If Z is the compressibility factor, then Van der Waal's equation at low pressure can be written as:

We start from the Van der Waals equation for one mole of a real gas

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

where $$P$$ is the pressure, $$V$$ is the molar volume, $$T$$ is the absolute temperature, $$R$$ is the gas constant, and the constants $$a$$ and $$b$$ account for intermolecular attraction and finite molecular size respectively.

At very low pressure the molar volume $$V$$ becomes very large. Because of this largeness, the correction term $$b$$ (which is a small volume) is negligible in comparison with $$V$$. So we drop $$b$$ but we keep the attractive term containing $$a$$, as its effect can still be significant even when the gas is dilute. Therefore the equation simplifies to

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

Now we expand the left-hand side:

$$P\,V + \frac{a}{V} = RT$$

We wish to obtain the compressibility factor $$Z$$, which is defined for one mole as

$$Z = \frac{P\,V}{R\,T}$$

Re-arranging the simplified Van der Waals expression to isolate $$P\,V$$ gives

$$P\,V = RT - \frac{a}{V}$$

Dividing every term of this equality by $$R\,T$$ we obtain

$$\frac{P\,V}{R\,T} = 1 - \frac{a}{V\,R\,T}$$

The left-hand side is precisely the definition of $$Z$$, so we can write

$$Z = 1 - \frac{a}{V\,R\,T}$$

This expression matches option B in the list provided.

Hence, the correct answer is Option B.