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Gas Laws

Gas Laws. Ideal Gas Law: PV = nRT where P = pressure V = volume n = moles T = temperature (K) R = gas constant. R = 0.08206 atm . L mol . K R = 62.36 L . torr mol . K. Gas Laws. The ideal gas law is used to describe the behavior of an ideal gas .

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Gas Laws

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  1. Gas Laws • Ideal Gas Law: PV = nRT where P = pressure V = volume n = moles T = temperature (K) R = gas constant • R = 0.08206 atm.L • mol.K • R = 62.36 L.torr • mol.K

  2. Gas Laws • The ideal gas law is used to describe the behavior of an idealgas. • Idealgas:hypothetical gas that obeys kinetic molecular theory and the ideal gas law

  3. Gas Laws • The ideal gas law is used in calculations for a specific sample of gas that has a constant T, P, V, and n. • i.e. no changes are being made to the sample of gas • If you know 3 of the 4 variables, you can calculate the other using the ideal gas law.

  4. Gas Laws Example: Calculate the volume of 1.00 mol of an ideal gas at 1.00 atm and 0.00oC. Given: P = 1.00 atm n = 1.00 mol T = 0.00oC + 273 = 273 K Find: V PV = nRT

  5. Gas Laws PV = nRT Solve for VV = nRT P V = 1.00 mol x 0.08206 atm.L x 273 K 1.00 atm mol.K V = 22.4 L

  6. Gas Laws • The temperature and pressure used in the previous problem are commonly used to report the properties of gases. • Standard Temperature and Pressure (STP): • 0oC and 1 atm. • KNOW THIS!!

  7. Gas Laws • Molar volume: • the volume one mol of a gas occupies (L/mole) • At STP, one mole of an ideal gas has a molar volume of 22.4 L: • 22.4 L 1 mol

  8. Gas Laws • The ideal gas law applies only to ideal gases. • Does not always accurately describe real gases • The molar volumes for many real gases at STP differ slightly from 22.4 L/mol. • In most cases, the differences between ideal gas behavior and real gas behavior is so small that we can ignore it

  9. Gas Laws Example: A weather balloon contains 4.75 moles of He gas. What volume does the gas occupy at an altitude of 4300 m if the temperature is 0oC and the pressure is 0.595 atm? Given: P = 0.595 atm T = 0oC = 273K n = 4.75 mol Find: V PV = nRT

  10. Gas Laws PV = nRT solve for V V = nRT P V = (4.75 mol) x (273K) x (0.08206 atm.L) (0.595 atm) mol.K V = 179 L

  11. Gas Laws Example: A used aerosol can contains 0.0173 mol of gas and has a volume of 425 mL. Calculate the pressure in the can if it is accidentally heated to 395oC. (Warning: Don’t do this!!) Given:V = 425 mL n = 0.0173 mol T = 395oC + 273 = 668K Find:P PV = nRT

  12. Gas Laws PV = nRT solve for P P = nRT V P = (0.0173 mol)(0.08206 atm.L) (668K) 425 mL mol K x 1000 mL 1 L P = 2.23 atm

  13. Gas Laws Example: A tire with an interior volume of 3.50 L contains 0.357 mol of air at a pressure of 2.49 atm. What is the temperature of the air in the tire in K? In oC? Given: P = 2.49 atm V = 3.50 L n = 0.357 mol Find: T (K) PV = nRT

  14. Gas Laws PV = nRT solve for T T = PV nR T = 2.49 atm x 3.50 L x mol.K 0.357 mol 0.08206 atm.L T = 297 K T = 297 – 273 = 24oC

  15. Gas Laws • The ideal gas law was useful in determining the properties of a specific sample of gas at constant T, P, V, and n. • We often need to know how a change in one (or more) properties impacts the other properties for a sample of a gas.

  16. Gas Laws • Combined Gas Laws: P1V1 = P2V2 T1 T2 This equation is true when the number of moles of a gas is constant.

  17. Gas Laws Special cases for the combined gas law: • At constant temperature (T1 = T2), P1V1 = P2V2 • At constant volume (V1 = V2), P1= P2 T1 T2 • At constant Pressure (P1 = P2), V1 = V2 T1 T2

  18. Gas Laws Example: A helium-filled balloon occupies 6.00 L at 19.5oC and 0.989 atm. What volume will the balloon occupy on top of Pike’s Peak if the pressure is 0.605 atm and the temperature is constant? Given: V1 = 6.00L P1 = 0.989 atm T1 = T2 = 19.5oC P2 = 0.605 atm Find: V2 Since T is constant, P1V1 = P2V2

  19. Gas Laws P1 V1 = P2 V2 (0.989 atm) x (6.00 L) = (0.605 atm) x V2 V2 = (0.989 atm) x (6.00 L) 0.605 atm V2 = 9.81 L

  20. P1V1 = P2V2 T1 T2 Gas Laws Example: Suppose a used aerosol can contains a gas at 0.989 atm at 23oC. If this can is heated to 425oC, what is the pressure inside the can? Given: P1 = 0.989 atm T1 = 23oC + 273 = 296K T2 = 425 + 273 = 698K V1 = V2 Find: P2

  21. Gas Laws Since V1= V2: P1 = P2 T1 T2 P2 = 0.989 atm (698K) 296K P2 = (0.989 atm) x 698K = 2.33 atm 296K

  22. Gas Laws--More Applications • The ideal gas equation can be used to determine either the density or the molar mass of a gas. d = P M RT Where d = density P = pressure T = temperature in K R = gas constant M = molar mass

  23. Gas Laws--More Applications • The density of a gas depends on: • pressure • temperature • molar mass • At constant temperature and pressure the densities of gas samples are directly proportional to their molar masses: • d = PM • RT

  24. d = P M • RT Gas Laws--More Applications • If molar mass and pressure are held constant, then the density of the gas will decrease with increasing temperature • Hot air rises because density is inversely proportional to temperature

  25. Gas Laws--More Applications Example: What is the density of helium gas at 1.00 atm and 25oC? Given: P = 1.00 atm T = 25 + 273 = 298K Find: d • d = PM • RT

  26. Gas Laws--More Applications • d = PM • RT M= molar mass of He = 4.00 g/mol d = (1.00 atm) 4.00 g mol.K (298K) 1 mol 0.08206 atm.L d = 0.164 g/L Notice: density of gases is usually in g/L instead of g/mL

  27. Gas Laws--More Applications Example: What is the average molar mass of dry air if it has a density of 1.17 g/L at 21oC and 740.0 torr? Given: P = 740.0 torr T = 21 + 273 = 294K d = 1.17 g/L Find:M • d = PM • RT

  28. M = dRT • P Gas Laws--More Applications M= 1.17 g 62.36 torr.L(294 K) mol.K 740.0 torr M = 29.0 g/mol • d = PM • RT

  29. Gas Laws--More Applications • Understanding the properties of gases is important because gases are often the reactants or products in a chemical reaction. • Often need to calculate the volume of gas produced or consumed during a reaction

  30. grams A grams A Moles A Moles A Molar mass Molar mass PV = nRT Molar ratio Molar ratio grams B grams B Moles B Moles B Molar mass Molar mass PV = nRT Gas Laws--More Applications Gas Data A PA, VA, TA Gas Data B PB, VB, TB

  31. Gas Laws--More Applications Example:The air bag in a car is inflated by nitrogen gas formed by the decomposition of NaN3: 2 NaN3(s)  2 Na (s) + 3 N2 (g) If an inflated air bag has a volume of 36 L and is to be filled with N2 gas at a pressure of 1.15 atm at a temperature of 26oC, how many grams of NaN3 must be decomposed?

  32. PV = nRT Gas Laws--More Applications Moles A Moles N2 Gas Data N2 Molar ratio Molar ratio grams B grams NaN3 Moles B Moles NaN3 Molar mass Molar mass

  33. PV = nRT Gas Laws--More Applications 36L N2, 1.15 atm 26oC Moles A Moles N2 Find Molar ratio Molar ratio Given grams B grams NaN3 Moles B Moles NaN3 Molar mass Molar mass

  34. Gas Laws--More Applications n = PV = (1.15 atm) ( 36 L) mol.K RT (299 K) 0.08206 atm.L n = 1.7 mol N2 g NaN3 = 1.7 mol N2 x 2 mol NaN3 x 65.0 g NaN3 3 mol N2 1 mol NaN3 g NaN3 = 74 g NaN3

  35. Gas Laws--More Applications Example: How many mL of oxygen gas can be collected at STP when 1.0 g of KClO3 decomposes: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

  36. PV = nRT Gas Laws--More Applications mL O2 at STP (1.0atm, 0oC) Moles A Moles O2 Given Molar ratio Molar ratio Find grams B grams KClO3 Moles B Moles KClO3 Molar mass Molar mass

  37. Gas Laws--More Applications Mol O2 = 1.0 g KClO3 x 1 mole KClO3x 3 mol O2 122.5 g 2 mol KClO3 mol O2 = 0.012 mol O2 V = nRT = (0.012 mol)0.0821 atm.L (273K) x 103 mL P 1.0 atm mol.K 1 L V = 270 mL

  38. Gas Laws – More Applications Example: What volume of CO2 at 125oC and 1.15 atm will be produced by the combustion of 1.00 L of C2H6 at 25oC and 1.00 atm? 2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (g) Given: VC2H6 = 1.00 L PC2H6 = 1.00 atm TC2H6 = 25oC = 298 K PCO2 = 1.15 atm TCO2 = 125oC = 398 K Find: VCO2

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