The van der Waals Equation

The van der Waals Equation - oxygen gas in 22.41 L at 0.0C...

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The van der Waals Equation The ideal gas equation is not much use at high pressures One of the most useful equations to predict the behavior of real gases was developed by Johannes van der Waals (1837-1923) He modified the ideal gas law to account for: o The finite volume of gas molecules o The attractive forces between gas molecules van der Waals equation: The van der Waals constants a and b are different for different gasses They generally increase with an increase in mass of the molecule and with an increase in the complexity of the gas molecule (i.e. volume and number of atoms) Substance a (L 2 atm/mol 2 ) b (L/mol) He 0.0341 0.0237 H 2 0.244 0.0266 O 2 1.36 0.0318 H 2 O 5.46 0.0305 CCl 4 20.4 0.1383 Example Use the van der Waals equation to calculate the pressure exerted by 100.0 mol of
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Unformatted text preview: oxygen gas in 22.41 L at 0.0C V = 22.41 L T = (0.0 + 273) = 273K a (O 2 ) = 1.36 L 2 atm/mol 2 b (O 2 ) = 0.0318 L /mol P = 117 atm- 27.1 atm P = 90 atm The pressure will be 90 atm, whereas if it was an ideal gas, the pressure would be 100 atm The 90 atm represents the pressure correction due to the molecular volume. In other words the volume is somewhat less than 22.41 L due to the molecular volume. Therefore the molecules must collide a bit more frequently with the walls of the container, thus the pressure must be slightly higher. The -27.1 atm represents the effects of the molecular attraction. The pressure is reduced due to this attraction....
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This note was uploaded on 11/22/2011 for the course CHEMISTRY CHM1025 taught by Professor Laurachoudry during the Fall '10 term at Broward College.

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The van der Waals Equation - oxygen gas in 22.41 L at 0.0C...

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