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IDEAL GAS LAW

IDEAL GAS LAW. Ideal Gas Law Derivation. Recall that P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 Also we learned that At O o C, and 1 atm (101.3 kPa) that 1 mole of gas occupies 22.4 L (molar volume). Ideal Gas Law Derivation. We can set one side of the equation under standard conditions ie.

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IDEAL GAS LAW

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  1. IDEAL GAS LAW

  2. Ideal Gas Law Derivation • Recall that P1V1 = P2V2 n1T1 n2T2 • Also we learned that • At OoC, and 1 atm (101.3 kPa) that 1 mole of gas occupies 22.4 L (molar volume)

  3. Ideal Gas Law Derivation • We can set one side of the equation under standard conditions ie. • n = 1 mole • V = 22.4 L • P = 101. 3 kPa • T = 0oC

  4. Ideal Gas Law Derivation P1V1 = P2V2 n1T1 n2T2 So…. (101.3kPa) (22.4L) = P2V2 (1mole) (273K) n2T2

  5. Ideal Gas Law Derivation 8.314 kPa x L = P V mol x K n T Overall…. PV = nRT 8.314 kPaL/mol K = R = Ideal Gas Law Constant

  6. 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.

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

  8. Additional R Values 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

  9. 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 (kPa) in the tank in the dentist office?

  10. 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 = ? ?

  11. 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)(8.314)(296 K) (20.0 L) = 351.91 kPa

  12. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 8.314 x 273.15 K V = 1 atm nRT P What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 101.3 kPa PV = nRT V = 30.6 L

  13. Learning Check G17 A 5.0 L cylinder contains oxygen gas at 20.0°C and 98 kPa. How many grams of oxygen are in the cylinder?

  14. Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (98 kPa)(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

  15. 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

  16. 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

  17. 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

  18. 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?

  19. 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

  20. 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

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