2. KMT and Gas Laws

3. Characteristics of Gases Gases expand to fill any container. –random motion, no attraction Gases are fluids (like liquids). –no attraction Gases have very low densities. –no volume = lots of empty space

4. Characteristics of Gases Gases can be compressed. –no volume = lots of empty space Gases undergo diffusion & effusion. Diffusion is the tendency of molecules and ions to move from areas of high concentration to areas of low concentration. The diffusion of gas through a small opening in an otherwise closed container is called effusion.

5. Kinetic-Molecular Theory (KMT) Based on the idea that particles of matter are always in motion Provided a model of “ideal gas” –An imaginary gas that perfectly fits all the assumptions of the KMT. FIVE (5) assumptions on KMT K.E. = (½) mv 2

6. Assumptions 1.Gases consist of large number of tiny particles that are far apart relative to their size. 2.Collisions between gas particles and between particles and container walls are elastic collisions (no loss in K.E.). 3.Gas particles are in continuous, rapid, random motion; K.E. = (½)mv 2

7. Assumptions 4.There are no forces of attraction or repulsion between gas particles. 5.The average kinetic energy of gas particles depends on the temperature of the gas. Temperature : Kinetic Energy

8. Pressure Which shoes create the most pressure?

9. Atmospheric Pressure Barometer –measures atmospheric pressure Mercury Barometer Aneroid Barometer Atmospheric pressure is measured with a barometer.

10. Pressure Units and Conversions Kilopascal = kPa; just 1000 times greater than a Pascal Relationships between pressure units: **important for conversions!!! 1 atm = 14.7 psi = 760 mm Hg = 29.9 in Hg = 760 Torr = 101,300 Pa = 101.3 kPa

11. Temperature ºF ºC K -459 32 212 -2730100 0 273 373 K = ºC + 273 Always use absolute temperature (Kelvin) when working with gases.

12. Boyle’s Law The pressure and volume of a gas are inversely related –at constant mass & temp P V P 1 V 1 =P 2 V 2

13. Example: A 5.0 L container of nitrogen gas is at a pressure of 1.0 atm. What is the new pressure if the volume is decreased to 500 mL, and the temperature remains constant? Formula needed Rearrange for unknown Plug-in to formula Include units and cancellation! Final answerUnit Unit conversion(s) needed:

14. V n Avogadro’s Law Equal volumes of gases contain equal numbers of moles –at constant temp & pressure –true for any gas

15. Avogadro’s Law Avogadro’s Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of moles. This means that the coefficients in the balanced equation stand for volume as well as moles, but only for the gases. twice as many molecules

16. Example 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Answer: this time I'll use V 1 n 2 = V 2 n 1 (5.00 L) (1.80 mol) = (x) (0.965 mol) x =

17. Example Consider the chemical equation: 2NO 2 (g) → N 2 O 4 (g) If 25 mL of NO 2 gas is completely converted to N 2 O 4 gas,under the same conditions, what volume will the N 2 O 4 occupy? Solution: Here, from the equation given, 2 moles of NO 2 (g) forms 1 mole of N 2 O 4 (g). So, V1 = 25 mL n1 = 2 mole V2 =? n2 = 1mole V2 = (V1 n2) / n1, V2 = (25 mL) (1 mole) / (2 moles) = 12.5 mL

18. Example A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.) Solution: 1.Convert grams of He to moles: 2.00 g / 4.00 g/mol = 0.500 mol 2. Use Avogadro's Law: V1/n1 = V2/n2 2.00 L / 0.500 mol = 2.70 L / x x = 0.675 mol 3. Compute grams of He added: 0.675 mol - 0.500 mol = 0.175 mol 0.175 mol x 4.00 g/mol = 0.7 grams of He added

19. V T Charles’ Law The volume and absolute temperature (K) of a gas are directly related –at constant mass & pressure

20. Example: A container of helium gas at 25°C in an expandable 500 mL container is heated to 80°C. What is the new volume if the pressure remains constant? Formula needed Rearrange for unknown Plug-in to formula Include units and cancellation! Final answerUnit Unit conversion(s) needed:

21. P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume

22. Example: A tank of propane gas at a pressure of 3.0 atm is cooled from 90°C to 30°C. What is the new pressure if the volume remains constant? Formula needed Rearrange for unknown Plug-in to formula Include units and cancellation! Final answerUnit Unit conversion(s) needed:

23. = kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

24. Example: A helium-filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36°C. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28°C, what will be the new volume of the balloon? Formula needed Rearrange for unknown Plug-in to formula Include units and cancellation! Final answerUnit Unit conversion(s) needed:

25. Ideal Gas Law An ideal gas is one whose particles take up no space and have no attractive forces for each other. In reality, gases do not always behave ideally, especially under conditions of high pressure and low temperature.

26. A. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K PV=nRT You don’t need to memorize these values!

27. Example PV=nRT, R=0.0821 L  atm/mol  K 1.32.0 g of oxygen gas is at a pressure of 760 mm Hg and a temperature of 0 ° C. What is the volume of the gas? 2.How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25 ° C? 3.What pressure will be exerted by 25 g of CO 2 at a temperature of 25 ° C and a volume of 500mL? 4.At what temperature will 5.00g of Cl 2 exert a pressure of 900 torr at a volume of 750 mL?

28. YOU MUST CONVERT: P into atm –Use pressure conversions V into L –Divide by 1000 if mL n into moles –divide by molecular weight T into K –Add 273 to Celcius

29. Example: 32.0 g of oxygen gas is at a pressure of 760 mm Hg and a temperature of 0 ° C. What is the volume of the gas? Formula needed Rearrange for unknown Plug-in to formula Include units and cancellation! Final answerUnit Unit conversion(s) needed:

30. Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2

31. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V PB =PB = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T P i ----- Partial pressure P T ------- Total pressure X i -------- Mole fraction

32. GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases.diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.effusion is the movement of molecules through a small hole into an empty container.

33. GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.

34. GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He

35. Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube.