Question

Relation between critical constant and vanderwaal constant

Answer 1

Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.

Thus the mathematical condition of critical point is,

(dP/dV)ᴛ = 0 and (d²P/dV²)ᴛ = 0

Van der Waals equation for 1 mole gas is

(P + a/V²)(V - b) = RT

or, P = RT/(V - b) - a/V²

Differentiating Van der Waals equation with respect to V at constant T,

We get Slope = (dP/dV)ᴛ = - {RT/(V - b)²} + 2a/V³

And Curvature = (d²P/dV²)ᴛ = {2RT/(V - b)³} - 6a/V⁴

Hence at the critical point,

- {RTc/(Vc - b)²} + 2a/Vc³ = 0

or, RTc/(Vc - b)² = 2a/Vc³

and {2RTc/(Vc - b)³} - 6a/Vc⁴ = 0

or, 2RTc/(Vc -b)³ = 6a/Vc⁴

Thus, (Vc - b)/2 = Vc/3

or, Vc = 3b