Gases. Gases and the Kinetic Molecular Theory. Chapter 5. Gases have different physical properties compared to liquids and solids. width.

Transcription

1 Gases Chapter 5 Gases and the Kinetic Molecular Theory 5.1 An Overview of the hysical States of Matter 5.2 Gas ressure and Its Measurement 5.3 The Gas Laws and Their Experimental Foundations 5.4 Further Applications of the Ideal Gas Law 5.5 The Ideal Gas Law and Reaction Stoichiometry 5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior 5.7 Real Gases: Deviations from Ideal Behavior Gases have different physical properties compared to liquids and solids 1. Gas are compressible 2. Gases exert pressure 3. Gases have low densities and low viscosities. 4. Gas volume changes with temperature. 5. Gas volume changes with pressure 6. Gases assume shape of container. Gas Liquid Solid Life would not exist without gases. Green plants consume and reduce CO2 to sugars and cellulose. During that process O2 in CO2 is reduced and released in the atmosphere. N2 is reduced to NO3 - by bacteria in the ground which is used by plants to make proteins. Some elements exist naturally on our planet as gases at 25 C and osphere. molecular gases H 2, N 2, O 2, F 2, Cl 2 atomic gases He,Ne,Ar,Kr,Xe ressure is the force exerted on a unit area of surface. Force = weight = mg mg width ressure = = Force Area = mg Area length Area = length x width dimensions Many covalent compounds are also gases: CO, CO2, NH3, NO, NO2 N2O, SO2, SO3, H2S, HCN kg m/s 2 m 2 kg = m s 2 = 1 ascal SI Unit of ressure

2 Chemists use various units to describe pressure. Memorize these. What is the pressure in atmospheres, pascals, mmhg, psi of a barometer that reads 739 torr? Unit atmosphere (atm) mm of (Hg) torr bar pascal(a) Atmospheric ressure 760 mm Hg* 760 torr* lb/in 2 (psi) 14.7 lb/in bar x 10 5 a KNOW How to use These # atmospheres = 739 torr X 760 torr 101,325 ascals ascals = atm X psi = atm X = atm = 9.85 X lbs/in 2 = 14.3 psi The pressure of the atmosphere is measured with a barometer. A glass tube closed at one end is filled with mercury & inverted into the same liquid creates a vacuum above the column. The weight of the earth s atmosphere applies a force (F= mg) that supports the weight of the mercury in the column to a height, h. vacuum in column At sea level = 760 mm Hg = Atmospheric pressure The pressure exerted on a column of liquid is proportional to the density of the liquid in the column and the height of the column. F = Area = = D g h mg Area = mg V / h = m g h V ressure = Density x GC x Height kg m/s 2 = kg/m 3 x m/s 2 x m m 2 kg m s 2 = 1 ascal SI Unit = D g h ressure exerted by the earth s atmosphere varies with elevation (height). fraction of (m) altitude (ft) 0 0 1/2 atm 5,486 18,000 1/3 atm 8,376 27,480 1/10 atm 16,132 52,926 1/100 atm 30, ,381 1/1000 atm 48, ,013 1/10000 atm 69, ,899 1/ , ,076 Mount Everest eak = 8848 m Water boils at ~72 C at the peak. Centuries of experiments have shown that 4 macroscopic variables are needed to completely describe the state of a gas : pressure, volume, temperature, number of moles. Boyle s law: V! 1 (at constant n and T) Charles law: V! T (at constant n and ) Avogadro s law: V! n (at constant and T) Consolidated into 1 equation V = n Ideal Gas Law!

3 ressure Consolidation of the historic laws. Boyle s law: 1 V = k (at constant n and T) Charles law: V = k1 T (at constant n and ) Avogadro s law: V = k2 n (at constant and T) Ideal Gas Law: V = k x k1 x k2 n T ressure (atm) Volume (L) V = n moles Temperature (Kelvin) = R n T Gas Constant = L atm mol -1 K -1 Boyles Law: At constant temperature (in Kelvin) and constant moles of gas, the pressure of a gas is inversely proportional to the gas volume..6.3 Final Condition 2 at V2 2 Volume 4! 1/V = k / V V = k Initial condition 1 at V1 Charles Law: At constant pressure and moles of gas, the volume of a of gas is directly proportional to the absolute temperature in kelvin. V! T V = k x T Tubing Avogadro s Law: At a constant temperature and pressure, the volume of a gas is proportional to the number of moles of the gas. V! n V = k n T= 25 C Hg Gas Volume (L) Volume (L) Low Temp High Temp Temp ( C) moles Temp ( C) gas Avogadro s Law says at constant and T, we can also use volumes as stoichiometric factors in chemical equations (as n => V) 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 constant temperature and constant pressure? NH3(g) + O 2 (g) NO(g) + H 2 O(g) Translation 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Balance Stoich from the equation Avagardro s Law Avogadro s Law 4 mole NH 3 4 mole NO At constant T and 1 volume NH 3 1 volume NO

4 Science has defined a standard laboratory temperature and pressure called ST. Memorize ST means! What is the volume occupied by 1 mole of any gas under conditions of ST? ST means: Temperature = K (0 C) ressure: = = 760 torr V = n V = n V = (1 mol)( L atm mol 1 K 1 )273.15K = 22.4 L Experiments show that at ST, one mole of any gas occupies 22.4 L. That s weird! Hereʼs is how the gas law consolidation looks mathematically. Boyle s law: 1 V = k (at constant n and T) Charles law: V = k1 T (at constant n and ) Avogadro s law: V = k2 n (at constant and T) N2 O2 Ideal Gas Law: V = k x k1 x k2 n T ressure (atm) Volume (L) V = n moles Temperature (Kelvin) = R n T Gas Constant = L atm mol -1 K -1 Mastering gas law problems. 1. You do not need to remember Boyle s Law, Charles Law, Avogadro s law to solve any problem...just V = n that it. 2. Memorize: V = n especially units 3. Watch for the got-cha sucker traps--almost always unit conversions traps. 4. When multiple variables are changing, place variables on one side, constants on the other. What is the volume (in liters) occupied by 49.8 g of HCl at ST? Remember that ST means: T = 0 0 C = K V = n V = n = 1 mol HCl n = 49.8 g x = 1.37 mol g HCl V = Latm 1.37 mol x x K molk V = 30.6 L

5 A bottle of chlorine gas occupies a volume of 946 ml at a pressure of 726 mmhg. What is the pressure of the gas (in mm Hg) if the gas is compressed at constant temperature to 154 ml in this problem n are all constants--put them on the right 2 = 1 x V 1 V 2 1 = 726 mmhg V 1 = 946 ml V = n = k V = constant and thus: 1 V 1 = 2 V 2 2 =? V 2 = 154 ml 726 mmhg x 946 ml = = 4460 mmhg 154 ml Announcements --Exam 2 Aug 29 6:00-7:30M Coverage Chapters 4-6. lease see blog for skipped material in Chapter 5 and 6. Chapter 5 roblems 2, 8, 14, 17, 22, 5.31*, 5.41*, 5.43*, 44*, 48, 5.52*, 5.54, 58, 69*, 70*, 72*, (rinciples of Chemistry) 2, 8, 11, 16, 19, 21, 24, 25, 27, 39, 41, 44, 46, 52, 56, 60*, 63, 69, 80,81 (4th Edition) Start Chapter 6 next class A sample of carbon monoxide gas maintained at constant pressure occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L? V 1 T 1 = V 2 T 2 = Constant V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 =? T 1 = 125 ( 0 C) (K) = K A bottle of gas occupies a volume of 50.0 L at 21.5 atm. Suppose the tank is connected to another empty tank, the valve opened to expand the gas into throughout the volume of the two tanks. The pressure at the gage is 1.55 atm. What is the volume of the second tank? Initial Condition T 2 = V 2 x T 1 V L x K = = 192 K 3.20 L A bottle of gas occupies a volume of 50.0L at 21.5 atm. Suppose the tank is connected to another empty tank, the valve opened to expand the gas into throughout the volume of the two tanks. The pressure at the gage is 1.55 atm. What is the volume of the second tank? 1. Identify constants = n, T, R and variables 2. Recall gas law: V = n R T = Constant 3. Toss constants to the right, variable to the left: V = n R T = Constant 4. Using the variables, now write the proportion: 1 V 1 = 2 V 2 V2 = 1 V 1 2 = 694 L Vtank = 694L L = 644L A sample of methane, CH 4, occupies 2.60 x 10 2 ml at 32 o C under a pressure of atm. At what temperature would it occupy 5.00 x 10 2 ml under a pressure of 1.20 x 10 3 torr? Identify and place constants to the right, variables to the left: V/T = n R = Constant 1 V 1 = 2 V 2 T1 T2

6 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)? V = n nr V = T = constant 1 2 = 1 x T 2 T 1 n, V and R are constant = 2 T 1 T 2 1 = 1.20 atm T 1 = 291 K 2 =? T 2 = 358 K = 1.20 atm x 358 K = 1.48 atm 291 K We can extend the usefulness of the gas law to molecular weight and density of a gas! Molar Mass (M ) V = n = MW m MW = m = V Density (d) from the Ideal Gas Equation d = m V = MW d m/v = d is the density of the gas in g/l m is the mass of the gas in g M is the molar mass of the gas Rate the gases according to density from lowest to highest assuming all of the gases are maintained at the ST? H2 Cl2 CO2 NH3 2.0 g/mol 71. g/mol 44. g/mol 17 g/mol density = m V = MW MW = A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? V = n = m MW MW = ( m V ) = d solving for MW d = density of gas ( 4.65 g 2.10 L )( L atm mol 1 K 1 ) (300.15K) = 54.6 g/mol 1.00 atm Find the density (in g/l) of CO 2 and the number of molecules (a) at ST (0 0 C and ) and (b) at room conditions (20. 0 C and 1.00 atm). V = n = m MW d = m V = (MW)() (44.01 g/mol)() d = L atm mol 1 K 1 = 1.96 g/l (273.15K) 1.96 g molecules CO 2 /L = X X L mol CO x10 23 molecules g CO 2 mol = 2.68x10 22 molecules CO 2 /L Find the density (in g/l) of CO 2 and the number of molecules (b) at room conditions (20. 0 C and 1.00 atm). V = n = m MW 1.83g L d = m V = (MW)() d = g/mol x atm*l x 293K mol*k = 1.83 g/l mol CO x10 23 molecules 44.01g CO 2 mol = 2.50x10 22 molecules CO 2 /L

7 1. Use the Gas Equation to get the expressions for molar and density : = pressure (atm) V = volume (L) m = mass gas (g) M = molar mass (g/mol) R = gas constant T = Temperature (K) V = n m V = M m M = V 2. Determine V flask : V flask = m H 2O / d H2O = ( g g) / ( g cm-3 ) = cm 3 = L 3. Determine m gas : m gas = m filled - m empty = ( g g) = g M = The ideal gas law ( g)( L atm mol -1 K -1 )(297.2 K) M = g/mol M = m V (0.974)( L) Dalton s Law of artial ressure: the total pressure of a mixture of gases is the sum of the pressures of the individual gases. Gas A + Gas B Mixture A+B a + b Total na V + nb V = (na + nb) = total V The partial pressure i in a mixture of gases in terms of the total pressure of gas, T and the mole fraction of each gas in the mixture. T A T = n A + n B + n C T = (n A + n B + n C ) A T = = n A n T = A B = n A n T n A n T n A n 1 + n 2 + n T = n B n T T i = X i T Dalton s Law Factor out /V Take ratio of A to T X i is called the mole fraction of the ith component. A sample of natural gas in a flask contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is measured to be 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? Dalton s Law says for mixtures of gases : X propane = i = X i T 3 gases mixed T = 1.37 atm We can calcuate the mole fraction of propane = propane = x 1.37 atm = STOICHIOMETRY WITH GASES: We can use the ideal gas to link moles of a substance to,v,t and n.,v,t of Gas A n = V moles of Gas A,V,T of Gas B What is the volume of CO 2 produced at constant 37 C and 1.00 atm when 5.60 g of glucose is completely consumed in the following reaction? C 6 H 12 O 6 (s) + 6O 2 (g) stoichiometric factors from balanced equation mole of Gas B n = V 6CO 2 (g) + 6H 2 O (l)

8 What is the volume of CO 2 produced at 37 C and 1.00 atm when 5.60 g of glucose is completely consumed in the following reaction? C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) At high temperatures sodium azide, NaN 3, decomposes with electrical current producing molecular nitrogen gas and elemental sodium as products. What volume of N 2 (g), measured at 735 mm Hg and 26 C, is produced when 70.0 g NaN 3 is decomposed. g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 1 mol C 5.60 g C 6 H 12 O 6 H 12 O 6 6 x 180 g C 6 H 12 O 6 V = n 6 mol CO x 2 = mol CO 1 mol C 6 H 12 O 2 6 Latm mol x x K molk = = 4.76 L 1.00 atm What is asked for? Translates words to balanced equations Determine moles of N 2 : n N 2 = 70.0 g NaN 3 1 mol NaN g N 3 /mol N 3 X 3 mol N 2 2 mol NaN 3 = 1.62 mol N 2 Determine volume of N 2 : n V = = 2 NaN 3 (s)! 2 Na(l) + 3 N 2 (g) (1.62 mol)( L atm mol -1 K -1 )(299 K) = 41.1 L (735 mm Hg) 1.00 atm 760 mm Hg Announcements --Exam 2 August 29 Homework 6:00-7:30M Chapter 4: 9, 11, 13, 15, 17, 19, 23, 27, 33, 39, 42, 43, 44, 51, 57, 59, 61, 66, 71, 83 Chapter 5: 2, 8, 14, 17, 22, 5.31*, 5.41*, 5.43*, 44*, 48, 5.52*, 5.54, 58, 69*, 70*, 72*, (rinciples of Chemistry) Chapter 6: 6, 8, 10, 12, 18, 23, 24, 27*, 36*, 39*,47*, 48*, 53*, 56*, 59* (rinciples of Chemistry) Excluded Topics Chapter 4/5/6 rinciple of Chemistry 2nd Edition Chapter 4: Nothing excluded. Chapter 5: Collecting A Gas Over Water p Relationship between KE and Temperature: p Effusion and Diffusion p The Ideal Gas Law in a Limiting-Reactant roblem The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at atm and 293K reacts with 17.0 g of potassium? Chapter 6: Constant Volume Calorimeter p Fossil Fuels and Climate Change p

9 What is asked for? Translates nomenclature to a balanced equation = atm Determine moles of Cl 2 : gram KCl =? 2K(s) + Cl 2 (g) 2KCl(s) V = 5.25 L T = 293K ncl2 = unknown n = V atm Cl = 2 x 5.25L = mol Cl2 atm*l x 293K mol*k 2 mol KCl mol KCl formed = mol Cl 2 = mol KCl formed 1 mol Cl 2 1 mol K 2 mol KCl mol KCl formed = 17.0 g X g K 2 mol K = mol KCl formed From Big (Macro) To Little (Micro) V = n gives a macroscopic description of the behavior of gases. How do we relate, V, n, T to atoms and molecules that we know exist on a microscopic level? The math and physics that bridges the macroscopic V = n to the microscopic world is called: Kinetic Molecular Theory Cl 2 is the limiting reactant. g KCl formed = mol KCl X g KCl mol KCl = 30.9 g KCl Big To Little Questions About Gases How do molecules make pressure in a container? molecular level? is pressure proportional to a gas mole fraction? What is temperature on a molecular scale? How do we explain Boyle s, Charle s, Avogadro s Laws on a How does Dalton s artial law arise microscopically or how Why does gas volume only depend on number of molecules (moles) and not the size of them? Shouldn t one mole of a larger molecule occupy more volume (Avogadro s Law)? Kinetic Molecular Theory of Gases ostulates: 1. The volume occupied by gas particles is negligible compared to the volume of container of the gas. 2. Gas molecules are in constant random motion. 3. Collisions among molecules are perfectly elastic, no kinetic energy is lost in collisions and remains constant. 4. Gas molecules exert neither attractive nor repulsive forces on one another (i.e. there are no intermolecular forces). The average kinetic energy of a gas is determined only by the absolute temperature of the gas. KE = Ek = 3 2 = 1 2 mu2 Gases at the same temperature have the same kinetic energy! KE = kinetic energy; R = 8.314J/K. mol T = Kelvin urms = 3! M Average speed u rms of a gas particle is inversely proportional to molar mass and proportional to T! Lighter gases have faster speeds for a given temperature! Questions: Molecular Origins of Gas Law 1. Origin of ressure--arises from gas molecules colliding with the container walls. The more molecules the more frequent the # of collisions => higher pressure. 2. Boyle s Law ( 1/V) Decreasing V increases collisions between gas particles and walls of container. 3. Charles Law (V T) Higher T at constant n and = faster particles velocity and higher KE. V must increase to reduce collisions of particles to keep constant. 4. Avogardro s Law (V n) If n increases at constant and T the more collisions unless volume increases. 6. Dalton s Law of artial ressure: Total of a fixed volume of gas depends only T and number of moles of particles.

10 A molecular description of Boyle s Law.! 1 V A molecular description of Charles s Law. V! T Constant T and n Higher temperature increases KE of gas molecules which would increase internal gas pressure. The volume must increase. If volume is decreased, then the frequency of collisions in the container increases and pressure increases (constant T). A molecular description of Avogadro s Law. V! n A molecular description of Dalton s law of partial pressures. Constant and T Each gas acts independently of one another. The more moles of the more collisions the higher the pressure. Additional molecules would increase collisions and therefore pressure---but external pressure is held constant experimentally---so the volume must increase. The ideal gas law works well at high T and low. Van der Waals Equation is used to predict the behavior of non-ideal gases at low T and high. The ideal gas law fails at high pressures and low temperatures. Why? Ideal Gas Law V = n 2 observed Vcontainer correction due to actual volume of molecules 1. At high pressure, the volume occupied by the gas becomes significant portion of total volume. (the assumption-gas takes up zero volume fails) 2. Intermolecular forces between gas particles alters elastic collisions reducing ideal correction for intermolecular attraction forces } } correction for intermolecular attraction forces (V nb) = n ( + an V2 ) } } Van der Waals Equation 2 (V nb) = n ( + an V2 ) measured Constant and n Vcontainer correction due to actual volume of molecules The van der Waals equation corrects the ideal gas law for two things: 1. a accounts for intermolecular forces of attraction 2. b accounts for volume of gas molecules

11 For 1 mole of ideal gas V = 1() n = V = 1.0 Deviations from Ideal Behavior Repulsive Forces Non-Ideal Behavior V Attractive Forces (atm) V/ > 1 Volume affects dominates Ideal Gas Line V/ < 1 IMF attraction forces dominates Graphic showing the effect of molecular volume that occurs at high pressure on the ideal gas volume. At low temperatures the KE of molecules is very low (speeds are low). Interaction between molecules attenuate pressure and force of collisions. Low ressure High ressure Low ressure High ressure Free Volume = V ideal Free Volume = V acutal < V ideal ideal = higher-- no molecular interactions acutal < ideal because of interactions, we add a factor to account for it.