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 0C 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. * Daltons 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 22.5C. Find the pressure of
the dry gas if the atmospheric pressure is 94.4 kPa. Look up
water-vapor pressure on p.R63 for 22.5C. 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 36C. Find its
volume at 94C. GIVEN: V 1 = 473 cm 3 T 1 = 36C = 309K V 2 = ? T 2 =
94C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 CHARLES LAW TT VV (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
BOYLES LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL A gas
pressure is 765 torr at 23C. At what temperature will the pressure
be 560. torr? GIVEN: P 1 = 765 torr T 1 = 23C = 296K P 2 = 560.
torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 GAY-LUSSACS LAW PP TT
(765 torr)T 2 = (560. torr)(296K) T 2 = 216 K = -57C A gas occupies
7.84 cm 3 at 71.8 kPa & 25C. Find its volume at STP. C.
Johannesson GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25C = 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 VV 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 Avogadros Principle: Numerical Values of the Gas
Constant Section 3 Gas Volumes and the Ideal Gas Law Chapter 11
GIVEN: P = ? atm n = mol T = 16C = 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 16C
& occupying 3.25 L. C. Johannesson GIVEN: V = ? n = 85 g T =
25C = 298 K P = kPa R = L kPa/mol K Find the volume of 85 g of O 2
at 25C 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 & 25C? 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 & 25C? GIVEN: P
= 103 kPa V = ? n = 1.26 mol T = 25C = 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 & 21C? GIVEN: P = 97.3
kPa V = 15.0 L n = ? T = 21C = 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 &
21C? 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.
