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Ideal Gas Law

Ideal Gas Law. The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = n R T R = ideal gas constant . Ideal Gases. Behave as described by the ideal gas equation; no real gas is actually ideal

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Ideal Gas Law

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  1. Ideal Gas Law The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = nRT R = ideal gas constant Lecture PLUS Timberlake 2000

  2. Ideal Gases • Behave as described by the ideal gas equation; no real gas is actually ideal • Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less • In real gases, particles attract each other reducing the pressure • Real gases behave more like ideal gases as pressure approaches zero. Lecture PLUS Timberlake 2000

  3. PV = nRT R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm mol-K Lecture PLUS Timberlake 2000

  4. Learning Check G15 What is the value of R when the STP value for P is 760 mmHg? Lecture PLUS Timberlake 2000

  5. Solution G15 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K Lecture PLUS Timberlake 2000

  6. Learning Check G16 Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office? Lecture PLUS Timberlake 2000

  7. Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L 20.0 L T = 23°C + 273 296 K n = 2.86 mol 2.86 mol P = ? ? Lecture PLUS Timberlake 2000

  8. Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol) = 2.64 x 103 mm Hg Lecture PLUS Timberlake 2000

  9. Learning Check G17 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? Lecture PLUS Timberlake 2000

  10. Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2 1 mol O2 Lecture PLUS Timberlake 2000

  11. Molar Mass of a gas What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = 0.00703 mol RT (0.0821 L-atm/molK) (303K) Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol Lecture PLUS Timberlake 2000

  12. Density of a Gas Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRT PV = nRT P = n RTV RTV RT V Lecture PLUS Timberlake 2000

  13. Substitute (1.00 atm ) mol-K = 0.0446 mol O2/L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O2 x 32.0 g O2 = 1.43 g/L 1 L 1 mol O2 Therefore the density of O2 gas at STP is 1.43 grams per liter Lecture PLUS Timberlake 2000

  14. Formulas of Gases A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas? Lecture PLUS Timberlake 2000

  15. Formulas of Gases Calculate Empirical formula 85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C 12.0 g C 14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H 1.0 g H Empirical formula = CH2 EF mass = 12.0 + 2(1.0) = 14.0 g/EF Lecture PLUS Timberlake 2000

  16. Using STP and density ( 1 L = 2.50 g) 2.50 g x 22.4 L = 56.0 g/mol 1 L 1 mol n = EF/ mol = 56.0 g/mol = 4 14.0 g/EF molecular formula CH2 x 4 = C4H8 Lecture PLUS Timberlake 2000

  17. Gases in Chemical Equations On December 1, 1783, Charles used 1.00 x 103 lb of iron filings to make the first ascent in a balloon filled with hydrogen Fe(s) + H2SO4(aq)  FeSO4(aq) + H2(g) At STP, how many liters of hydrogen gas were generated? Lecture PLUS Timberlake 2000

  18. Solution lb Fe  g Fe  mol Fe  mol H2 L H2 1.00 x 103 lb x 453.6 g x 1 mol Fe x 1 mol H2 1 lb 55.9 g 1 mol Fe x 22.4 L H2 = 1.82 x 105 L H2 1 mol H2 Charles generated 182,000 L of hydrogen to fill his air balloon. Lecture PLUS Timberlake 2000

  19. Learning Check G18 How many L of O2 are need to react 28.0 g NH3 at24°C and 0.950 atm? 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Lecture PLUS Timberlake 2000

  20. Solution G18 Find mole of O2 28.0 g NH3 x 1 mol NH3 x 5 mol O2 17.0 g NH3 4 mol NH3 = 2.06 mol O2 V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 L P 0.950 atm Lecture PLUS Timberlake 2000

  21. Summary of Conversions with Gases Volume A Volume B Grams A Moles A Moles B Grams B Atoms or Atoms or molecules A molecules B Lecture PLUS Timberlake 2000

  22. Daltons’ Law of Partial Pressures The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mmHg 20.95% O2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO2 0.2 mmHg PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2 Total Pressure 760 mm Hg Lecture PLUS Timberlake 2000

  23. Learning Check G19 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 2) 9.14 3) 0.109 Lecture PLUS Timberlake 2000

  24. Solution G19 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 Lecture PLUS Timberlake 2000

  25. Partial Pressure Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. PT = P1 + P2 + P3 + ..... Lecture PLUS Timberlake 2000

  26. Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. STP P = 1.00 atm P = 1.00 atm 0.50 mol O2 + 0.20 mol He + 0.30 mol N2 1.0 mol He Lecture PLUS Timberlake 2000

  27. Health Note When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents. Lecture PLUS Timberlake 2000

  28. Learning Check G20 A 5.00 L scuba tank contains 1.05 mole of O2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank? Lecture PLUS Timberlake 2000

  29. Solution G20 P = nRT PT = PO + PHe V 2 PT = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L (K mol) = 7.19 atm Lecture PLUS Timberlake 2000

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