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Compressibility factor and van der Waals equation for temp

  1. May 2, 2015 #1
    The pressure exerted on the walls of the container by a real gas is less compared to an ideal gas. This is due to the attractive forces of the gas pulling the molecules back towards the rest of the gas molecules. However, there is also a relationship whereby at lower temperatures, the z is even lesser than at higher temperatures. I am trying to prove that using the van der Waals equation but I am having some trouble with that.

    (P+a(n/v)^2 )(V-nb)=nRT

    I wanted to compare 2 of the same gases at the same pressure but at different temperature would give us two different volumes. I tried to resolve V from the van der Waals equation but due to the cube on the equation I'm not sure how to evaluate the volume of the 2 gases at different temperatures.

    PV/nRT = z = V/V-nb - (an/RTV)

    Using the van der Waals equation, the above equation gives us the compressibility factor. I wanted to substitute the volume we obtain from the first equation and temperature to confirm that z is lower at lower temperatures. However because of the problem mentioned I am unable to do so.

    Does anyone know how to use the van der Waals equation to explain why at lower temperatures the z is lower?
  2. jcsd
  3. May 4, 2015 #2
    You could numerically solve the van der Waals cubic equation for V, and if it yields three real solutions you can choose the right one based on an approximate result gotten from a simpler EOS like the Ideal Gas Law. If it yields only one real solution, choosing the right result should be a no-brainer.
  4. May 8, 2015 #3
    Would it show that as temperature decreases the compressibility factor falls below 1 more than compared to when a higher temperature is used?
  5. May 8, 2015 #4
    I guess you could obtain several different values of V for different temperatures with the help of a spreadsheet, then calculate each correspondig Z value and plot them against temperature. That'll give you some insight.
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