1. Gases Chang Chapter 5

2. Chapter 5 Outline Gas Characteristics Pressure The Gas Laws Density and Molar Mass of a Gas Dalton’s Law of Partial Pressure Kinetic Molecular Theory of Gases Molecular Effusion and Diffusion Deviations from Ideal Gas Behavior

3. Gases Characteristics Which substances are most likely to be gases at 25 o C and 1 atm? –Not ionic compounds –Most likely molecular compounds Properties of Gases –Assume shape and volume of container –Easy to compress –Mix with other gases completely –Have low densities

4. Pressure Pressure = Force = m ∙ a Area area »Units = Pascals (Pa) = Newtons m2m2 –Atmospheric pressure varies with altitude –Pressure is measured using a barometer –Units of Pressure 1 atm = 760 torr = 101,325 Pa = 760 mmHg (UNIT FACTORS)

5. The Gas Laws Boyle’s Law (1661) –Relates Pressure and Volume of Gases –P  1/V or PV = k 1 or P = k 1 /V –The pressure of a fixed amount of gas at constant temperature is inversely proportional to the volume of the gas.

6. The Gas Laws Charles’ and Gay-Lussac’s Law –Relates Temperature and Volume of Gases –V  T or V = k 2 ∙T (T must be in K) –The volume of a fixed amount of gas at constant pressure is directly proportional to the absolute temperature of the gas

7. The Gas Laws Avogadro’s Law –Relates the Volume of a Gas with the Amount of Gas –V  n or V=k 3 ∙n –At constant pressure and temperature the volume of gas is directly proportional to the number of moles of gas present

8. The Gas Laws Combing all 3 gas laws yields the IDEAL GAS EQUATION –PV = k 1 and V = k 2 ∙T and V = k 3 ∙n PV = k 1 k 2 k 3 nT k 1 k 2 k 3 = constant = R = 0.8206 L∙atm/mol∙K PV = nRT

9. The Gas Laws Example: What volume of gas would be occupied by 1.00 mol N 2 at STP? STP = standard temperature (0 o C) and pressure (1 atm) UNIT FACTOR!!!!

10. The Gas Laws Example: A sample of CH 4 in a 2.00 L vial at 25 o C and 0.750 atm is compressed and heated until T=75 o C and the new volume is 100L. What is the new pressure?

11. Density and Molar Mass of Gas PV = mRT therefore n/V = P/RT n= m/ M (substitute) m/ M V = P/RT Therefore: m/V = P M /RT = d = density Therefore: M = dRT/P = molar mass

12. Dalton’s Law of Partial Pressure “the total pressure of a mixture of gases is the sum of the pressures that would exist if each gas were present by itself” Consider a mixtuer of gases A, B, and C –P A =n A RT/V, P B =n B RT/V, and P C =n C RT/V –P Total =P A +P B +P C =n A RT/V + n B RT/V + n C RT/V = (n A +n B +n C )(RT/V) = n Total (RT/V)

13. Dalton’s Law of Partial Pressure A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15 L vessel at 0 o C. What is the particle pressure of each gas and the total pressure in the vessel?

14. Dalton’s Law of Partial Pressure

15. Example: If 50mL of gas are collected at a pressure of 750.0 torr at 25 o C, how many grams of Zinc Reacted? Zn (s) + HCl (aq)  H 2(g) + Zn 2+ (aq) + Cl - (aq)

16. Dalton’s Law of Partial Pressure

17. Kinetic Molecular Theory of Gases Assumptions Distance between particles is very large compared to the size of individual particles Gas molecules undergo constant, random, rapid motion and experience frequent collisions with other particles Gas particles do not experience any attractive or repulsive forces w.r.t. other molecules The average kinetic energy per particle is: KE=1/2 mu 2 where m is mass and u 2 is the mean square speed – Any gas at the same temperature has the same KE –KE  T or 1/2mu 2  T

18. Kinetic Molecular Theory of Gases Results –Gas pressures result from collisions between particles and the container walls –Absolute temperature of a gas is really a measure of the average KE of the particles High T: more KE, more frequent collisions Low T: less KE, less frequent collisions –Many observations can be explained by the Kinetic Molecular Theory of Gases EX: gasses are easily compressed, P  T

19. Molecular Effusion and Diffusion Distribution of Molecular Speeds

20. Molecular Effusion and Diffusion All gases at the same temperature have the same KE –KE=1/2mu 2 A lighter gas (small m) must have a greater speed, u –u= =  because M is in the denominator, a less massive gas will be faster Note: T = K, M = kg/mol, R= 8.314 J/mol∙K

21. Molecular Effusion and Diffusion Graham’s Law of Effusion Effusion= the process by which a gas under pressure escapes through a pinhole to an area a lower pressure The effusion rate of a gas is inversely proportional to the square root of the M of the gas Assuming we have two gases at the same T and P in identical containers, How 235 U and 238 U are separated

22. Molecular Effusion and Diffusion Example: An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O 2 at the same T. What is the molar mass of the unknown gas and identify it.

23. Molecular Effusion and Diffusion Graham’s Law of Diffusion Diffusion= the gradual mixing of molecules of different gases by virtue of their kinetic properties Diffusion is a slow process It turns out that the rate of diffusion is also inversely proportional to the molar mass so the same equation applies

24. Deviations of Gases from Ideal Behavior All gases deviate from ideal behavior in one way or another –Real gases deviate from ideal behavior at high pressures, and low temperatures The van der Waals Equation corrects the Ideal Gas Equation to account for this non-ideal behavior***** a and b are van der Waals constants determined experimentally for each gas, found in table 5.4

25. Deviations from Ideal Gas Behavior Example: Estimate the real pressure exerted by 1.000 mol of Cl 2(g) in 22.4 L at 0.0 o C. Compare this to the idealized pressure.