Chemistry II

Chemistry IIUnit 1 GasesThe Nature of GasesObjectives:Describe the assumption of the kinetic theory as it applies to gases.Interpret gas pressure in terms of kinetic theoryDefine the relationship between Kelvin temperature and average kinetic energy.Explain why gases are easier to compress than solids or liquids.Describe the 3 factors that affect gas pressure.The Kinetic Theory

Properties of Gases

Expand to fill their containertake the shape of their containerlow densityCompressibleCompressibility measures how much the volume of matter decreases under pressure.mixtures of gases are always homogeneousFluids (flow)

Gas pressureResults from collisions of gas particles with an object.In empty space where there are no particles, there is no pressure and is called a vacuum.Atmospheric pressure (air pressure): due to atoms and molecules in air. Barometer: used to measure atmospheric pressure.

Units for measuring pressure:Pascal (Pa)Standard atmosphere (atm)Millimeters of mercury (mmHg) 1 atm = 760 mmHg = 101.3 kPa1kpa = 1000 paStandard pressure: 1 atmFactors affecting gas pressureAmount of gas Volume TemperatureStandard temperature : 0C (273K)

Converting between units of pressureA pressure gauge records a pressure of 450 kPa. What is the measurement expressed in atmospheres and millimeters of mercury?For converting to atm:450 kpa x 1 atm = 4.4 atm 1013.kPaFor converting to mmHg:

450kPa x 760 mmHg = 3.4 x 103 mmHg 101.3 kPa What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg?

The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm?

What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg? 51.3 kPa, 0.507 atm

The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm?33.7 kPa is greater than 0.25 atmKinetic energy and temperatureThe Kelvin temperature of a substance is directly proportional to the average kinetic energy of the particles of the substance.Directly proportional means that temperatures increases as the average kinetic energy increases or decreases as the average kinetic energy decreases.K = C + 273

~78%The Atmosphere an ocean of gasesmixed together Composition nitrogen (N2).. oxygen (O2) argon (Ar)... carbon dioxide (CO2) water vapor (H2O). Trace amounts of:

~21%~1%~0.04%~0.1%He, Ne, Rn, SO2, CH4, NxOx, etc. Depletion of the Ozone LayerO3 depletion is caused by chlorofluorocarbons (CFCs). Uses for CFCs: refrigerants Ozone (O3) in upper atmosphereblocks ultraviolet (UV) light from Sun.UV causes skin cancer and cataracts. O3 is replenished with each strike of lightning. aerosol propellants -- banned in U.S. in 1996

CFCs

12Reaction_to_Air_Pressure_Below_Sea_Level.asf

Classwork:Read pages 103-105Do problems 1-6

Gas LawsObjectivesDescribe the relationships among the temperature, pressure, and volume of a gasUse the gas laws to solve problemsBoyles Law : Pressure and VolumeStates that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure.If pressure increases, volume decreases; if pressure decreases, volume increases.

P1 x V1 = P2 x V2 P: pressure1: initial condition V: volume2: final condition

Using Boyles LawA balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)?

Using Boyles LawA balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)?P1=103 kPaV1= 30.0LP2= 25.0kPa V2=?

P1V1= P2V2 V2= P1V1 = (103 kPa)(30.0L) = 124 L P2 (25.0 kPa)Since pressure decreases, you expect volume to increase.

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)?

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)?P1= 205 kPa V1= 4.00LP2=?V2=12.0L

P1V1=P2V2 P2= P1V1 = (205 kPa)( 4.00 L) = 68.3 kPa V2(12.0L)

Classwork: p 121 # 2-6Charless Law: Temperature and VolumeStates that the temperature of an enclosed gas varies directly with the volume at constant pressure.As temperature increases, volume increases.

V1 = V2T1 T2V1: initial volumeV2: final volumeT1: initial temperatureT2: final temperatureTemperature has to be in Kelvin scale.22Volume and Temperature

As a gas is heated, it expands.This causes the density of thegas to decrease.

Using Charless LawA balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ?

Using Charless LawA balloon inflated in a room at 24C has a volume of 4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ? V1 = V2 V1= 4.00LV2= ?T1 T2 T1= 24C +273= 297 K T2= 58C + 273 = 331 KV2= V1T2 = (4.00L)(331K) = 4.46 L T1 (297K)Since temperature increases, you expect the volume to increase. Classwork: p124 # 11, 12 (a-c), 13

Combined Gas LawDescribes the relationship among the pressure, temperature and volume, when the amount of gas is constant.P1V1 = P2V2 T1 T2Standard temperature and pressure (STP): 0C, 1 atmUseful conversions:1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L

Using the combined gas law:The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)?

Using the combined gas law:The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)?P1= 153 kPaV1= 30.0 LT1= 313KP2= 1 atm=101.3kPa V2= ? T2= 0C= 273K P1V1 = P2V2 T1 T2V2= P1V1T2 = (153 kPa)(30.0L)(273K)= 39.5 L P2T1(101.3kPa)(313K)Classwork: p 126 #14(a,b), 15 (c,d), 16

Ideal GasesObjectivesCompute the value of an unknown using the ideal gas law.Compare and contrast real and ideal gases.Avogadros Principle and Molar VolumeAvogadros principle states that equal volume of all gases, measured under the same conditions of pressure and temperature, contain the same number of particles.At STP, the volume of one mole of gas is 22.4 (this is called the molar volume)

From last year, to convert between grams and moles of a substance we used its molar mass.Avogadros Law

30Converting between moles and grams:How many moles are 98.32g CO2?

Calculate molar mass CO2 (use periodic table)Molar mass= 12 + (2 x 16) =44.0 g/1 mol

To convert grams to moles:

98.32g x 1 mol = 2.23 mol CO2 44g Using Avogadros PrincipleHow many grams of carbon dioxide, CO2, will occupy a volume of 500.0 mL at STP?V= 500.0mL =0.5 LMolar mass CO2= 44g/molAt STP, molar volume is : 22.4L/1 mol

0.5 L x 1 mol x 44 g = 0.982g CO2 22.4L 1 mol

Classwork: p 132 # 1 (a-c), 3 (a-d)The ideal gas lawConsiders that amount of gas varies.New variablen: number of moles of gas (mol)Ideal gas constant (R)R= 8.314 L kPa (when pressure is measured in kPa) mol KR= 0.0821 L atm (when pressure is measured in atm)mol K PV= nRT(T must be in Kelvin)

Using the ideal gas law1. A deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many kilograms of CH4 does the cavern contain?Using the ideal gas lawA deep underground cavern contains 2.24x106 L of methane gas (CH4) at a pressure of 1500 kPa and a temperature of 315K. How many moles of CH4 does the cavern contain?P= 1500kPa V= 2.24x106 L R= 8.314 L kPa n= ? T= 315 K mol K

PV=nRTn= PV = (1500 kPa) (2.24x106 L ) (mol K) RT (8.314 L kPa)(315K)Classwork: p141 # 1-4Ideal gas variationsFor calculating molar massM= mRT M: molar mass (g/mol) PV m: mass (g)For calculating densityD= MP D: density (g/L) RTSample problemWhat is the molecular mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP?

Sample problemWhat is the molecular mass of sulfur dioxide, SO2, if 300 mL of the gas has a mass of 0.855 g at STP?M= ? V= 300mL=0.3 L T=273K, P= 101.3kPa m=0.855g

M= mRT = (0.855g )x(8.314 L kPa)x(273K) PV (mol K ) (101.3kPa)(0.3L)

= 63.8g/mol

Classwork: p 133 #5,6, 9Gases: Mixtures and MovementsObjectives:Relate the total pressure of a mixture of gases to the partial pressures of the component gases.Explain how the molar mass of a gas affects the rate at which the gas diffuses and effusesDaltons LawPartial pressure: the contribution each gas in a mixture makes to the total pressure.Daltons law of partial pressure: at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component of the gases.P total= P1 + P2 + P3 +

Grahams LawDiffusion: the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout.Effusion: a gas escapes through a tiny hole inits container.Particles of lower molar mass diffuse and effuse faster than gases of higher molar mass.

Using Daltons and Grahams Laws:Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10kPa, 0.040kPa, and 0.94kPa respectively?Ptotal = 101.3 kPa PN2= 79.10kPa PCO2= 0.040kPa Pothers= 0.94kPa PO2= ? Ptotal = PN2 + PCO2 + PO2 + Pothers

PO2 = Ptotal- (PN2+ PCO2+ PO2) = 101.3kPa-(79.1kpa+0.040kPa+0.94kPa)= 21.22 kPa2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature.

2. A nitrogen,N2, molecule travels at about 505 m/s at room temperature. Find the velocity of a helium, He, atom at the same temperature.

rateHe /rateN2= (molar mass N2/molar massHe)

rateHe = rateN2 (molar mass N2/molar massHe)= 505m/s (28/4)= 1336 m/s

Classwork: daltons and grahams law handout.