Presentation on theme: "Chapter 5: the Gaseous state"— Presentation transcript:
1Chapter 5: the Gaseous state Vanessa Prasad-PermaulValencia Community CollegeCHM 1045
2GAS LAWS: PRESSURE AND MEASUREMENT Elements that exist as gases at 250C and 1 atmosphere
3GAS LAWS: PRESSURE AND MEASUREMENT Physical Characteristics of GasesGases 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.
4GAS LAWS: PRESSURE AND MEASUREMENT a) Gas is a large collection of particles moving at random throughout a volumeb) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure
5GAS LAWS: PRESSURE AND MEASUREMENT HOW IS PRESSURE DEFINED?The force the gas exerts on a given area of the container in which it is contained.The SI unit for pressure is the Pascal, Pa.Pressure =ForceAreaIf you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).
6GAS LAWS: PRESSURE AND MEASUREMENT Barometer760mmmercuryHeight is proportionalto the barometric pressureUnits of Pressure1 pascal (Pa) = 1 kg/m·s21 atm = 760 mmHg = 760 torr1 atm = 101,325 Pa1 bar = 105 Pa1 atm = lb/in2Hg is used instead of H2O :more densebetter visibilityaccuracy
7GAS LAWS: PRESSURE AND MEASUREMENT EXERCISE 5.1A GAS CONTAINER HAD A MEASURED PRESSURE OF57kPa. CALCULATE THE PRESSURE IN UNITS OF ATMAND mmHg.First, convert to atm (57 kPa = 57 x 103 Pa).57 x 103 Pa x atm = = atmx 105 PaNext, convert to mmHg.57 x 103 Pa x mmHg = = 4.3 x 102 mmHg
8GAS LAWS: EMPIRICAL GAS LAWS You can predict the behavior of gases based on the following properties:PressureVolumeAmount (moles)Temperature* If two of these physical properties are held constant,it is possible to show a simple relationship between the other two…
9GAS LAWS: EMPIRICAL GAS LAWS Boyle’s experiment:A manometer to study the relationship betweenpressure (P) & Volume (V) of a gasHgHgAs P (h) increasesV decreases
10GAS LAWS: EMPIRICAL GAS LAWS BOYLE’S LAWthe volume of a sample of gas at a given temperaturevaries inversely with the applied pressureSo….For a given amount of gas (n)&@ constant temperature (T) :If pressure (P) increases, the volume (V) of the gas decreasesP1 * V1 = P2 * V2
12GAS LAWS: EMPIRICAL GAS LAWS Boyle’s LawP a 1/VP * V = constantP1 * V1 = P2 * V2Constant temperatureAny given amount of gas
13Application of Boyle’s law gives GAS LAWS: EMPIRICAL GAS LAWSEXERCISE 5.2A volume of carbon dioxide gas equivalent to 20.0 L was 23oC and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm?P1 * V1 = P2 * V2Application of Boyle’s law givesV2 = V1 x P1 = L x 1.00atm = = LP atm
14Charles’ Law: relating volume and temperature GAS LAWS: EMPIRICAL GAS LAWSCharles’ Law: relating volume and temperatureAs T increasesV increases
16GAS LAWS: EMPIRICAL GAS LAWS Charles’ LawVariation of gas volume with temperatureV a TFor a given amount of gasat constant pressureTemperature mustbe in KelvinV / T = constantT (K) = t (0C)
17GAS LAWS: EMPIRICAL GAS LAWS CHARLES’ LAWthe volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperatureSo….For a given amount of gas (n)&@ constant pressure (P) :If temperature (T) increases, the volume (V) of the gas increasesV V2T1 = T2
18GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.3A chemical reaction is expected to produce 4.3dm3 of oxygen at 19oC and 101kPa. What will the volume be at constant pressure and 25oC?First, convert the temperatures to the Kelvin scale.Ti = ( ) = 292 KTf = ( ) = 298 KFollowing is the data table.Vi = dm3 Pi = 101 kPa Ti = 292 KVf = ? Pf = 101 kPa Tf = 298 KApply Charles’s law to obtainVf = Vi x Tf = dm3 x 298K = = dm3Ti K
19COMBINED GAS LAW Relating Volume, Temperature and Pressure GAS LAWS: EMPIRICAL GAS LAWSCOMBINED GAS LAWRelating Volume, Temperature and PressureTaking Boyle’s Law and Charles’ Law:The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure:V = constant x T or PV = constant (for a given amount of gas)P TP1V1 = P2V2T T2
20Use Kelvins for temp, any pressure, any volume Combined Gas LawWe can combine Boyle’s and charles’ law to come up with the combined gas lawUse Kelvins for temp, any pressure, any volume
21EXERCISE 5.4 A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa. GAS LAWS: EMPIRICAL GAS LAWSEXERCISE 5.4A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa.Suppose the gas in the balloon is heated to 35oC and thepressure is now 102.8kPa, what is the volume of the gas?First, convert the temperatures to kelvins.Ti = ( ) = 297 KTf = ( ) = 308 KFollowing is the data table.Vi = dm3 Pi = kPa Ti = 297 KVf = ? Pf = kPa Tf = 308 KApply both Boyle’s law and Charles’s law combined to getVf = Vi x Pi x Tf =5.41 dm3 x 101.5kPa x 308K = = 5.54 dm3Pf Ti kPa K
22AVOGADRO’S LAW relating volume and amount GAS LAWS: EMPIRICAL GAS LAWSAVOGADRO’S LAW relating volume and amountequal volumes of any two gases at the sametemperature & pressure contain the same number of moleculesV a number of moles (n)V = constant x nV1/n1 = V2/n2Constant temperatureConstant pressure
24V1 = V2 n1 n2 The Volume–Amount Law (Avogadro’s Law): At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present.Use any volume and molesV1 = V2n n2
25GAS LAWS: EMPIRICAL GAS LAWS Avogadro’s Number = one mole of any gas contains the same number of molecules x 1023Must occupy the same volume at a given temperature and pressureThe conditions 0 0C and 1 atm are called standard temperature and pressure (STP).Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.
26IDEAL GAS LAW GAS LAWS: THE IDEAL GAS LAW Boyle’s law: V a (at constant n and T)1PCharles’ law: V a T (at constant n and P)Avogadro’s law: V a n (at constant P and T)V anTPV = constant x = RnTPR is the molargas constantPV = nRT
27IDEAL GAS EQUATION GAS LAWS: THE IDEAL GAS LAW PV = nRT PV nT=(1 atm)(22.414L)(1 mol)( K)R = L • atm / (mol • K)R = J / (mol • K)R = cal / (mol • K)*The units of pressure times volume are the units of energyjoules (J) or calories (cal)
281 mole of an ideal gas occupies 22.414 L at STP The Ideal Gas LawIdeal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.1 mole of an ideal gas occupies L at STPSTP conditions are K and 1 atm pressureThe gas constant R = L·atm·K–1·mol–1P has to be in atmV has to be in LT has to be in K
29Or, expressing this as a proportion, GAS LAWS: THE IDEAL GAS LAWEXERCISE 5.5Show that the moles of gas are proportional to the pressure forconstant volume and temperatureUse the ideal gas law, PV = nRT, and solve for n:n = PVRTn = V x PNote that everything in parentheses is constant. Therefore, you can writen = constant x POr, expressing this as a proportion,n P
30GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.6What is the pressure in a 50.0L gas cylinder that contains3.03kg of oxygen at 23oC?T = 23oC + 273K = 296KV = 50.0LR = L . atm/K . moln = 3.03kg x 1000g x 1mol = mol O21 kg gP = ?PV= nRTP = nRT = mol x L . atm x 296KV K . mol = 46.0 atm50.0L
31Density and Molar Mass Calculations: The Ideal Gas LawDensity and Molar Mass Calculations:You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.
32GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.7Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium?VariableValueP752 mmHg x 1 atm = atm760mmHgV1 L (exact number)T( ) = 294 Kn?32
33GAS LAWS: THE IDEAL GAS LAW Using the ideal gas law, solve for n, the moles of helium.n = PV = atm x 1L =RT L . atm/ K . mol x 294KmolNow convert mol He to grams.mol He x g He = g He = g/L1.00 mol He1 LTherefore, the density of He at 21°C and 752 mmHg is g/L.The difference in mass between one liter of air and one liter of He:mass air − mass He = g − g = = g difference
34GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.8A sample of a gaseous substance at 25oC and atm has a density of 2.26 g/L. What is the molecular mass of the substance?Variable ValueP atmV L (exact number)T ( ) = 298 Kn ?From the ideal gas law, PV = nRT, you obtainn = PV = atm x 1 L = molRT L . atm/K . Mol x 298K3434
35GAS LAWS: THE IDEAL GAS LAW Dividing the mass of the gas by moles gives you the mass per mole (the molar mass).Molar mass = grams of gas = g = g/molmoles of gas molTherefore, the molecular mass is 64.1 amu.
36Dalton’s Law of Partial Pressures GAS LAWS: GAS MIXTURESDalton’s Law of Partial PressuresV and T are constantP1P2Ptotal = P1 + P2
37Dalton’s Law of Partial Pressures For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB).Mole fraction is related to the total pressure by:
38Each gas obeys the ideal gas law. GAS LAWS: GAS MIXTURESEXERCISE 5.10A 10.0L flask contains 1.031g O2 and 0.572g CO2 at 18oC. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture?Each gas obeys the ideal gas law.1.031 g O2 x 1 mol = mol O232.00gP = nRT = mol x L.atm/K.mol x 291K = atmV L0.572 g CO2 x 1 mol = mol CO244.01gP = nRT = mol x L.atm/K.mol x 291K = atm
39The total pressure is equal to the sum of the partial pressures: GAS LAWS: GAS MIXTURESThe total pressure is equal to the sum of the partial pressures:PT = PO2 + PCO2 = atm atm == atmThe mole fraction of oxygen in the mixture isMole fraction O2 = PO2 = atm = =PT atm0.712 x 100% = mole % of O2 in this gas mixture
40Kinetic Molecular Theory This theory presents physical properties of gases in terms of the motion of individual molecules.Average Kinetic Energy Kelvin TemperatureGas molecules are points separated by a great distanceParticle volume is negligible compared to gas volumeGas molecules are in constant random motionGas collisions are perfectly elasticGas molecules experience no attraction or repulsion
42Average Kinetic Energy (KE) is given by: U = average speed of a gas particleR = J/K molm = mass in kgMM = molar mass, in kg/molNA = x 1023
43The Root–Mean–Square Speed: is a measure of the average molecular speed. Taking square root of both sides gives the equation
44Example 17:Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C.
45Maxwell speed distribution curves. Kinetic Molecular TheoryMaxwell speed distribution curves.
46Graham’s LawDiffusion is the mixing of different gases by random molecular motion and collision.
47Graham’s LawEffusion is when gas molecules escape without collision, through a tiny hole into a vacuum.
48Graham’s Law: Rate of effusion is proportional to its rms speed, urms. For two gases at same temperature and pressure:
49Behavior of Real GasesAt higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.
50Behavior of Real GasesThe volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.
51IntermolecularAttractions Behavior of Real GasesCorrections for non-ideality require van der Waals equation.ExcludedVolumeIntermolecularAttractions
52Example 1: Boyle’s LawA sample of argon gas has a volume of 14.5 L at 1.56 atm of pressure. What would the pressure be if the gas was compressed to 10.5 L? (at constant temperature and moles of gas)
53Example 2: Charles’ LawA sample of CO2(g) at 35C has a volume of 8.56 x10-4 L. What would the resulting volume be if we increased the temperature to 85C? (at constant moles and pressure)
54Example 3: Avogadro’s Law 6.53 moles of O2(g) has a volume of 146 L. If we decreased the number of moles of oxygen to 3.94 moles what would be the resulting volume? (constant pressure and temperature)
55Example 4:Combined Gas Law Oxygen gas is normally sold in 49.0 L steel containers at a pressure of atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?
56Example 5: Gas LawsAn inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, 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?
57Example 6: Ideal Gas LawSulfur hexafluoride (SF6) 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°C.
58What is the volume (in liters) occupied by 7.40 g of CO2 at STP? Example 7: Ideal Gas LawWhat is the volume (in liters) occupied by 7.40 g of CO2 at STP?
59What is the molar mass of a gas with a density of 1.342 g/L at STP? Example 8: Density & MMWhat is the molar mass of a gas with a density of g/L at STP?
60Example 9: Density & MMWhat is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP?
61Example 10: Density & MMThe density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass?
62Example 11: DaltonExactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25oC. What are the partial pressures of each gas and the total pressure?
63Example 12: DaltonA sample of natural gas contains 6.25 moles of methane (CH4), moles of ethane (C2H6), and moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases?
64Example 13: Mole Fraction What is the mole fraction of each component in a mixture of g of H2, g of N2, and 2.38 g of NH3?
65Example 14: Partial Pressure On a humid day in summer, the mole fraction of gaseous H2O (water vapor) in the air at 25°C can be as high as Assuming a total pressure of atm, what is the partial pressure (in atm) of H2O in the air?
66Gas Stoichiometry & Example In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles.Example: Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)
67Example 15:All of the mole fractions of elements in a given compound must add up to?1001502
68Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq) Example 16:Hydrogen gas, H2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react?Zn(s) HCl(aq) H2(g) + ZnCl2(aq)
69Example 18:Under the same conditions, an unknown gas diffuses times as fast as sulfur hexafluoride, SF6 (MM = 146 g/mol). What is the identity of the unknown gas if it is also a hexafluoride?
70Example 19: DiffusionWhat are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne?
71Deviations result from assumptions about ideal gases. Behavior of Real GasesDeviations result from assumptions about ideal gases.1. Molecules in gaseous state do not exertany force, either attractive or repulsive, onone another.2. Volume of the molecules is negligibly smallcompared with that of the container.
72Example 20: Ideal Vs. Van Der Waals Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) usingthe ideal gas equation(b) the van der Waals equation. (a = 4.17, b = )
73Example 21: Ideal Vs. Van Der Waals Assume that you have mol of N2 in a volume of L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation.For N2, a = 1.35 L2·atm mol–2, and b = L/mol.