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E13: The Gas Laws

E13: The Gas Laws. Properties of Gases. All gases are influenced by Volume Temperature Pressure Amount Gases respond to changes in these variables in predictable ways. TEMPERATURE. Is measured in Kelvin (K) for gases

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E13: The Gas Laws

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  1. E13: The Gas Laws

  2. Properties of Gases All gases are influenced by • Volume • Temperature • Pressure • Amount Gases respond to changes in these variables in predictable ways.

  3. TEMPERATURE • Is measured in Kelvin (K) for gases • The Kelvin scale is based on the Celsius scale, but zero is set at Absolute Zero (the theoretical coldest possible temp) instead of at the freezing point of H2O like ˚C. • Absolute Zero (0 K) is -273.15˚C • Therefore, we can easily convert measurements between Celsius and Kelvin by adding or subtracting that value: • TempKelvin ≃ Temp˚Celsius + 273

  4. TempK  Temp˚C + 273 We don’t use ˚F in science because it is such a complicated conversion! To go between Kelvin & ˚C, simply add or subtract 273!

  5. TempK  Temp˚C + 273 TEMPERATURE PRACTICE • Water changes to steam at 100.00˚C. • What temperature is this in Kelvin? 100˚C + 273 = 373 K • A gas liquefies at -178˚C. • What temperature is this in Kelvin? -178˚C + 273 = 95 K

  6. VOLUME Gases will spread to fill the size and shape of their container. • Measured in L, mL, or cubic meter values If you need to convert, use these known equalities: 1L = 1dm3 = 1000 cm3 = 1000 mL

  7. CONVERSIONS • You MUST get used to canceling out by using CONVERSION FACTORS (equalities) • Given unwanted unit x equal value of desired unit equal value of unwanted unit Example: 200mL = ? dm3 200mL x _1dm3 = .2 dm3 1000mL • Get in the habit of WRITING UNITS to keep track of what you have cancelled. • Don’t forget sig figs!

  8. Equalities: 1L = 1dm3 = 1000 cm3 = 1000 mL Volume Conversion P.P. 3. A gallon jug of milk holds 3.78 L. • How many mL is this? 3.78L x 1000 mL = 3780 mL 1 1 L 4. A balloon contains 347 cm3 of air. • How many liters is this? 347cm3 x 1L = .347L 1 1000cm3 NOTE: look at the data for sig figs. Known equalities do not limit sig figs.

  9. PRESSURE • The amount of FORCE exerted on a given area. • Based on Pascals (Pa) but MANY different units can be used. • Useful conversion equalities: 101300 Pa= 101.3kPa = 1atm = 760mmHg = 760.torr = 14.7psi = 1 bar All these known equalities mean we can move between all of these units by setting up more conversion factors.

  10. 101300 Pa=101.3kPa=1atm=760mmHg=760.torr =14.7psi= 1 bar Pressure Conversion P.P. 5. A radio announcer reports an atmospheric pressure of 99.6 kiloPascals (kPa). • What is the pressure in atmospheres (atm)? 99.6 kPa x 1atm = .983 atm 1 101.3 kPa

  11. 101300 Pa=101.3kPa=1atm=760mmHg=760.torr =14.7psi= 1 bar Pressure Conversion P.P. 6. Express a pressure of 729 mmHg in atmospheres. 729mmHg x 1atm = .959 atm 1 760.mmHg

  12. KINETIC THEORY of MATTER • SOLIDS: FIXED volume & shape. Tightly packed atoms (most dense state). Molecules vibrate in place. • LIQUIDS: FIXED volume, FLUID shape. Molecules vibrate AND have limited rotation. • GASES: NO FIXED VOLUME or SHAPE. Least densely packed atoms, take the shape of container. Molecules vibrate, rotate, & travel freely. As TEMPERATURE goes UP Kinetic ENERGY goes Up (warm molecules move more!)

  13. “IDEAL” GASES: • Gases consist of small particles with mass • Compared to the size of their container, the volume of the gas particles is tiny. • The particles can collide without losing energy (= “elastic”) • Collisions w/ container walls cause PRESSURE • The speed of their movement is directly related to temperature (↑T ↑kinetic energy)

  14. STP:Standard Temp. & Pressure • Unless otherwise stated, gases will be assumed to be at Standard Temperature and Pressure (STP). • STP is based on air pressure at sea level (760.mmHg) and zero ˚C (= 273K) S.T.P. = 273 K & 760. mmHg

  15. BOYLE’S LAW:The inverse relationship between Volume & Pressure We explored this in E11: • As you pressed the syringe, you decreased the volume inside. • It got harder to push, because as thevolume decreased, the pressure of the gas inside increased. • Inverse relationship: V↓P↑ (or V↑P↓) Boyle’s Law assumes that temperature isn’t changed (is constant).

  16. SOLVING GAS LAW PROBLEMS • If you’re given a problem that involves more than one gas variable (vol., pressure, temp., or amount), you’ll need to figure out which are involved and how they will influence each other. • STEP 1: While reading the problem fill this in: (BEFORE) (AFTER) V1 = V2 = P1 = P2 = T1 = T2 = • You may need to convert unitsbefore moving on (like units should match, always use K for temp)

  17. BOYLE’S LAW P.P. #1 A sample gas occupies a volume of 2.00L at 27˚C and 1.00 atm pressure. What will thevolume be at same temperature but a pressure of 2.75 atm? STEP 1: Identify knowns & unknowns. (BEFORE) (AFTER) V1 = 2.00 L V2 = ? P1 = 1.00 atm P2 = 2.75 atm T1 = 27˚C T2 = 27˚C

  18. SOLVING GAS LAW PROBLEMS • STEP 1: Identify knowns & unknowns, convert. • STEP 2:Identify relationship involved Is it V-P (Boyle’s Law)? T-V (Charles Law)? or is it T-P (Guy Lussac’s Law) • STEP 3: Set up problem • STEP 4: Solve & check against known relationship.

  19. Boyle’s Law P.P. #1 STEP 1: Identify knowns & unknowns (BEFORE) (AFTER) V1 = 2.00L V2 = ? P1 = 1.00atm P2 = 2.75atm T1 = 27˚C T2 = same • STEP 2: Identify relationship involved Is it V-P (Boyle’s Law)? YES!V ↓ P ↑ T-V (Charles Law)? T V or is it T-P (Guy Lussac’s Law) T P

  20. SOLVING GAS LAW PROBLEMS • STEP 1: Identify knowns & unknowns, convert. • STEP 2: Identify relationship involved • STEP 3:Set up problem • You can use Algebraic method or conversions • I will use conversions, because in upcoming concepts you will HAVE to be familiar with it. • The conversion fraction is set up to cause the expected result. • STEP 4: Solve & check against the known relationship. Also check your sig figs!

  21. Boyle’s Law P.P. #1 • STEP 1: Identify knowns & unknowns V1 = 2.00L V2 = ? P1 = 1.00atm P2 = 2.75atm • STEP 2:Identify relationship:V↓ P↑ • STEP 3:Set up problem • Put your “lonely” known in front (V1 here) • Arrange the paired variable in a fraction to match the known relationship (V↓ P↑) by putting the larger value on the top or bottom of the fraction, whichever the ARROW of the unknown points to!(in this case, V should go ↓, so we put the larger P value on bottom: 1.00 atm / 2.75 atm)

  22. Boyle’s Law P.P. #1 • STEP 1: Identify knowns & unknowns V1 = 2.00L V2 = ? P1 = 1.00atm P2 = 2.75atm • STEP 2:Identify relationship: V↓ P↑ • STEP 3:Set up problem 2.00L x 1.00 atm = .727 atm 2.75 atm STEP 4: Solve & check against the known relationship. Also check your sig figs!

  23. BOYLE’S LAW P.P. #2 A balloon is inflated to a volume of 12.6L on a day when atmospheric pressure is 674mmHg. What’s th volume on a day when the pressure drops to 651mmHg? (Temp is constant) (BEFORE) (AFTER) V1 = 12.6 L V2 = ? P1 = 674mmHg P2 = 651mmHg Relationship = inverse: V ↑ P ↓ (so we will put larger P on top to make volume increase)

  24. BOYLE’S LAW P.P. #2 V1 = 12.6 L V2 = ? Relationship: P1 = 674mmHg P2 = 651mmHgV ↑ P ↓ Solve for V2: 12.6L x 674mmHg = 8,492.4L =13.0 L 1 651mmHg651 4) Check your answer! Did it go up or down as expected? Units? SIG FIGS should match given DATA 3) Do the math: cancel out units and multiply 2) Follow the arrow!V ↑ V2 should go UP, so put larger P on top of fraction! 1) Put lonely known over 1

  25. HOMEWORK: • BACK OF NOTESHEET:Do #1-3! • WORKBOOK: Read pgs. 65 & 66: Charles’ Law & Guy Lussac’s Law Do1st Practice Problem pg.65 & pg.66

  26. BOYLE’S LAW problems • A sample of CO2 gas occupies a volume of 8.75L at 0.940 atm. If the temperature remains constant, calculate the volume when pressure is 1.00atm. V1= V2= V P P1= P2= ________ x __________ = 1

  27. BOYLE’S LAW problems • At STP a balloon has a volume of 1.75L. When the balloon is moved to higher elev., the gas expands to 1900mL. What is the air pressure of the higher elevation? V1= V2= P1= P2= V P ________ x __________ = 1

  28. BOYLE’S LAW problems 3. What will the new volume be if a 4.25mL sample of gas collected at 0.75 atm is subjected to a new pressure of 1.25 atm? V1= V2= P1= P2= V P ________ x __________ = 1

  29. CHARLES’ LAW NOTES In the E12 computer simulation lab, you raised the temperature of a confined gas, and recorded its effects on the volume. There is a DIRECT relationship between temperature and volume = Charles’ Law When the temperature increases, the gas expanded (volume also increases). T↑ V↑

  30. CHARLES’ LAW The same steps are followed when solving a Charles’ Law problem (dealing with volume & temp at a constant pressure) as you do for Boyle’s Law. Just remember to convert temps to Kelvin! Also Boyle’s Law is the only gas law that is INVERSE (V↑ P↓) Charles Law & the upcoming Guy-Lussac’s Law are DIRECT relationships. (T↑ V↑)

  31. CHARLES’ LAW example 10.L of H2 gas at 20.˚C was heated to 100˚C. What was the new volume? V1= 10.L V2= ?T↑ V↑ T1= 20.˚C= 293K T2= 100˚C = 373K __10.L___ x ___293K____ = 7.9L(2s.f. in data) 1 373K

  32. GUY-LUSSAC’S LAW • The DIRECT relationship between gas temperature and pressure. • As temperature increases, pressure also increases (T↑ P↑). • Again, don’t forget to convert any ˚C to Kelvin by adding 273 before using!

  33. GUY-LUSSAC’S example A sample of Ne gas with a pressure of 600.mmHg and a temperature of 100.˚C was allowed to drop to half the pressure. What was the new temperature in K? ˚C? P1= 600.mmHg P2= 300.mmHg T1= 100.˚C= 373K T2= ? T↓ P ↓ __373K___ x _300.mmHg = 187 K = 460.˚C 1 600.mmHg (3s.f. in all given data) Unknown ↓, follow the arrow!

  34. COMBINED GAS LAW The previous 3 laws (Boyle’s, Charles’, & G-L) require one variable to remain constant. Obviously, all 3 main variables can influence eachother. If temperature is changed it will usually not only influence volume, but also pressure. The relationship among all 3 variables is the Combined Gas Law: P1V1= P2V2 T1 T2 This can also be solved by conversion…. We’ll multiply by 2 fractions, 1 for each known pair

  35. COMBINED GAS LAW practice 10. L of Ar gas with a pressure of 200.torr & a temperature of 400.K was altered to 450. torr and 600.K. What’s the new volume? V1 = 10.L V2 = ? P1 = 200.torr P2 = 450.torr T1 = 400. K T2 = 600. K P↑V↓ T↑V↑ 10. L x 200.torr x 600. K = 1,200,00L = 6.7 L 1 450.torr 400. K 180,000 (2 s.f. in 1 data) Make 2 separate fractions, follow the arrow for the basic gas law that governs the unknown variable & that known.

  36. Homework: • Questions on back of notes: #4-12 • READ about Dalton’s Law in workbook, try to do questions #13-15. • KEEP IN MIND: THE GAS LAWS TEST IS THIS WEDNESDAY, 10/19! Tomorrow we’ll finish notes & do practice test in class. Start studying!

  37. DALTON’S LAW • Sometimes, the best way to isolate a gas is by using water. This, however, can influence the gas’ pressure. • Look for words like “over water” or “wet” • To account for that, we must subtract the water vapor pressure (using a chart). • Then we can just solve it as a combined gas law problem!

  38. Water Vapor Pressure Chart(pg.21 in little white booklet)

  39. DALTON’S LAW practice • 15 L of Cl2 gas was collected over water at a temperature of 14.0˚C and a pressure of 752mmHg. What would the dry volume be at STP? P1 = 752mmHg P2 =stp= mmHg V1(wet)= 15L - = V2(dry)= ? T1 = 14.0˚C + = K T2 =stp= K P↑V__ T↓V__ ___L x _______mmHg x _____ K = 14 L 1 mmHg K (2 s.f. in 1 data)

  40. DALTON’S LAW practice • 15 L of Cl2 gas was collected over water at a temperature of 14.0˚C and a pressure of 752mmHg. What would the dry volume be at STP? P1 = 752mmHg P2 =stp= 760mmHg V1(wet)= 15L - 11.99 (chart)= 3L V2(dry)= ? T1 = 14.0˚C +273= 287K T2 =stp= 273 K P↑V↓ T↓V↓ 3 L x 752mmHg x 273. K = 2.8 L 1 760mmHg 287. K (2 s.f. in 1 data)

  41. DO NOW! • E13 GAS LAWS PRACTICE TEST: ALL • Workbook page 69 • Use booklet page 20 &21 for reference. • Show work!!! • Keep units straight so you can cancel out. • CIRCLE FINAL ANSWERS • CHECK answers at front desk now or online tonight • Mhschemistry.wordpress.com • E13 GAS LAWS TEST = TOMORROW!!! • E13 problems & pract test will be collected

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