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Ch. 10 & 11 - Gases III. Ideal Gas Law (p. 334-335, 340-346)

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Presentation on theme: "Ch. 10 & 11 - Gases III. Ideal Gas Law (p. 334-335, 340-346)"— Presentation transcript:

1 Ch. 10 & 11 - Gases III. Ideal Gas Law (p , )

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

3 UNIVERSAL GAS CONSTANT
A. Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: PV nT PV T V n = k = R UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values!

4 UNIVERSAL GAS CONSTANT
A. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values!

5 C. Ideal Gas Law Problems
Calculate the pressure in atmospheres of mol of He at 16°C & occupying L. 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

6 C. Ideal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and kPa. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm3kPa/molK WORK: 85 g 1 mol = 2.7 mol 32.00 g = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3

7 Gas Density and Molar Mass
Density, D = m/V Let M stand for molar mass M = m/n n= PV/RT substituting in for n M = m so PVM = mRT PV/RT M = mRT = m RT = DRT PV V P P So PM = DRT

8 IV. Gas Stoichiometry at Non-STP Conditions (p. 347-350)
Ch. 10 & 11 - Gases IV. Gas Stoichiometry at Non-STP Conditions (p )

9 A. Gas Stoichiometry 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.

10 B. Gas Stoichiometry Problem
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? CaCO3  CaO CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters.

11 B. Gas Stoichiometry Problem
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = dm3kPa/molK WORK: PV = nRT (103 kPa)V =(1.26mol)(8.315dm3kPa/molK) (298K) V = 30.3 dm3 CO2

12 B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K) n = mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

13 B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al O2  2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

14 V. Two More Laws (p. 322-325, 351-355) Read these pages first!
Ch. 10 & 11 - Gases V. Two More Laws (p , ) Read these pages first!

15 Ptotal = P1 + P2 + ... B. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. Ptotal = P1 + P When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

16 Dalton’s Law Therefore, when collecting a gas over water by water displacement, the pressure of the gas is found by subtracting the vapor pressure of water at the collection temperature from the total atmospheric pressure. Pgas = Patm – Pwater

17 Dalton's Law PTotal = P1 + P2 + P3 + P4 + P5 ... For each P = nRT/V
PTotal = n1RT + n2RT + n3RT V V V In the same container R, T and V are the same. PTotal = (n1+ n2 + n3+...)RT V PTotal = (nTotal)RT V

18 Dalton’s Law - mole fraction
Ratio of moles of the substance to the total moles. symbol is Greek letter chi c c1 = n1 = P nTotal PTotal so P1 = c1PTotal Sum of all mole fractions must equal one c1 + c2 + c = 1

19 B. Dalton’s Law 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. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH kPa PH2 = 91.7 kPa Look up water-vapor pressure on p.899 for 22.5°C. Sig Figs: Round to least number of decimal places.

20 B. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr
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? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. GIVEN: Pgas = ? Ptotal = torr PH2O = 42.2 torr WORK: Ptotal = Pgas + PH2O 742.0 torr = PH torr Pgas = torr Look up water-vapor pressure on p.899 for 35.0°C. Sig Figs: Round to least number of decimal places.

21 C. Graham’s Law Diffusion
Spreading of gas molecules throughout a container until evenly distributed. Effusion Passing of gas molecules through a tiny opening in a container

22 KE = ½mv2 C. Graham’s Law 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 KE = ½mv2

23 C. Graham’s Law 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.

24 C. Graham’s Law Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Kr diffuses times faster than Br2.

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

26 C. Graham’s Law An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign.

27 Ideal Gas Law & Gas Stoichiometry
PV = nRT R = Latm/molK = dm3kPa/molK Ideal Gas Law & Gas Stoichiometry 1) Work out each problem on scratch paper. 2) Click ANSWER to check your answer. 3) Click NEXT to go on to the next problem. CLICK TO START

28 QUESTION #1 How many grams of CO2 are produced from 75 L of CO at 35°C and 96.2 kPa? 2CO + O2  2CO2 ANSWER

29 V = 26.6 dm3/mol ANSWER #1 Find the new molar volume: n = 1 mol V = ?
P = 96.2 kPa T = 35°C = 308 K R = dm3kPa/molK PV = nRT V = 26.6 dm3/mol BACK TO PROBLEM CONTINUE...

30 = 120 g CO2 ANSWER #1 (con’t) 2CO + O2  2CO2 75 L CO 1 mol CO 26.6 L
44.01 g CO2 1 mol CO2 = 120 g CO2 BACK TO PROBLEM NEXT

31 QUESTION #2 How many moles of oxygen will occupy a volume of 2.5 L at 1.2 atm and 25°C? ANSWER

32 n = 0.12 mol ANSWER #2 n = ? V = 2.5 L P = 1.2 atm T = 25°C = 298 K
R = Latm/molK PV = nRT n = 0.12 mol BACK TO PROBLEM NEXT

33 QUESTION #3 What volume will 56.0 grams of nitrogen (N2) occupy at 96.0 kPa and 21°C? ANSWER

34 V = 50.9 dm3 ANSWER #3 V = ? n = 56.0 g = 2.00 mol P = 96.0 kPa
T = 21°C = 294 K R = dm3kPa/molK PV = nRT V = 50.9 dm3 BACK TO PROBLEM NEXT

35 QUESTION #4 What volume of NH3 at STP is produced if 25.0 g of N2 is reacted with excess H2? N2 + 3H2  2NH3 ANSWER

36 = 40.0 L NH3 ANSWER #4 N2 + 3H2  NH3 25.0 g N2 1 mol N2 28.02 g 2 mol
BACK TO PROBLEM NEXT

37 QUESTION #5 What volume of hydrogen is produced from 25.0 g of water at 27°C and 1.16 atm? 2H2O  2H2 + O2 ANSWER

38 V = 21.2 L/mol ANSWER #5 Find the new molar volume: n = 1 mol V = ?
P = 1.16 atm T = 27°C = 300. K R = Latm/molK PV = nRT V = 21.2 L/mol BACK TO PROBLEM CONTINUE...

39 H2 ANSWER #5 (con’t) 2H2O  2H2 + O2 25.0 g H2O 1 mol H2O 18.02 g
BACK TO PROBLEM NEXT

40 QUESTION #6 How many atmospheres of pressure will be exerted by 25 g of CO2 at 25°C and L? ANSWER

41 P = 28 atm ANSWER #6 P = ? n = 25 g = 0.57 mol T = 25°C = 298 K
V = L R = Latm/molK PV = nRT P = 28 atm BACK TO PROBLEM NEXT

42 QUESTION #7 How many grams of CaCO3 are required to produce 45.0 dm3 of CO2 at 25°C and 2.3 atm? CaCO3 + 2HCl  CO2 + H2O + CaCl2 ANSWER

43 V = 11 L/mol ANSWER #7 Find the new molar volume: n = 1 mol V = ?
P = 2.3 atm T = 25°C = 298 K R = Latm/molK PV = nRT V = 11 L/mol BACK TO PROBLEM CONTINUE...

44 = 410 g CaCO3 ANSWER #7 CaCO3 + 2HCl  CO2 + H2O + CaCl2 45.0dm3 CO2
1 mol CO2 11 dm3 1 mol CaCO3 CO2 g CaCO3 1 mol = 410 g CaCO3 BACK TO PROBLEM NEXT

45 QUESTION #8 Find the number of grams of CO2 that exert a pressure of 785 torr at 32.5 L and 32°C. ANSWER

46 n = 1.34 mol  59.0 g CO2 ANSWER #8 n = ? P = 785 torr = 1.03 atm
V = 32.5 L T = 32°C = 305 K R = Latm/molK PV = nRT n = 1.34 mol 59.0 g CO2 BACK TO PROBLEM NEXT

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