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

C. Johannesson CHARACTERISTICS OF GASES Gases can be compressed. no volume = lots of empty space Gases undergo diffusion & effusion. random motion

Pressure Pressure is the force created by the collisions of molecules with the walls of a container -is equal to force/unit area -is equal to force/unit area 1 standard atmosphere (atm)  1 standard atmosphere (atm)  kPa (kilopascals)  14.7 lbs/in2  760 mm Hg (millimeters of mercury)  760 torr

MEASURING PRESSURE The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

AN EARLY BAROMETER The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

Temperature Gas law problems involving temperature require that the temperature be in KELVINS! Lord Kelvin Kelvins ( o K) = o C + 273

STANDARD TEMPERATURE AND PRESSURE “STP” Either of these:  273 Kelvin (273 K)  0  C Either of these:  273 Kelvin (273 K)  0  C And any one of these: 1 atm  1 atm  kPa  14.7 lbs/in 2 (psi)  760 mm Hg  760 torr And any one of these: 1 atm  1 atm  kPa  14.7 lbs/in 2 (psi)  760 mm Hg  760 torr

GAS LAWS

BOYLE’S LAW Pressure is inversely proportional to volume when temperature is held constant.

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

CHARLES’S LAW The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. ( P = constant ) Temperature MUST be in KELVINS!

GAS LAW PROBLEMS A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: V 1 T 2 = V 2 T 1 CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

GAY LUSSAC’S LAW The pressure and temperature of a gas are directly related, provided that the volume remains constant. Temperature MUST be in KELVINS!

GAS LAW PROBLEMS A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 T 2 = P 2 T 1 GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

THE COMBINED GAS LAW The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

GAS LAW PROBLEMS A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. 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 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =( kPa) V 2 (298 K) V 2 = 5.09 cm 3 P  T  VV COMBINED GAS LAW

IDEAL GASES Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases consist of tiny particles that are far apart relative to their size. Collisions between gas particles and between particles and the walls of the container are elastic collisions No kinetic energy is lost in elastic collisions

IDEAL GASES IDEAL GASES (CONTINUED) Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion There are no forces of attraction between gas particles The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

IDEAL GAS LAW PV = nRTPV = nRT P = pressure in atm V = volume in liters n = moles R = proportionality constant = L atm/ mol·   L kPa/mol-k T = temperature in Kelvins Holds closely at P < 1 atm

REAL GASES DO NOT BEHAVE IDEALLY Real gases DO experience inter-molecular attractions Real gases DO have volume Real gases DO NOT have elastic collisions

DEVIATIONS FROM IDEAL BEHAVIOR Likely to behave nearly ideally Gases at high temperature and low pressure Small non-polar gas molecules Likely not to behave ideally Gases at low temperature and high pressure Large, polar gas molecules

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

GAS STOICHIOMETRY PROBLEM What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 1 mol CaCO g CaCO 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.

GAS STOICHIOMETRY PROBLEM What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = dm 3  kPa/mol  K

GAS STOICHIOMETRY PROBLEM How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = mol O 2 GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = dm 3  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 

GAS STOICHIOMETRY PROBLEM How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 2 mol Al 2 O 3 3 mol O 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.