1. Gas Laws Accelerated Chemistry Chapter 12 Molecular Composition of Gases

2. Gas Laws 12.112.212.312.42 First, Let’s Review Chapter 11… The coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometric Calculations

3. Gas Laws 12.112.212.312.43 Stoichiometric Calculations From the mass of Substance A you can use the ratio of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

4. Gas Laws 12.112.212.312.44 Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

5. Gas Laws 12.112.212.312.45 Let’s Review… How Many Cookies Can I Make? –You can make cookies until you run out of one of the ingredients –Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat) How Many Cookies Can I Make? –You can make cookies until you run out of one of the ingredients –Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat) Limiting Reactants

6. Gas Laws 12.112.212.312.46 How Many Cookies Can I Make? In this example the sugar would be the ________________, because it will limit the amount of cookies you can make

7. Gas Laws 12.112.212.312.47 Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

8. Gas Laws 12.112.212.312.48 Limiting Reactants In the example below, the O 2 would be the ______________________

9. Gas Laws 12.112.212.312.49 12.1 Volume - Mass Relationships of Gases Last chapter, we related the volume and the mass of a gas: –_____________________________ Note: a “stoich” problem can be recognized by the fact that information from one substances is given and the problem is asking about a different substance. –_____________________________ Let’s compare the volumes of gases in two example problems…

10. Gas Laws 12.112.212.312.410 …Reaction Stoichiometry at standard conditions Example 1 - How many L of Oxygen are needed to react with 50.0L of Hydrogen at STP? 2H 2(g) + O 2(g) ---> 2 H 2 O (g) –But...we can skip steps 1 and 3. –Why? All gases take up the same amount of space at STP.

11. Gas Laws 12.112.212.312.411 Example 2 - Calculate the volume of one mole of H 2 at 20ºC and 1000 torr. …The Combined Gas Law at nonstandard conditions In both examples, the relationship between the volume and mass of a gas can be compared because…

12. Gas Laws 12.112.212.312.412 Avogadro’s Principle 1 mole of gas at STP = ____________ “Equal volumes of gases at equal temperatures and equal temperatures and equal pressure contain an equal number of molecules.”

13. Gas Laws 12.112.212.312.413 In general… The combined gas law, P o T n V o = P n T o V n, is used for changing conditions. But, a new gas law, the Ideal Gas Law, can be introduced when you have problems containing: –one set of _________________ –solving for _________________ –calculating ____________________________ –calculating _________________ –involving _______________________________

14. Gas Laws 12.112.212.312.414 12.2 The Ideal Gas Law A relationship between pressure, volume, temperature and the # of moles of a gas A new formula that can help us solve all kinds of gas law problems more easily. PV = nRT.

15. Gas Laws 12.112.212.312.415 The Ideal Gas Law Again, the Ideal Gas Law, can be introduced when you have problems containing: –one set of conditions –solving for grams –solving for moles –calculating molecular weight (molar mass) –calculating density –involving stoichiometry and non-STP conditions Let’s try sample problems for each of the above conditions.

16. Gas Laws 12.112.212.312.416 One Set of Conditions Example: Calculate the volume of 1.00 mole of Hydrogen at 20.0 ˚C and 1000 torr Step 1: Convert your pressure to atm. Step 2: Write the ideal gas law and derive Volume. Step3: Using V = nRT/P, we have

17. Gas Laws 12.112.212.312.417 Solving for Grams Example: Calculate the number of grams of helium in a 6.0 liter cylinder at 27˚C and 800 torr. Step 1: Convert your pressure to atm. Step 2: Write the ideal gas law and derive for moles. PV = nRT n = PV/RT 800 torr x 1 atm / 760 torr = 1.05 atm Step 3: Calculate the moles. n = (1.05 atm)(6.0 L)/(.0821 L atm/mol K)(300K) n = 0.26 moles Step 4: Convert to grams..26 moles x 4.0 g/mole = 1.04 g He

18. Gas Laws 12.112.212.312.418 Solving for Moles Example: A sample of CO 2 in a 10.0 L container at 293K exerts a pressure of 50,000 torr. How many moles of CO 2 are in your sample? Step 1: Convert your pressure to atm. Step 2: Write the ideal gas law and derive for moles. Step 3: Calculate the moles.

19. Gas Laws 12.112.212.312.419 Calculating M.W. Example: If 18.0 grams of a gas at 380 torr and 546.0 K occupies 44.8 L, what is the molecular weight of the gas? Step 2: Identify the m.w. formula. So, plug in 18.0 grams into the formula. Step 3: Now, use PV = nRT to solve for n (the number of moles). Step 1: Convert your pressure to atm. Step 4: Plug this value into the molecular weight (m.w.) equation.

20. Gas Laws 12.112.212.312.420 Step 2: If it is assumed that one mole of CO 2 is present, then the mass can be calculated from the periodic table and the volume at STP is 22.4 L. Calculating Density at STP Example: Find the density of carbon dioxide at STP. Step 1: Identify the density formula.

21. Gas Laws 12.112.212.312.421 Step 2: Assume one mole of CO 2. Thus, CO 2 weighs 44.0 grams. Plug this into the Density formula. Calculating Density (not at STP) Example: Find the density of carbon dioxide at 546.0 K and 4.00 atm. Step 1: Identify the density formula. Step 3: Because the conditions are non-STP values, 22.4 L can’t be used. So, use PV=nRT and solve for V (assume one mole). Step 4: Plug this value into the density equation.

22. Gas Laws 12.112.212.312.422 Deviations from Ideal Behavior Real gases do not behave according to the KMT - Why? –Real gases have molecules that occupy space –Real gases have attractive and repulsive forces Ideal gases conform exactly to the KMT –no such gas exists –gases only behave close to ideally at low P and high T. –At low T and high P, gases deviate greatly from ideal behavior. There is an equation that enables us to account for deviations in behavior - it contains correction factors that are specific for a gas - Van der Walls equation – you’ll see it in College Chemistry Some gases are close to ideal (if they are small and nonpolar): –H 2, He, Ne –O 2 and N 2 are not too bad –NH 3, H 2 O are not even close to ideal

23. Gas Laws 12.112.212.312.423 Deriving the Ideal Gas Law Let’s derive the ideal gas law and gas constant... –Volume is proportional to 1/P (as P is reduced, the V increases) –V is proportional to T (as T increases, the V increases) –V is proportional to n (as more moles are added, the V increases)

24. Gas Laws 12.112.212.312.424 12.3 Stoichiometry of Gases At STP, there is nothing new here. The only thing new is that 1 mole = 22.4 L must be adjusted if not at STP. Problems for Stoichiometry of Gases include converting: –_________________________

25. Gas Laws 12.112.212.312.425 Grams to Liters Example: How many liters of O 2 are generated when 50.0 grams of sodium chlorate decomposes at 0.950 atm and 20.0˚C? Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2: Plug this value into the appropriate step of the stoichiometry problem. 2NaClO 3(s) + heat ---> 2NaCl (s) + 3O 2(g)

26. Gas Laws 12.112.212.312.426 Liter to Grams Example: If a lawn mower engine generates 555.0 L CO 2 on a lovely Sunday afternoon (.996 atm and 37.00 C 0 - well it is kind of hot today...), how many grams of octane were consumed? Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2: Plug this value into the appropriate step of the stoichiometry problem. 2C 8 H 18(l) + 25O 2(g) ---> 16CO 2(g) + 18H 2 O (g)

27. Gas Laws 12.112.212.312.427 Liter to Liter (at STP) Example: How many L of CO 2 can be made from the combustion of 2.00 L of propane (C 3 H 8 ) at STP? –Again...we can skip steps 1 and 3. –Since, all gases take up the same amount of space at STP. 4C 8 H 3(l) + 18O 2(g) ---> 16CO 2(g) + 6H 2 O (g)

28. Gas Laws 12.112.212.312.428 Liter to Liter (not at STP) Example: How many L of carbon dioxide can be made from the combustion of 2.00 L of propane (C 3 H 8 ) at 500. K and 3.00 atm? 4C 8 H 3(l) + 18O 2(g) ---> 16CO 2(g) + 6H 2 O (g) Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2 : Plug this value into the appropriate step of the stoichiometry problem. –Again...we can skip steps 1 and 3.

29. Gas Laws 12.112.212.312.429 12.4 Effusion and Diffusion Graham’s Law of Diffusion - (English - 1824) - Graham noticed that gases with low densities diffuse ____________ than gases with higher densities “Under the same conditions of temperature and pressure, gases diffuse at a rate ____________ proportional to the square roots of their ____________ ____________ “

30. Gas Laws 12.112.212.312.430 Graham’s Law Formulas

31. Gas Laws 12.112.212.312.431 Gas “A” has a kinetic energy of ½ mv 2 and Gas “B” has a kinetic energy of ½ mv 2. At the same temperature, the two gases will have the same kinetic energy. Derive Graham’s Law from Gas A ½ mv 2 = Gas B ½ mv 2

32. Gas Laws 12.112.212.312.432 In other words…. If CO 2 molecules travel at 200.0 mph, how fast do H 2 molecules go?

33. Gas Laws 12.112.212.312.433 If He atoms travel at 800.0 mph, how fast do nitrogen molecules go? A second example….

34. Gas Laws 12.112.212.312.434 Citations Textbook –Nicholas D. Tzimopoulos, H. Clark Metcalfe, Williams, Castka, Holt Modern Chemistry. Austin, Texas: Holt Rinehart & Winston, 1993.