1. Gases Chapter 10 AP Chemistry

2. I. Characteristics of Gases air 78% Nitrogen 21% Oxygen normal solids & gases become vapors composed of non- metals simple formulas low molar mass expand to fill container spontaneously molecules are far apart – have no I.F. most important properties – tempt- volume, # of molecules

3. II. Pressure Atmospheric Pressure P = F/A force/area SI = measured in Pascal's Pa = 1N/m^2 one Newton/meter sq 1N = 1kg-m/s^2 Atmos. Pressure = 1x10^5 Pa or 1x10^2 kPa

4. Measuring Atmospheric Pressure Based on a principle developed by Evangelista Torricelli in 1643, the Mercurial Barometer is an instrument used for measuring the change in atmospheric pressure Mercurial Barometer

5. Pressure Units 1 atmos = 760 mmHg = 760 torr you got it! 1 torr = 1 mmHg 1 atmos = 1.01325 x 10^5 Pa = 101.325 kPa

6. Robert Boyle (1627-1691)

7. Gases are Kool So as pressure increases the volume decreases The value of the constant depends on the temperature and amount of the gas

8. Jacques Charles (1746-1823)

9. Gases are Kool Increase the temp increase the vol

10. Joseph-Gray-Lussac (1788-1823)

11. Example

12. Amadeo Avogadro Using Lussas’s Law

13. Double the number of moles of a gas at the same T&P doubles the volume Later it was found that at STP ( one atm and 0 deg C) one mole (6.02 x 10^23 molecules) of any gas occupies exactly 22.4 liters of that gas.

15. Ex 1. One mole of any gas occupies? PV = nRT V = nRT/P Avogadro’s # = 22.4 l

16. Ex 2. using the Ideal Eq A sample of KNO 3 is heated producing O 2 gas in a 750 ml flask under 2.8 atms of pressure. If the temperature is 53.6 deg C how many moles are produced?

17. Combined Gas Law formula P 1 V 1 N 1 =N 1 R 1 T 1 = P 2 V 2 N 2 =N 2 R 2 T 2 If the moles are held constant we can determine how changes in tempt, pressure, and volume will effect the gas. Using the combined gas law formula P 1 V 1/ T 1 = P 2 V 2/ T 2

18. Ex 1 combined Gas Law If one liter of air at room temp 25 deg C and one atmos is compressed to a volume of 3.3 ml at a pressure of 1000 atm. What is the new temp of the sample?

19. Density – Molar Mass and the Ideal Gas Eq. Derive M= dRT/P and d = MP/RT Given D= mass/vol N = m (mass)/M (molar mass)

20. Ex. Density prob. What is the density of carbon tetrachloride at 714 torr and 125 deg C?

21. Gas Mixtures and Partial Pressures

22. Combining formulas P total = P 1 + P 2 + P 3 …. = P 1 = n1 RT/V 1 + n2 RT/V 2 ….. = P t = n t = RT/V At cons. temp and volume the total pressure of a gas sample is determined by the number of moles of a gas present – whether a single gas or a mixture.

23. Ex prob. A gaseous mixture made of 10g of oxygen and 10g of methane is placed in a 10 liter vessel at 25 deg c. What is the total pressure in the vessel?

24. Partial Pressures and Mole Fractions The ratio of partial pressure of one component of a gas to the total pressure is P 1 = n 1 RT/V = n1 Mole P t = n t RT/V nt Fraction Mole fraction of gas 1 denoted X 1 The partial pressure of a gas is equal to its mole fraction times the total pressure X 1 = P 1 or P 1 = X 1 P t P t

25. Ex Prob. A synethic atmosphere is created by blending 2 mol percent CO 2, 20 mol percent O 2 and 78 mol percent N 2. If the total pressure is 750 torr calc the partial pressure of the oxygen component.

26. Collecting Gas Over Water Its an old trick but it just might work! A sample of KClO 3 is decomposed producing O 2 gas over water. The volume of the gas collected is 0.25 l at 25 deg C and 765 torr. How many mole of O 2 are collected?

27. Collecting Gas over Water Pressure of Water Table TempPressure Torr 04.58 2523.76 3542.2 65187.5 100760

28. Adjusting for Water Pressure Pt = 760 torr = PO 2 + PH 2 O PO 2 = 765 – 23.76 = 741.2 PO 2 = 741.2/760 = 0.975 atm N1= PV/RT 0.975atm * O.25 l K*mol (273 +25 k) ).0821 L* atm = 9.96 x 10^-3 moles How many grams of KClO 3 were decomposed?

29. Kinetic Molecular Theory (KMT) A. A gas consists of molecules in constant random motion Ek (energy kinetic) = ½ m X speed^2 B. KMT & Ideal Gases (5 postulates) 1.Size of gas molecules are negligible compared to the average distance between them. 2.Molecules travel in random patterns in straight lines in all directions at various speed. Properties are same in all directions.

30. (KMT postulates) 3. Intermolecular forces are weak except when they collide. Molecules travel with unchanging speeds until they collide with each other or the sides of the container. 4. Collisions between molecules are elastic ( no energy is lost) 5. The average Ke of a molecule is proportional to the absolute temperature.

31. C. The ideal gas law from KMT Pressure in KMT depends upon the collisions frequency and the force exerted – p is port freq x average force Average force = mass and average speed mu (momentum) Freq = is also proportional to the average speed mu ( faster means more collisions per unit time) Freq is inversely proportional to the gas volume Inc. vol decrease freq collisions

32. Freq is directly proportional to the N number of molecules in the gas vol. P is port ( mu x 1/v x N) x mmu PV is port (Nmmu^2) or Mmu^2 = ke port T So if the average KE of a molecule of mass m and average speed mu is ½ mmu^2 then PV port to NT also postulate 5 and N is port to moles of gas then PV = nRT

33. Molecular Speeds Diffusion and Effusion Molecules are constant random motion, therefore molecular speed varies over a range of values. At any temp molecular speed varies widely but most are close to the average speed corresponding to the max in the distribution curve Note the distribution of molecules at diff speeds is temp dep (called a maxwell distribution

34. Diffusion and Effusion The root-mean square (rms) molecular speed, mu, is a type of average molecular speed equal to the speed of a molecule having average molecular kinetic energy Mu =

35. Diffusion and Effusion R = molar gas constant (careful units must match) T = absolute tempt M = molar mas of the gas Diffusion – is the process whereby a gas spreads out through another gas to occupy the space uniformly - molecules move chaotically (randomly) - molecules never travel very far in one direction

36. Diffusion and Effusion - complicated to calculate rate Effusion – the process in chich a gas flows through a small hole in a container - Graham’s law of effusion – at cons. tempt & pressure the rate of effusion of gas molecules is inversely proportional to the square root of the molecular weight of the gas - Rate of effusion depends upon cross sec. of whole, # mol per unit vol, average molecular speed

37. Graham’s Law Because mu (average speed) = the square root of 3RT/M M = molar mass Then Graham’s Law Rate of effusion is port 1/ square root of the molar mass Calc. the ration of effusion rates of molecules of CO 2 and SO 2.