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I. Physical Properties

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Real vs. Ideal Gases: b Ideal gas = an imaginary gas that conforms perfectly to all the assumptions of the kinetic-molecular theory We will assume that the gases used for the gas law problems are ideal gases.

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Real Gases vs. Ideal Gases: b Real gas = a gas that does not behave completely according to the assumptions of the kinetic- molecular theory. b All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.

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Causes of non-ideal behavior: The kinetic-molecular theory is more likely to hold true for gases whose particles have NO attraction for each other.

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Causes of non-ideal behavior: 1. High pressure (low volume): Space taken up by gas particles becomes significant Intermolecular forces are more significant between gas particles that are closer together

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Causes of non-ideal behavior: 2. Low temperature: Gas particles move slower so intermolecular forces become more important

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b Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences. b theory developed in the late 19 th century to account for the behavior of the atoms and molecules that make up matter KMT (Kinetic Molecular Theory)

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based on the idea that particles in all forms of matter are always in motion and that this motion has consequences KMT (Kinetic Molecular Theory) SOLID LIQUID GAS

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can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them KMT (Kinetic Molecular Theory)

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B. Kinetic Molecular Theory – Ideal gas b KMT describing particles in an IDEAL gas: have no volume. have elastic collisions. are in constant, random motion. dont attract or repel each other. have an avg. KE directly related to Kelvin temperature.

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C. Kinetic Molecular Theory - Real Gases b KMT describing particles in a REAL gas: have their own volume attract each other b Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

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D. Characteristics of Gases- using KMT b Gases expand to fill any container. b KMT - gas particles move rapidly in all directions without significant attraction or repulsion between particles

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D. Characteristics of Gases- using KMT b Gases are fluids (like liquids). KMT - No significant attraction or repulsion between gas particles; glide past each other

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D. Characteristics of Gases- using KMT b Gases have very low densities. KMT - particles are so much farther apart in the gas state sodium in the solid state: sodium in the liquid state: sodium in the gas state:

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b Gases can be compressed. KMT - gas particles are far apart from one anther with room to be squished together D. Characteristics of Gases- using KMT

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b Gases undergo diffusion & effusion. KMT – gas particles move in continuous, rapid, random motion D. Characteristics of Gases- using KMT Effusion

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E. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 b Always use Kelvin temperature when working with gases.

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F. Pressure Which shoes create the most pressure?

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F. Pressure b Why do gases exert pressure? Gas particles exert a pressure on any surface with which they collide! More collisions = increase in pressure!

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F. Pressure b Barometer = measures atmospheric pressure b The height of the Hg in the tube depends on the pressure The pressure of the atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure! Mercury Barometer

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F. Pressure b Pressure UNITS 101.3 kPa (kilopascal) 1 atm 760 mm Hg 760 torr

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G. STP Standard Temperature & Pressure 0°C (exact)1 atm (exact) 273 K101.3 kPa 760 mm Hg (exact) 760 torr (exact) STP

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II. The Gas Laws Gases

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A. Boyles Law P V PV = k

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A. Boyles Law b The pressure and volume of a gas are INVERSELY related at constant mass & temp P V

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A. Boyles Law Real life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.

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A. Boyles Law Ex: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.

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A. Boyles Law As the volume increases, the pressure decreases!

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V T B. Charles Law

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V T b The volume and temperature (K) of a gas are DIRECTLY related at constant mass & pressure

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B. Charles Law b Real life application: Bread dough rises because yeast produces carbon dioxide. When placed in the oven, the heat causes the gas to expand, and the bread rises even further.

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B. Charles Law b As temperature decreases, the volume of the gas decreases Liquid nitrogens temp. is about 63K or -210 ºC or -346 ºF!

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B. Charles Law b As temperature decreases, the volume of the gas decreases

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B. Charles Law b NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!

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P T C. Guy-Lussacs Law

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P T b The pressure and temperature (K) of a gas are DIRECTLY RELATED at constant mass & volume

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C. Guy-Lussacs Law b When the temp. of a gas increases (KE increases) and gas particles move faster and hit container walls more frequently and collisions are more forceful

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C. Guy-Lussacs Law b Real life application: The air pressure inside a tire increases on a hot summer day.

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= k Boyles PV P Guy Lussacs T V Charles T PV T D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

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E. EXAMPLE Problems b You need your calculators!

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C. Johannesson GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: Gas Law Problems 1.) A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. T V V 2 = (473 cm 3 )(367 K) (309 K) V 2 = 562 cm 3

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C. Johannesson GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: V 2 = P 1 V 1 P 2 Gas Law Problems 2.) A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLES LAW P 1 V 1 = P 2 V 2 P V V 2 = (150.kPa)(100.mL) 200.kPa V 2 = 75.0 mL

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C. Johannesson GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.3 kPa T 2 = 273 K WORK: V 2 = P 1 V 1 T 2 P 2 T 1 V 2 = (71.8 kPa)(7.84 cm 3 )(273 K) (101.3 kPa) (298 K) V 2 = 5.09 cm 3 Gas Law Problems 3.) A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P T V

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C. Johannesson GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: T 2 = P 2 T 1 P 1 Gas Law Problems 4.) A gas pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? P T T 2 = (560. torr)(296K) 765 torr T 2 = 217 K

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III. Standard Molar Volume, Gas Densities & Molar Mass Gases

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A. Standard Molar Volume b The volume occupied by one mole of a gas at STP 22.41410 L/mol or about 22.4 L/mol b In other words, one mole of any ideal gas at STP will occupy 22.4 L

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B. Example Problems-standard molar volume 1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP? 0.068 mol O 2 22.4 L O 2 1 mol O 2 = 1.52 L O 2

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B. Example Problems-standard molar volume 1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP? 0.068 mol O 2 22.4 L O 2 1 mol O 2 = 1.52 L O 2

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B. Example Problems-standard molar volume 2.) A chemical reaction produced 98.0 mL of SO 2 at STP. What mass (in grams) of the gas was produced? 98.0 mL SO 2 1 L SO 2 1 mol O 2 = 0.280 g SO 2 22.4 L SO 2 1000 mL SO 2 1 mol SO 2 64.064 g SO 2

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C. Density b The ratio of an objects mass to its volume. b D = M V b The volume (and density) of a gas will change when pressure and temperature change. Densities are usually given in g/L at STP

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C. Density b For an ideal gas: Density (at STP) = molar mass standard molar volume

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D. Example Problems- density 1.) What is the density of CO 2 in g/L at STP? 44.009 g CO 2 1 mol CO 2 1 mol CO 2 22.4 L CO 2 = 1.96 g/L CO 2

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D. Example Problems- density 2.) What is the molar mass of a gas whose density at STP is 2.08 g/L. 2.08 g 22.4 L 1 L 1 mol = 46.6 g/mol

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IV. More Gas Laws: Ideal Gas Law & Avogadro's Law Gases

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V n A. Avogadros Law b Equal volumes of gases contain equal numbers of moles at constant temp & pressure

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PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L atm/mol K R=8.315 dm 3 kPa/mol K = R You dont need to memorize these values! Boyles, Charles, and Avogadros laws are contained within the ideal gas law! Merge the Combined Gas Law with Avogadros Law:

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B. Ideal Gas Law GAS CONSTANT R=0.0821 L atm/mol K R=8.315 dm 3 kPa/mol K PV=nRT You dont need to memorize these values! P= pressure V = volume n = moles T = temperature (in K) R = the ideal gas constant

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C. Johannesson GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L atm/mol K WORK: P = nRT V P =(0.412 mol)(0.0821 L atm/mol K) (289K) 3.25L P = 3.01 atm C. Ideal Gas Law Problems 1.) Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. PV = nRT

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C. Johannesson GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3 kPa/mol K C. Ideal Gas Law Problems 2.) Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. PV = nRT = 2.7 mol WORK: 85 g O 2 1 mol O 2 = 2.7 mol 31.998 g O 2 O 2 V = nRT P V =(2.7 mol)(8.315 dm 3 kPa/mol K) (298K) 104.5 kPa V = 64 dm 3

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C. Johannesson GIVEN: V = 1.00 L n = ? T = 28°C = 301 K P = 98.7 kPa R = 8.314 L kPa/mol K C. Ideal Gas Law Problems 3.) At 28 C and 98.7 kPa, 1.00 L of an unidentified gas has a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRT WORK: n = PV RT (98.7 kPa) (1.00L) (8.314 L kPa/mol K) (301 K) n = 0.0394 mol m = 5.16g 0.0394 mol = = 131 g/mol

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Gases IV. Two More Laws: Daltons Law & Grahams Law

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A. Daltons Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 + P 3...

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A. Daltons Law P total = P 1 + P 2 + P 3...

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A. Daltons Law When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 AND water vapor.

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C. Johannesson GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa A. Daltons Law 1.) Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure on gas formula sheet for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

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C. Johannesson GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr 2.) A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? Look up water-vapor pressure on gas formula sheet for 35.0°C. Sig Figs: Round to least number of decimal places. A. Daltons Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the gas and water vapor.

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B. Grahams Law b Diffusion = Spreading of gas molecules throughout a container until evenly distributed. b Effusion = Passing of gas molecules through a tiny opening in a container

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B. Grahams Law KE = ½mv 2 b Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. Larger m smaller v

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B. Grahams Law b KE = kinetic energy b m= molar mass b v = velocity KE = ½mv 2

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C. Johannesson B. Grahams Law b Grahams Law b Grahams Law = Rate of effusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas As speed to Gas Bs speed.

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1.) Determine the relative rate of effusion for krypton and bromine. Kr effuses 1.381 times faster than Br 2. C. Grahams Law Problems The first gas is Gas A and the second gas is Gas B. Relative rate mean find the ratio v A /v B.

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C. Johannesson 2.) A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? C. Grahams Law Problems Put the gas with the unknown speed as Gas A.

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C. Johannesson 3.) An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. C. Grahams Law Problems The first gas is Gas A and the second gas is Gas B. The ratio v A /v B is 4.0. Square both sides to get rid of the square root sign.

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Gases VI. Gas Stoichiometry at Non-STP Conditions

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A. Gas Stoichiometry b Moles Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law b Non- STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv.

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B. Volume Mass Stoich. b Thinking: Gas volume A moles A moles B mass B b If gas A is at STP: ? L A 1 mol A ? mol B molar mass(g) B 22.4 L A ? mol A 1 mol B Mole Ratio

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B. Volume Mass Stoich. b If gas A is NOT at STP: Use PV = nRT to find moles of A ? mol A ? mol B molar mass (g) B ? mol A 1 mol B

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C. Johannesson WORK: n = PV RT n= (97.3 kPa) (15.0 L) (8.315 dm 3 kPa/mol K ) (294K) n = 0.597 mol O 2 B. Volume Mass 1.) How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm 3 kPa/mol K 4 Al + 3 O 2 2 Al 2 O 3 15.0 L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT

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2 mol Al 2 O 3 3 mol O 2 B. Volume Mass 1.) How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2 2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.

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1 mol SO 2 22.4L SO 2 B. Volume Mass 2.) What mass of sulfur is required to produce 12.6 L of sulfur dioxide at STP according to the equation: S 8 (s) + 8 O 2 (g) 8 SO 2 (g) 12.6 L SO 2 = 18.0 g S 8 256.582 g S 8 1 mol S 8 12.6L ? g 1 mol S 8 8 mol SO 2

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C. Mass Volume b Thinking: Mass A moles A moles B gas volume B b If gas B is at STP: ? g A 1 mol A ? mol B 22.4 L B molar mass (g) A ? mol A 1 mol B

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C. Mass Volume b If gas B is NOT at STP: ? g A 1 mol A ? mol B molar mass (g) A ? mol A Then use PV = nRT to find volume of B

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C. Mass Volume 1.) What volume of chlorine gas at STP is needed to react completely with 10.4 g of sodium to form NaCl? 2 Na + Cl 2 2 NaCl 10.4 g ? L 10.4 g Na 22.990 g Na 22.4 L Cl 2 1 mol Cl 2 2 mol Na 1 mol Cl 2 1 mol Na = 5. 07 L Cl 2

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1 mol CaCO 3 100.09g CaCO 3 C. Mass Volume 2.) What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.26 mol CO 2 CaCO 3 CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

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C. Johannesson WORK: V = nRT P V = (1.26mol)(8.315dm 3 kPa/mol K) (298K) (103 kPa) V = 30.3 dm 3 CO 2 C. Mass Volume 2.) What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm 3 kPa/mol K

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