Question

How to find order of Van Der Waal constants a and b?

Answer 1

Molecules are assumed to be a hard rigid spheres and in areal gas, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1 mole gas mole gas molecules (Vi) = (V-b)where v is the molar volume of the gas and b is the volume correction factor.
Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.

Let us take r is the radius of the molecule and σ=2r is the diameter using the molecule as a rigid sphere. Then it can be shown that the volume correction term or effective volume of one mole gas molecules,

b=4×N₀ 4/3 πr₃

Answer 2

$\mathfrak{The\:Answer\:Is}$


$\underline{Volume\:Correction:}$

Molecules are assumed to be a hard rigid spheres and in areal gas, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1 mole gas mole gas molecules (Vi) = (V-b) where v is the molar volume of the gas and b is the volume correction factor.
Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.

Let us take r is the radius of the molecule and σ=2r is the diameter using the molecule as a rigid sphere. Then it can be shown that the volume correction term or effective volume of one mole gas molecules,

b=4×N₀ 4/3 πr₃


$\underline{Pressure\:Condition:}$

Pressure of the gas is developed due to the wall collision of the gas molecules.
But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas(Pi).
Thus, Pi 〉P
or, Pi = P+Pa
Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular attraction in a gas, greater is the magnitude of Pa.


$\boxed{Hope\:This\:Helps}$