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Boyle’s law:  V a        (at constant n and T) Va nT nT nT P P P V = constant x  = R 1 P Ideal Gas Equation Charles’ law:  VaT(at constant n and P) Avogadro’s law:  V a  n(at constant P and T) R is the gas constant PV = nRT
The Universal Gas Constant 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. PV = nRT R = PV/nT At STP R = 		 (1 atm)(22.414L)     (1 mol)(273.15 K) R = 0.082057 L•atm                         		     mol • K
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
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 =
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
Ideal Gas Equation

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Ideal Gas Equation

  • 1. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT
  • 2. The Universal Gas Constant 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. PV = nRT R = PV/nT At STP R = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L•atm mol • K
  • 3. 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
  • 4. 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 =
  • 5. 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