Ch Gases

 To describe a gas fully you need to state 4 measurable quantities:  Volume  Temperature  Number of molecules  pressure

Which shoes create the most pressure?

Click below to watch the Visual Concept. Chapter 11 Section 1 Gases and Pressure Visual Concept

Relationship Between Pressure, Force, and Area Chapter 11 Section 1 Gases and Pressure

 Gas pressure is caused by collisions of the gas molecules with each other and with surfaces with which they come into contact.  The pressure exerted by a gas depends on volume, temperature, and the number of molecules present.  The greater the number of collisions of gas molecules, the higher the pressure will be.

 Barometer  measures atmospheric pressure Mercury Barometer Aneroid Barometer

Click below to watch the Visual Concept. Visual Concept Chapter 11 Section 1 Gases and Pressure

 Manometer  measures contained gas pressure U-tube ManometerBourdon-tube gauge

Units of Pressure Pg 344

 Sample Problem A  The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in  a. millimeters of mercury (mm Hg) and  b. kilopascals (kPa)

Standard Temperature & Pressure 0°C 273 K 1 atm kPa -OR- STP

* The pressure of each gas in a mixture is called the partial pressure of that gas. * John Dalton, the English chemist who proposed the atomic theory, discovered that the pressure exerted by each gas in a mixture is independent of that exerted by other gases present. * Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of the component gases. P atm = P gas + P H20

Click below to watch the Visual Concept. Visual Concept Chapter 11 Section 1 Gases and Pressure

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 H kPa P H2 = 91.7 kPa  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.R63 for 22.5°C. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

 The pressure and volume of a gas are inversely related  at constant mass & temp P V PV = k

 The volume and absolute temperature (K) of a gas are directly related  at constant mass & pressure V T

 The pressure and absolute temperature (K) of a gas are directly related  at constant mass & volume P T

= kPV PTPT VTVT T P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

 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: P 1 V 1 T 2 = P 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

 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 T 2 = P 2 V 2 T 1 BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

 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 V 1 T 2 = P 2 V 2 T 1 GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(296K) T 2 = 216 K = -57°C

 A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. C. Johannesson 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

Ch Gases

V n  The law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.  also indicates that gas volume is directly proportional to the amount of gas, at a given temperature and pressure

Click below to watch the Visual Concept. Visual Concept Chapter 11 Section 3 Gas Volumes and the Ideal Gas Law

PV=nRT Use Table 11-1 for numerical values of R (p. 342) Merge the Combined Gas Law with Avogadro’s Principle:

Numerical Values of the Gas Constant Section 3 Gas Volumes and the Ideal Gas Law Chapter 11

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  Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.

C. Johannesson GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = L  kPa/mol  K  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) (8.315) (298) kPa mol L  kPa/mol  K K V = 64 L

Ch Gases

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

1 mol CaCO g CaCO 3  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.

C. Johannesson WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 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 = dm 3  kPa/mol  K

C. Johannesson WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = mol O 2  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 = dm 3  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  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.