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GAS LAWS

GAS LAWS. Elements that exist as gases at 25 0 C and 1 atmosphere. Pressure. Remember: Pressure = Force / area or mass x gravity / area Force = mass x gravity. Force. Area. Barometer. Pressure = . Units of Pressure. 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr

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GAS LAWS

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  1. GAS LAWS

  2. Elements that exist as gases at 250C and 1 atmosphere

  3. Pressure Remember: • Pressure = Force / area or mass x gravity / area • Force = mass x gravity

  4. Force Area Barometer Pressure = Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

  5. Measuring Pressure Devices used: • Barometer-Invented by Torricelli • Pressure Gauge • Tire Gauge • Sphygmomanometer

  6. Units of Pressure • Standard atmospheric pressure (1 atm) = 760 mm Hg or 760 torr (1 mm Hg = 1 torr) • In the United States we use inches in atmospheric pressure. • Pascal- used in science as a unit of pressure (Pa). • Kilopascals are used as well (KPa).

  7. Pressure Column height measures Pressure of atmosphere • 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg * = 14.7 pounds/in2 (psi) *HD only = 101.3 kPa (SI unit is PASCAL) * HD only = about 34 feet of water! * Memorize these!

  8. Practice • How many torrs are in 4.5 atm? • The barometer reads 1,760 torr, how many atm? • I have a reading of 3700 torr, convert to atm. • You have 3 atm, how many torr?

  9. Boyle’s Law • Shows the relationship between Pressure and Volume. • The Law assumes that the temperature does not change. • Proposed by Robert Boyle.

  10. Boyle’s Law P1V1 = P2V2 • P1= Pressure 1 • V1= Volume 1 • P2= Pressure 2 • V2= Volume 2 • This formula works with any unit of Pressure (torr, Pa) • Any unit of volume works (mL, L). • Make sure you have the same units on both sides! If not, change one to match the other. • If one side has mL and the other has L, convert one to match the other.

  11. Boyle’s Law • A gas measured a volume of 100 mL under pressure of 740 torr. What would be the volume under a pressure of 780 torr with constant temperature? Step 1: Figure out what you have. • P1 = 740 torr • V1= 100 mL • P2= 780 torr • V2= X

  12. Boyle’s Law • Step 2: Plug the numbers into the formula P1V1 = P2V2 (740 torr)(100 mL) = (780 torr)(X) • Step 3: Solve for x 74000 = 780X X= 94.87 mL

  13. Boyle’s Law • So the new volume in this case is 94.87 mL • Since we increased the pressure, the volume is decreased. • The formula is a proportion. If something increases, something else will decrease!

  14. P1V1 = P2V2 Practice • A sample of gas is confined to a 100 mL flask under pressure of 740 torr. If the same gas were transferred to a 50 mL flask, what’s the new pressure? P2 = P1V1/V2 P1 = V1 = P2 = V2 = 740 torr 100 ml X 50 ml P2 = (740)(100)/50 P2 = 74000/50 P2 = 1480 torr

  15. P1V1 = P2V2 Practice 2. You are given a gas that you measure under a pressure of 720 Pa. When the pressure is changed to 760 Pa, the volume became 580 mL. What is the first volume? V1 = P2V2/P1 P1 = V1 = P2 = V2 = 720 Pa X 760 Pa 580 ml V1 = (760)(580)/720 V1 = 440800/720 V1 = 612.2 ml

  16. P1V1 = P2V2 Practice • A pressure on 134 mL of air is changed to 1200 torr at a constant temperature, if the new volume is 45 mL what is the original pressure? P1 = P2V2/V1 P1 = V1 = P2 = V2 = X 134 ml 1200 torr 45 ml P1 = (1200)(45)/134 P1 = 54000/134 P1 = 402.9 torr

  17. P1V1 = P2V2 Practice 4. An amount of Oxygen occupies 2 L when under pressure of 680 torr. If the volume is increased to 3 L what is the new pressure? P2 = P1V1/V2 P1 = V1 = P2 = V2 = 680 torr 2 L X 3L P2 = (740)(100)/50 P2 = 74000/50 P2 = 1480 torr

  18. Reminders • Check your units • Temperature must remain constant in Boyle’s Law. • P= Pressure (torr, Pa, KPa). • V= Volume (mL, L). “If you don’t exhale on your way up Cookie, your lungs will explode! Robert De Niro as Chief Saturday

  19. When the temperature increases, the volume increases.

  20. As T increases V increases

  21. Charles Law • Shows a relationship between Volume and Temperature. One thing increase, so does the other. Both behave the same way. • Pressure remains constant. • V= Volume • T= Temperature

  22. Few things first… • ALL TEMPERATURES MUST BE CONVERTED TO KELVIN! • The Law only works with Kelvin • If you have Celsius, make sure you convert it to Kelvin. • Remember Kelvin = Celsius + 273

  23. Temperature Conversion Practice 85 + 273 = 358K Convert 85°C to K. ____________________ Convert 376K to °C. ____________________ Convert 154K to °C. ____________________ Convert -65°C to K. ____________________ Convert 0°C to K. _____________________ Convert 0K to °C. _____________________ 376 - 273 = 103 °C 154 - 273 = -119 °C 208K -65 + 273 = 273K 0 + 273 = 0 - 273 = -273 °C

  24. Few Practice Problems V1/T1 = V2/T2 You heat 100 mL of a gas at 25 C to 80 C. What is the new volume of the gas? V2 = V1T2/T1 V2 = (100)(353)/298 V1 = T1 = V2 = T2 = 100 ml 25ºC X 80ºC + 273 = 298K V2 = 35300/298 V2 = 118.5 ml + 273 = 353K

  25. Few Practice Problems V1/T1 = V2/T2 You have 200 mL of a gas at 55 C and you freeze it to 0 C. What is the new volume? V2 = V1T2/T1 V2 = (200)(273)/328 V1 = T1 = V2 = T2 = 200 ml 55ºC X 0ºC + 273 = 328K V2 = 54600/328 V2 = 166.5 ml + 273 = 273K

  26. Few Practice Problems V1/T1 = V2/T2 A 150 mL sample of gas is at 125 K. After cooling the new volume is 80 mL. What is the new temperature? T2 = V2T1/V2 T2 = (80)(125)/150 V1 = T1 = V2 = T2 = 150 ml 125K 80 ml X T2 = 10000/150 T2 = 66.67K

  27. Gay-Lussac’s Law P1/T1 = P2/T2 • Shows relationship between pressure (P) and temperature (T). If Temperature increases, so does the pressure. • Based on Charles’ Law. • Volume remains constant. • Like Charles’ Law, all temperature units must be in KELVIN. • Used in pressure cookers and autoclaves.

  28. Practice P1/T1 = P2/T2 A soda bottle with a temperature of 25ºC and 3 atm of pressure was put into a freezer with a temperature of –1°C. What is the pressure on the bottle inside the freezer? P2 = P1T2/T1 P1 = T1 = P2 = T2 = 3 atm 25ºC X -1ºC P2 = (3)(272)/298 + 273 = 298K P2 = 816/298 P2 = 2.74 atm + 273 = 272K

  29. Combined Gas Law • Combines Boyle’s Law, Gay-Lussac’s Law, and Charles’ Law. • Shows the relationship of Pressure, Volume and Temperature.

  30. Combined Gas Law • Derived from Boyle’s, Gay-Lussac, and Charles Law. • Used when both temperature and pressure changes. • Can be used to find a constant. • If two things increase, one thing will decrease. • Math was used to derive the law.

  31. Combined Gas Law (P, V, T) Combined Gas Law:nR= PV T

  32. 1. A gas has a volume of 800.0 mL at -23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? P1V1 = P2V2 T1 T2 P1 = V1 = T1 = P2 = V2 = T2 = 300 torr 800 ml -23ºC 600 torr X 227.0ºC V2 = P1V1T2 T1 P2 + 273 = 250K V2 = (300)(800)(500) (250)(600) V2 = 800 ml + 273 = 500K

  33. Remember this… • When you are working with Charles’ Law and Combined Gas Law, make sure you have the proper units! • The laws help us predict the behavior of gases under different circumstances.

  34. How is this possible? KC-135 Jet Rail Car 

  35. Avogadro’s Law V1/n1 = V2/n2 • Allows us to calculate the number of moles (n) of a gas with in a volume (V). • Pressure and Temperature remains the same. • If volume increases, so does the number of moles.

  36. Practice V1/n1 = V2/n2 In 150 mL sample, there are 2.5 grams of Cl2 gas, if I double the volume to 300 mL how much Cl2 gas in moles, is in the new sample? n2 = V2 n1/ V1 V1 = n1 = V2 = n2 = 150 ml 2.5g 300 ml x n2 = (300)(0.147)/150 n2 = 44.1/150 n2 = 0.294 m ÷ 17g/m = 0.147 m

  37. Ideal gas law • Formula: PV= nRT • P= Pressure in atmospheres • V= Volume in Liters • n= number of moles of a gas • R= constant, 0.0821 L • atm / mol • K • T= Temperature in Kelvin

  38. Ideal Gas Law • An ideal gas is a gas that obeys Boyle’s and Charles’ Law. • Based on the Kinetic Theory. • Shows the relationship between Pressure, Volume, number of moles and Temperature. • Make sure everything has the proper unit. • Can be used to figure out density of gas or determine the behavior of a tire.

  39. Practice PV = nRT How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure? ÷ 760 mmHg/atm = 0.987 atm P = V = n = R = 0.0821 Latm/molK T = 750.0 mm Hg 890.0 ml X 21.0ºC ÷ 1000 ml/L = 0.89 L n = PV/RT n = (0.987)(0.89)/(0.0821)(294) n = 0.878/24.1 n = 0.036 m + 273 = 294 K

  40. And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

  41. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) PV = nRT R = 0.082057 L • atm / (mol • K)

  42. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 273.15 K V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT V = 30.6 L

  43. Low density High density GAS DENSITY 22.4 L of ANY gas AT STP = 1 mole

  44. m V PM = RT dRT P Density (d) Calculations m is the mass of the gas in g d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance d is the density of the gas in g/L M =

  45. What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mol CO2 g C6H12O6 mol C6H12O6 mol CO2V CO2 x 1 mol C6H12O6 1 mol C6H12O6 x 180 g C6H12O6 L•atm mol•K nRT 0.187 mol x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.187 mol CO2 V = = 4.76 L

  46. Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  47. Deviations from Ideal Gas Law • Real molecules have volume. The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves. • There are intermolecular forces. An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions. • Otherwise a gas could not condense to become a liquid.

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