For a real gas at $$25^\circ C$$ temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Van der Waal's constant 'b' should be _____ $$\times 10^{-2}$$ L mol$$^{-1}$$. (Given $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$)

The van der Waals equation for a real gas is:

$$\mathrm{\left(P + \frac{an^2}{V^2}\right)(V-nb)=nRT}$$

For one mole of gas:

$$\mathrm{n=1}$$

Therefore,

$$\mathrm{\left(P + \frac{a}{V^2}\right)(V-b)=RT}$$

At very high pressure, volume becomes very small.

Hence, the intermolecular attraction term $$\mathrm{\frac{a}{V^2}}$$ becomes negligible compared to $$\mathrm{P}$$.

Thus,

$$\mathrm{P(V-b)\approx RT}$$

$$\mathrm{PV-Pb=RT}$$

Dividing throughout by $$\mathrm{RT}$$:

$$\mathrm{\frac{PV}{RT}-\frac{Pb}{RT}=1}$$

Since:

$$\mathrm{Z=\frac{PV}{RT}}$$

we get:

$$\mathrm{Z-\frac{Pb}{RT}=1}$$

$$\mathrm{Z=1+\frac{Pb}{RT}}$$

Given:

$$\mathrm{Z=2}$$

$$\mathrm{P=99\ bar}$$

$$\mathrm{R=0.083\ L\ bar\ K^{-1}\ mol^{-1}}$$

$$\mathrm{T=25^\circ C=298\ K}$$

Substituting values:

$$\mathrm{2=1+\frac{99\times b}{0.083\times298}}$$

$$\mathrm{1=\frac{99b}{24.734}}$$

$$\mathrm{b=\frac{24.734}{99}}$$

$$\mathrm{b\approx0.25\ L\ mol^{-1}}$$

Expressing in the form $$\mathrm{\times10^{-2}\ L\ mol^{-1}}$$:

$$\mathrm{0.25 = 25\times10^{-2}}$$

$$\boxed{\mathrm{25}}$$