constants that can be looked up in an appendix or table. Corrected Pressure Corrected Volume


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