Why Does the Van der Waals Equation Adjust Pressure Up and Volume Down?

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SUMMARY

The Van der Waals equation, represented as (P + a)(V - b) = nRT, modifies the ideal gas law to account for real gas behavior. The adjustment of pressure upwards compensates for intermolecular attractions that reduce the effective pressure exerted by gas molecules. Conversely, the volume is adjusted downwards to reflect the finite volume occupied by gas particles, which is not considered in the ideal gas law. This dual adjustment allows for a more accurate representation of gas behavior under non-ideal conditions.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV = nRT)
  • Familiarity with the Van der Waals equation
  • Knowledge of intermolecular forces and their effects on gas behavior
  • Basic concepts of thermodynamics and gas properties
NEXT STEPS
  • Study the derivation and implications of the Van der Waals equation
  • Explore the concept of real gases versus ideal gases
  • Investigate the effects of temperature and pressure on gas behavior
  • Learn about other equations of state for gases, such as the Redlich-Kwong equation
USEFUL FOR

Students in chemistry or physics, educators teaching thermodynamics, and researchers studying gas behavior in various conditions will benefit from this discussion.

Qube
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Homework Statement



Why does the Van der Waals equation adjust pressure up and volume down?

Homework Equations



(P + a)(V-b) = nRT (overly simplified, but it gets the point across).

The Attempt at a Solution



Okay. Volume refers to total free volume in a container in the context of the gas law. I can understand why volume is adjusted down relative to the ideal gas law equation (PV = nRT). Gases aren't point masses and actually occupy some volume.

However, why is pressure adjusted up? Gas molecules may attract and participate in inelastic collisions. This would seem to decrease the actual pressure. Why then would pressure need to be adjusted up?
 
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Qube said:
Gas molecules may attract and participate in inelastic collisions. This would seem to decrease the actual pressure.
If you were to measure the pressure of a gas with ##a = 0##, and then, with everything else the same, with ##a > 0##, how would the measurements differ?
 

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