Question

When vander walls constant a, b for gases (a) and (b) are given. Then which factor confirm liquification?

Answer 1

Here is the Van der Walls’s equation [P+a(nV)2](Vn−b)=RT[P+a(nV)2](Vn−b)=RT

Where P is the pressure, V is the volume R is gas constant and T is the temperature, n is number of moles and a and b are constants

Now taking dimensional analysis, of [P+a(nV)2][P+a(nV)2] we know the fact that quantities whose dimension are same can be added.

We will talk in terms of units only and not going in dimensions for time being

Unit of Pressure is atm or pasals

Unit of Volume is litre or cubic meter

Unit of number of moles is mol

Hence unit of constant a should be such that [a(nV)2][a(nV)2] have unit that of pressure i.e. either atm or pascal only

∴∴ Unit of constant aa will be atm×(litremol)2atm×(litremol)2

Hence constant a depends on volume, pressure and number of moles. As a fact number of moles in given volume of gas remains constant.

Now talking about other quantity (Vn−b)(Vn−b)

Quantities that have same dimensions can be subtracted from each other. Here too we will only talk about the units of both the quantities and not dimensions

Hence dimension of constant b should be similar to that of the ratio (V/n)

Unit of ratio (V/n) is litremollitremol

∴∴ units of constant b have to be litremollitremol

The constant a provides a correction for the inter molecular forces. Constant b is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules.