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Ideal Gas Law & Gas Stoichiometry

Ideal Gas Law & Gas Stoichiometry. Ideal Gas Law. P V = n R T P = Pressure (atm) V = Volume (L) T = Temperature (K) n = number of moles R is a constant, called the Ideal Gas Constant

malcolm-tyson
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Ideal Gas Law & Gas Stoichiometry

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  1. Ideal Gas Law & Gas Stoichiometry

  2. Ideal Gas Law P V = n R T • P = Pressure (atm) • V = Volume (L) • T = Temperature (K) • n = number of moles • R is a constant, called the Ideal Gas Constant • Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. • R = 0.0821 L atm / mol K

  3. PV = nRT • Calculate the number of moles of a gas contained in a 3.0 L vessel at 300.0K with a pressure of 1.50 atm

  4. PV = nRT • n = ? • V = 3.0 L • T = 300.0 K • P = 1.50 atm • PV = nRT • (1.50 atm)(3.0 L) = n (0.0821L atm / mol K)(300.0 K) • n = 0.18 mol

  5. Example 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? Note: 1atm = 760 mm Hg

  6. Example n = 2.86 mol • V = 20.0 L • T = 23 °C = 273 + 23 = 296 K • P = ? • PV = nRT • (P)(20.0 L) = (2.86) (0.0821L atm / mol K)(296 K) • P = 3.5 atm • P = 2600 mm Hg

  7. Permutations of the Ideal Gas Law PV = mRT M • P = Pressure (atm) • V = volume (L) • m = mass of the gas (g) • R = 0.0821 L atm / mol K • T = Temperature (K) • M = molecular mass

  8. Example • What is the pressure 2.0 g of nitrogen gas in a 5.0 L container at 300.0 K? • P = ? • m = 2.0 g • V = 5.0 L • T = 300.0K • M = 28.02 g/mol

  9. Example PV = mRT M P(5.0) = (2.0)(0.0821)(300.0) 28.04 P = 2.8 atm

  10. Permutations of the Ideal Gas Law P = DRT M • P = pressure (atm) • D = density (g/L) • R = 0.0821 L atm / mol K • T = temperature (K) • M = molecular mass

  11. Example • What is the molar mass of a gas that has a density of 1.40 g/L at STP? • NOTE – STP is standard temperature and pressure • At STP temperature is 273 K and pressure is 1.00 atm

  12. Example • P = 1.00 atm • D = 1.40 g/L • R = 0.0821 L atm / mol K • T = 273 K • M = ? P = DRT M 1.00 = (1.40)(0.0821)(273) M M = 31.4 g/mol

  13. Avogadro’s Principle • Avogadro’s Principle – equal volumes of gases at equal temperature and pressure contain the same number of particles • Molar volume – the volume of gas that 1 mole of a substance occupies at STP • At STP 1 mol of a gas = 22.4 L • New conversion factor at STP ONLY! 1 mol 22.4 L

  14. Example • Calculate the volume 0.881 mol of a gas will occupy at STP. • 0.881 mol x 22.4 L = 19.7 L 1 mol (You could also have worked this out with the ideal gas law equation)

  15. Example • Calculate the volume that 2.000 kg of methane would occupy at STP. • 2.000 kg x 1x10 3g x 1 mol x 22.4 L = 1kg 16.05g 1 mol • 2791 L CH4

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