For 1 mol of gas, the plot of pV vs p is shown below. p is the pressure and V is the volume of the gas.

What is the value of compressibility factor at point A?

For one mole of a real gas, the Van der Waals equation is

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

Expanding the equation,

$$PV-Pb+\frac{a}{V}-\frac{ab}{V^2}=RT.$$

Dividing throughout by (RT),

$$\frac{PV}{RT}-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

Since the compressibility factor is defined as

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

we obtain

$$Z-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

At low pressures, the molar volume (V) is very large, so terms containing (\frac{1}{V^2}) can be neglected. Therefore,

$$Z\approx 1-\frac{a}{RTV}.$$

Point A corresponds to the low-pressure region where intermolecular attractive forces dominate, causing the gas to be more compressible than an ideal gas ((Z<1)). The deviation from ideal behaviour is represented by the attractive term (-\frac{a}{RTV}).

Hence, the compressibility factor at point A is

$$\boxed{Z=1-\frac{a}{RTV}}.$$

Therefore, the correct answer is

$$\boxed{1-\frac{a}{RTV}}.$$