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- 1. Boyle’s law: V a (at constant n and T)<br />Va<br />nT<br />nT<br />nT<br />P<br />P<br />P<br />V = constant x = R<br />1<br />P<br />Ideal Gas Equation<br />Charles’ law: VaT(at constant n and P)<br />Avogadro’s law: V a n(at constant P and T)<br />R is the gas constant<br />PV = nRT<br />
- 2. The Universal Gas Constant<br />The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).<br />Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.<br />PV = nRT<br />R = PV/nT<br />At STP R = (1 atm)(22.414L)<br /> (1 mol)(273.15 K)<br />R = 0.082057 L•atm mol • K<br />
- 3. 1 mol HCl<br />V = <br />n = 49.8 g x <br />= 1.37 mol<br />36.45 g HCl<br />1.37 mol x 0.0821 x 273.15 K<br />V =<br />1 atm<br />nRT<br />L•atm<br />P<br />mol•K<br />What is the volume (in liters) occupied by 49.8 g of HCl at STP?<br />T = 0 0C = 273.15 K<br />P = 1 atm<br />PV = nRT<br />V = 30.6 L<br />
- 4. m<br />V<br />PM<br />=<br />RT<br />dRT<br />P<br />Density (d) Calculations<br />m is the mass of the gas in g<br />d = <br />M is the molar mass of the gas<br />Molar Mass (M ) of a Gaseous Substance<br />d is the density of the gas in g/L<br />M =<br />
- 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:<br />C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)<br />6 mol CO2<br />g C6H12O6 mol C6H12O6 mol CO2V CO2<br />x<br />1 mol C6H12O6<br />1 mol C6H12O6<br />x<br />180 g C6H12O6<br />L•atm<br />mol•K<br />nRT<br />0.187 mol x 0.0821 x 310.15 K<br />=<br />P<br />1.00 atm<br />Gas Stoichiometry<br />5.60 g C6H12O6<br />= 0.187 mol CO2<br />V = <br />= 4.76 L<br />

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