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1. Gases and Liquids

2. Topics • Properties of gases • Gas laws • Gas Stoichiometry • Diffusion / Effusion, defined • Liquids - properties

3. The Properties Of Gases • Gases exert pressure • Gases can be compressed • A gas takes up the volume of the container it is in. • Gases have no free surfaces • Gases are fluid

4. Free Surface... A free surface is one that does not have to be contained by a wall of a vessel. A gas must be contained on all sides or it will flow out of the container.

5. Gases exert pressure Pressure is defined as the force per unit area. Gas particles are always colliding with each other and with the walls of the container in which they are held.

6. The collisions with the walls of the container produce pressure. Gas also gets less dense when heated

7. Why does a gas get less dense when heated ? When matter is heated, its particles begin to vibrate more. They have more kinetic energy.

8. The particles get farther apart when they possess more kinetic energy. The more kinetic energy they possess, the more the particles will overcome the intermolecular forces that hold them close to each other.

9. Since density = mass per unit of volume The density decreases because the particles move farther apart, thus decreasing the amount of mass in a given unit of volume.

10. Gases can be compressed Gas is compressible because the particles are far apart. Ex. A scuba tank may have 58 L of O2 and He compressed into a 5 L tank. The pressure in the tank would then be 11.7 atm.

11. Atmospheric pressure is measured with a barometer. • Some of the units associated with the measurement of pressure are: • mm Hg (millimeters of mercury) • torr • atm. (atmospheres) • bar (we will not use this one)

12. Converting from one unit to another 1 atm. = 760 mm Hg = 760 torr Ex.1 Convert 752.1 mm Hg to atm. 752.1 mmHg x 1 atm. =0.989= 0.99 760 mmHg atm.

13. Boyle’s Law - This gas law states that as the pressure on a gas is increased, its volume decreases. Pressure is inversely proportional to volume.

14. P 1 V 1 = P 2 V 2 If the pressure of a gas is changed while the temperature is held constant, the new volume can be calculated, using the above equation, (Boyle’s law). Or… If the volume is changed the newpressure can be calculated.

15. Ex. 2 Boyle’s Law A sample of Chlorine gas occupies 946 mL at a pressure of 726 mmHg. What is the pressure of the gas if the volume is reduced to 154 mL at constant temperature ? 946 mL x 726 mmHg = P2 x 154 mL P2 = 4.46 x 10 3 mm Hg

16. Charles’s Law - This gas law states that as the temperature of a gas is increased, the volume of the gas increases when pressure is constant. Volume is directly proportional to temperature.

17. V 1 / T 1 = V 2 / T 2 If the temperature of a certain volume of gas is changed while the pressure is held constant, the new volume can be calculated using the above equation, (Charles’s law). If the volume is changed the newtemperature can be calculated.

18. Ex. 3 Charles’s Law 452 mL of F2 is heated from 22.0o C to 187o C at constant pressure. What is the final volume ? PROBLEM ! ALL GAS CALCULATIONS MUST BE DONE USING ABSOLUTE TEMPERATURE,KELVIN.

19. Convert the o C to kelvin... K = 22.0 oC + 273 = 295 K K = 187 oC + 273 = 460 K

20. V 1 / T 1 = V 2 / T 2 452 mL / 295 k = V2 / 460 k V2 = 705 mL

21. The combined gas law - This is a combination of the last two laws and one other law. P1 V1/ T1 = P 2 V 2/ T 2

22. Ex. 4 Combined gas law - A small bubble with volume of 2.1 mL at the bottom of a lake where the pressure is 6.4 atm and the temperature is 8.0oC rises to the surface. The surface pressure is 1.0 atm. and the temperature is 25oC. Calculate the new volume.

23. It is helpful to write out the variables... P1 = 6.4 atm. P2 = 1.0 atm. V1 = 2.1 mL V2 = ? T1 = 8.0 + 273 T2 = 25 + 273 T1 = 281 k T2 = 298 k

24. Rearrange to solve for V2 V 2 = V 1 x P 1 x T2 P 2 x T 1 V2 = 2.1 atm. x 6.4 atm x 298 k 1.0 atm. x 281 k V2 = 14 mL

25. THE IDEAL GAS LAW - This equation explains how a gas would behave if it were an ideal or perfect gas. Real gases are not ideal but at high temperatures and low pressure they follow this equation fairly closely.

26. The Ideal Gas Law - The ideal gas law accounts for moles, pressure, volume, temperature, and uses a proportionality constant called the gas constant, R.

27. P V = n R T The gas constant, R, has the value, 0.0821 L atm / mole k These units will cancel with the undesired units in the calculation.

28. Ex. 5 Ideal Gas Law - Calculate the pressure (in atm) exerted by 1.82 moles of SF6 gas in a steel vessel with a volume of 5.43 L at 69.5oC. Note: When using the ideal gas law, you must use units of L, atm., and K to match the units in R.

29. Ex. continued... Write out the variables... P = ? n = moles = 1.82 moles V = 5.43 L T = 69.5oC +273 =342.5 k R = 0.0821 L atm / mole k

30. Rearrange to solve for P. P = nRT V P=(1.82 mol)(0.0821 L atm /mol k)(342 k) 5.43 L P = 9.42 atm

31. Dalton’s Law Of Partial Pressures This law states that the sum of the individual pressures of the gases in a mixture of gases is equal to the total pressure of the mixture. PT = P1 + P2 +... Pn Pi = xi (PT) and...

32. P i = xi (P T) In this equation the ‘ i ’ stands for the individual gas, the ‘ x ’ stands for the mole fraction of the individual gas in the mixture. mole fraction is the desired gas moles, divided by the total moles of gas.

33. Ex. 6 Dalton’s Law of Partial Pressures P i = x i (P T ) and Xi = moles i moles i + j + k +…

34. Ex. 6 continued... A mixture of gases contains 4.46 moles of Ne, 0.74 moles Ar, and 2.15 moles Xe. Calculate the partial pressures of the gases if the total pressure is 2.00 atm.

35. First calculate the mole fraction XNe = nNe= 0.607 nNe +n Ar + n Xe The same procedure is followed to find the mole fractions of Ar and Xe. They are 0.10 and 0.293 respectively.

36. Finally, Pi = Xi (PT) PNe = 0.607 x 2.00 atm = 1.21 atm PAr = 0.10 x 2.00 atm = 0.20 atm PXe = 0.293 x 2.00atm = 0.586 atm

37. PT = P1 + P2 + ...Pn Ex. 7 Calculate the partial pressure of oxygen gas collected through water if the total pressure is 762 mm Hg and the water vapor pressure is 22.4 mm Hg. POxygen = PT - P water = 740 mm Hg

38. Gas Stoichiometry An important piece of information; At STP, 1 mole of gas occupies 22.4 L

39. STP STP is defined as ... STANDARD TEMPERATURE AND PRESSURE STANDARD PRESSURE IS EQUAL TO 1 atm. STANDARD TEMPERATURE IS EQUAL TO 0oC OR 273 k.

40. 22.4 Lxgrams mole liter With this information, the molarmass of a gas at STP can be calculated, given the density. or The density can be calculated, given the molar mass.

41. Ex. 8 Molar Mass of a Gas The density of a gas at STP is equal to 0.761 g/L. Calculate its molar mass. 0.761 g x 22.4 L = 17.04 g / mole Liter mole This gas would be ammonia.

42. Molar Mass and Density If given the quantity, in grams, the Ideal gas law can be used to calculate the molar mass of a gas when not at STP. Rearrange... n = PV / RT Solve for n, then divide grams /n

43. n = PV / RT Ex. 9 Calculate the molar mass of a 0.100 g sample of a gas that fills 0.0470 L at 298 k and 0.993 atm. n = 0.993 atm x 0.0470 L =0.00191 0.0821 L atm 298 k mol mole k

44. Ex. 9 continued... 0.100 g = 52.4 g/mol 0.00191 mole

45. Gas Stoichiometry At STP, the coefficients in a balanced chemical equation are related to the volumes of gas that react.

46. Ex. 10Stoichiometry Calculate the volume of O2 at STP, required for the complete combustion of 2.64 L of acetylene (C2H2) at STP. Remember the first step in stoichiometry is to write the balanced equation.

47. 2C2 H2 (g) + 5 O2(g) ----->4CO2(g) + 2H2O(l) volume of O2 = 2.64 L C2 H2 x 5 L O2 2 L C2 H2 = 6.60 L O2

48. Ex. 11 Stoichiometry Sodium Azide (NaN3) is used in some automobile air bags. The NaN3 is decomposed when the collision impact triggers the reaction. Calculate the volume of N2 produced at 21oC and1.08 atm for 60.0 g of NaN3.

49. 2 NaN3(s) ----> 2 Na (s) + 3 N2(g) Since the reaction is NOT at STP we must use the Ideal Gas Law. We were given T and P and grams, but not moles, n. We must first find the moles in order to use the Ideal gas law to find volume.

50. moles N2 = =60.0 g NaN3x 1 mol NaN x 3 molN2 65.02g NaN3 2mol NaN3 = 1.38 mol N2 continued...