1. Chapter 11.1

2. Pressure Macro-Scale Pressure is the amount of force exerted over a given area  Familiar unit is “pounds per square inch” or psi (tire pressure)  This is how many pounds of force something pushes on every 1 square inch of surface contact it has with an object  Pressure can be increased by: 1.Increasing the force 2.Decreasing the amount of surface area

3. Atmospheric Pressure Atmospheric pressure is the amount of pressure exerted on an object by all the atmosphere directly above that object  Remember that air is made of particles, which have mass. That massive amount of atmosphere is pushing down on you at all times

4. STP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

5. Measuring Pressure Kilopascals:  The standard metric unit for measuring pressure. Atmospheres:  A convenient unit where the pressure at 0°C at sea level is equal to 1.00 atm. mmHg:  A unit based on the amount of liquid mercury that can be pushed up a tall column by atmospheric pressure.

6. Pressure conversions 760mmHg = 1atm= 14.7 lb/in 2 =101.3kPa = 760Torr = 101,300Pa All of the above represent the force exerted by atmospheric pressure at sea level.

7. Pressure conversion problems Convert 750. mmHg to atm 750. mmHg 1 atm 760 mmHg = 0.987 atm

8. Devices that Measure Pressure  Barometer  measures atmospheric pressure Mercury Barometer Aneroid Barometer

9. Devices that Measure Pressure Manometer  A device that uses a fluid in the bottom of a U- shaped tube to measure differences in pressure. Gas sealed in a container with an unknown pressure that we want to measure. Fluid in the U-shaped tube. Fluid level on the side containing the gas of unknown pressure (height in units of mm). End open to the air. Pressure is equal to atmospheric pressure.

10. Devices that Measure Pressure ΔP When pressure is greater on one side of the fluid, that side of the fluid is pushed down and the fluid on the side with less pressure goes up. The difference between the two heights is equal to the difference in pressure. If we know the atmospheric pressure and the pressure difference (ΔP), we can calculate the unknown pressure of the gas sealed in the container using the following formula: P 1 = P 2 + ΔP, where ΔP = h 2 – h 1. h1h1 h2h2

11. Devices that Measure Pressure  Consider the following example:  What is the pressure of the gas in the sealed part of the manometer? P 1 = ? P 2 = 730 mmHg h 1 = 84 mmHg h 2 = 127 mmHg 1.We know the gas inside the sealed bulb has a greater pressure than the atmospheric pressure because it is pushing the fluid down on its side of the U-tube. 1.The change in height is equal to h 2 – h 1. So ΔP = 127 mmHg – 84 mmHg = 43 mm Hg. 3.The pressure of the gas in the sealed glass bulb is equal to the atmospheric pressure plus the change in pressure. So this means that P 1 = 730 mmHg + 43 mmHg = 773 mmHg

12. Closed Manometer  The pressure in the chamber is equal to h.

14. A. Boyle’s Law P V PV = k

15. A. Boyle’s Law  The pressure and volume of a gas are inversely related  at constant mass & temp P V PV = k

16. A. Boyle’s Law

17. V T B. Charles’ Law

18. V T  The volume and absolute temperature (K) of a gas are directly related  at constant mass & pressure

19. B. Charles’ Law

20. P T C. Gay-Lussac’s Law

21. P T  The pressure and absolute temperature (K) of a gas are directly related  at constant mass & volume

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

23. GIVEN: V 1 = 473 cm 3 T 1 = 36.0°C = 309K V 2 = ? T 2 = 94.0°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems  A gas occupies 473 cm 3 at 36.0°C. Find its volume at 94.0°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

24. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems  A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

25. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25.0°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 E. Gas Law Problems  A gas occupies 7.84 cm 3 at 71.8 kPa & 25.0°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

26. GIVEN: P 1 = 765 torr T 1 = 23.0°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems  A gas’ pressure is 765 torr at 23.0°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

27. = kPV PTPT VTVT T D. REVIEW Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

28. Gases

29. V n A. Avogadro’s Principle  Equal volumes of gases contain equal numbers of moles  at constant temp & pressure  true for any gas

30. PV T VnVn PV nT A. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K = R You don’t need to memorize these values! Merge the Combined Gas Law with Avogadro’s Principle:

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

32. GIVEN: P = ? atm n = 0.412 mol T = 16.0°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm C. Ideal Gas Law Problems  Calculate the pressure in atmospheres of 0.412 mol of He at 16.0°C & occupying 3.25 L.

33. GIVEN: V = ? n = 85 g T = 25.0°C = 298 K P = 104.5 kPa= 1.03 atm R = 0.082 L  atm/mol  K C. Ideal Gas Law Problems  Find the volume of 85 g of O 2 at 25.0°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (1.03)V=(2.7) (0.082) (298) atm mol L*atm/mol  K K V = 85.5 L

34. Gases

35. B. Dalton’s Law  The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

36. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa B. Dalton’s Law  Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

37. GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr  A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? Look up water-vapor pressure for 35.0°C. Sig Figs: Round to least number of decimal places. B. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

38. C. Graham’s Law b Diffusion  Spreading of gas molecules throughout a container until evenly distributed. b Effusion  Passing of gas molecules through a tiny opening in a container

39. C. Graham’s Law KE = ½mv 2  Speed of diffusion/effusion  Kinetic energy is determined by the temperature of the gas.  At the same temp & KE, heavier molecules move more slowly.  Larger m  smaller v

40. C. Graham’s Law  Graham’s Law  Rate of diffusion of a gas is inversely related to the square root of its molar mass.  The equation shows the ratio of Gas A’s speed to Gas B’s speed.

41.  Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. C. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

42.  A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? C. Graham’s Law Put the gas with the unknown speed as “Gas A”.

43.  An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. C. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.

44. A. Gas Stoichiometry  Moles  Liters of a Gas:  STP - use 22.4 L/mol  Non-STP - use ideal gas law  Non- STP  Given liters of gas?  start with ideal gas law  Looking for liters of gas?  start with stoichiometry conv.

45. What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC?  CaCO 3  CaO + CO 2

46. How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C?  4 Al + 3 O 2  2 Al 2 O 3