Think About This… Gas Atmosphere This is a U-Tube Manometer. The red stuff is a liquid that moves based on the pressures on each end of the tube. Based on the position of the red liquid, which has a higher pressure, the Gas or the Atmosphere? How do you know?

Think About This… Gas Atmosphere Which has a higher pressure in this U- Tube, the Gas or the Atmosphere? How do you know?

Think About This… Gas Atmosphere What about this U- Tube? Which has the higher pressure, the Gas or the Atmosphere? How do you know?

Follow along in your text Chapter 12 Section 1 Pages The Physical Properties of Gases

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 average kinetic energy directly related to Kelvin temperature.

Real Gases b Particles in a REAL gas… have their own volume attract each other b Gas behavior is most ideal… at high temperatures at low pressures in nonpolar, basically covalent, atoms/molecules

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 effusion & diffusion. random motion

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

What is Pressure? Which shoes create the most pressure?

Pressure Equipment b Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

Pressure Equipment b Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

b KEY UNITS AT SEA LEVEL 101,325 Pa kPa 1 atm 760 mm Hg 760 torr 14.7 psi 1.01 bar *These are all equal to each other!*PressurePressure

STP is the Norm! Standard Temperature & Pressure 0°C273 K 1 atm kPa -OR- STP

Example: How many torr equals 5.00 psi?ClosureClosure 1.How many Pascals equals 7.3 atm? 2.How many mmHg equals 19.3 barr? 3.How many atms equals 96.3 kPa? 5.00 psi 760 torr 14.7 psi = 259 torr

Warm-Up : Complete the following pressure conversions ,650 Pa to mm Hg atm to psi ,000 torr to kPa psi to barr

Follow along in your text Chapter 12 Section 2 Pages The Gas Laws

Meet the Variables b P = pressure exerted by the gas b T = temperature in kelvins of the gas b V = total volume occupied by the gas b n = number of moles of the gas b k = symbolizes anything constant

Boyle’s Law V P PV = k

Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp P 1 V 1 = P 2 V 2 b When the temp & number of particles remains the same, this is the equation

GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Boyle’s Law Problem b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. Remember BOYLE’S LAW! PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

V T Charles’ Law

b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V 1 V 2 T 1 T 2 = b If all conditions are kept constant, then the equation looks like this

GIVEN: V 1 = 473 mL T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Charles’ Law Problem b A gas occupies 473 mL at 36°C. Find its volume at 94°C. Remember CHARLES’ LAW! TT VV (473 mL)(367 K)=V 2 (309 K) V 2 = 562 mL

P T Gay-Lussac’s Law

b The pressure and absolute temperature (K) of a gas are directly related b At constant mass & volume, the equation looks like this P 1 P 2 T 1 T 2 =

GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gay-Lussac’s Problem b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? Remember GAY-LUSSAC’S LAW! PP TT (765 torr)T 2 = (560. torr)(296K) T 2 = 217 K = -38°C

Warm Up: b DO NOT SOLVE! Just determine which Gas Law you should use. 1. What is the new volume if a 5.0L container of gas at 23 °C is heated to 70 °C? 2. If a gas at STP is heated to 900 K, what is the resulting pressure? mL of a gas at 1 atm is compressed to 5 mL. How much pressure was applied? 4. A 2.0 L container at 4.0 atm and 0.0 °C is heated to 50. °C at constant volume. What is the resulting pressure? 5. A 35 L container at 23 °C is expanded to 50 L to obtain a pressure of 4.0 atm. What was the original pressure if temperature was constant?

Combined Gas Law b This is a combination of the 3 main gas laws: Boyle’s Law, Charles’ Law & Gay-Lussac’s Law b Pressure is opposite temperature & volume P V T

= k PV T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

GIVEN: V 1 = 7.84 mL 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 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 mL)(273 K) =( kPa) V 2 (298 K) V 2 = 5.09 mL Combined Gas Law Problem b A gas occupies 7.84 mL at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

Warm Up: Stoichiometry Quiz #5 How many grams of CO 2 are produced from 75 L of CO at STP? ___CO + ___O 2 → ___CO 2

Follow along in your text Chapter 12 Section 3 Pages The Ideal Gas Law

PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.314 L  kPa/mol  K R=62.4 L  mmHg/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

Ideal Gas Law UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.314 L  kPa/mol  K R=62.4 L  mmHg/mol  K PV=nRT

Ideal Gas Law PVM=mRT If you are given the mass of a gas, you can use this equation instead of converting mass to moles first. mass molar mass

Ideal Gas Law d=PM/RT If you are given the molar mass of a gas, you can use this equation to find the density molar mass density

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

GIVEN: V = ? m = 85 g M O 2 = 32 g/mol T = 25°C = 298 K P = kPa R = L  kPa/mol  K Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and kPa. WORK: PVM = mRT V= (85 g)(8.314)(298 K) (104.5 kPa)(32 g/mol) V = 64 L

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 conversion

1 mol CaCO g CaCO 3 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 non-STP 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)(8.314 L  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 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 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 non-STP ? 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 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.

Closure:Closure: How many grams of CO 2 are produced from 75L of CO at 35 ºC and 96.2 kPa? ___CO + ___O 2 → ___CO 2

Warm Up: Solve using the Ideal Gas Law & Stoichometry What mass of NH 3 is created when 20.0 L of N 2 is combined with excess H 2 at 42 °C and 2.50 atm?

More Gas Laws! Follow along in your text Chapter 12 Sections 2 & 3 Pages

V n Avogadro’s Law b Equal volumes of gases contain equal numbers of moles at constant temp & pressure 22.4 L/mole 6.02 x particles/mole

Dalton’s Law of Partial Pressure 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 P 1 = (P total ) (% of Gas1)

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 Dalton’s Law of Partial Pressure b Hydrogen gas is collected over water. The water vapor has a pressure of 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. 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.

Graham’s Law of Diffusion b Diffusion Spreading of gas molecules from high density to low density until even all over. b Effusion Passing of gas molecules under pressure through a tiny opening

Graham’s Law of Diffusion 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.  “The smaller the mass, the faster the gas!”

Graham’s Law of Diffusion 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. v √m

b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses times faster than Br 2. Graham’s Law of Diffusion 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? Graham’s Law of Diffusion Put the gas with the unknown speed as “Gas A”.