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E13: The Gas Laws

Learn about volume, temperature, pressure, and amount as they affect gases. Explore conversions, kinetic theory of matter, STP conditions, Boyle's Law, and solving gas law problems.

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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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