Ch. 10 Gases

Characteristics of Gases b Gases expand to fill any container. random motion, no attraction b Gases are fluids (like liquids). no attraction b Gases have very low densities. no volume = lots of empty space

Characteristics of Gases b Gases can be compressed. no volume = lots of empty space b Gases undergo diffusion & effusion. random motion

Kinetic Molecular Theory b Model used to predict behavior of a gas on a molecular level Assumes gas is an ideal gas b Ideal gas- a model that effectively describes the behavior of a real gases at conditions close to standard temperature and pressure

Kinetic Molecular Theory b Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight- line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature. DO NOT EXIST!!!

Real Gases b 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

TemperatureTemperature ºF ºC K K = ºC b Always use absolute temperature (Kelvin) when working with gases.

PressurePressure Which shoes create the most pressure?

PressurePressure Total force exerted by the moving particles of a gas as they collide with the walls of a container

PressurePressure b Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

PressurePressure b Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

PressurePressure b KEY UNITS AT SEA LEVEL kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

STPSTP Standard Temperature & Pressure 0°C K 1 atm kPa -OR- STP

STPSTP Standard Temperature & Pressure 1 mole of a gas at STP = molar volume 1 molar volume of any gas is 22.4 L 1 mole of any gas = 22.4 L STP

A. Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp Can write PV = k OR P V P 1 V 1 = P 2 V 2

A. Boyle’s Law b Check out the following link: Or this… Or this…

V T Charles’ Law b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

Charles’ Law b A demonstration of Charles’ Law… And…

P T Gay-Lussac’s Law b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

Gay-Lussac’s Law b This is demonstrated by… Wk Or…

Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

GIVEN: V 1 = 473 cm 3 T 1 = 36°C = K V 2 = ? T 2 = 94°C = K WORK: V 1 /T 1 = V 2 /T 2 Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 560 cm 3

GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 = P 2 V 2 Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = kPa T 2 = 273 K WORK: P 1 V 1 /T 1 = P 2 V 2 /T 2 (71.8 kPa)(7.84 cm 3 )(273 K) =( kPa) V 2 (298 K) V 2 = 5.1 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

Ch. 10 & 11 - Gases Two More Laws

B. Dalton’s 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 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.

GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H kPa P H2 = 91.7 kPa B. Dalton’s Law b 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 p.859 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.

GIVEN: P gas = ? P total = torr P H2O = 42.2 torr WORK: P total = P gas + P H2O torr = P H torr P gas = torr b A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas? Look up water-vapor pressure on p.859 for 35.0°C. Sig Figs: Round to least number of decimal places. B. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

C. Graham’s 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

C. Graham’s 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

C. Graham’s Law b Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed.

b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses times faster than Br 2. C. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

b 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. Graham’s Law Put the gas with the unknown speed as “Gas A”.

b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. C. Graham’s Law 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.

Ideal Gas Law Ch. 10 & 11 - Gases

V n A. Avogadro’s Principle b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

PV T VnVn PV nT A. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.314 L  kPa/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

A. Ideal Gas Law UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.314 L  kPa/mol  K PV=nRT

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

GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = L  kPa/mol  K C. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol g PV = nRT (104.5)V=(2.7) (8.314) (298) kPa mol L  kPa/mol  K K V = 64 L

Ch. 10 & 11 - Gases Gas Stoichiometry

A. Gas Stoichiometry Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conv.

1 mol CaCO g CaCO 3 B. Gas Stoichiometry Problem b 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 g? L Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

WORK: PV = nRT (103 kPa)V =(1mol)(8.345L  kPa/mol  K )(298K) V = 1.26 L CO 2 B. Gas Stoichiometry Problem b 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 = L  kPa/mol  K

WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.314 L  kPa/mol  K ) (294K) n = mol O 2 B. Gas Stoichiometry Problem b 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 = L  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O L ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT 

2 mol Al 2 O 3 3 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O g Al 2 O 3 1 mol Al 2 O L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.