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**Hint:**Compressibility factor is also known as gas deviation factor. This is a correction factor which describes the deviation of real gas from ideal gas.

**Complete step by step answer:**

The real gases obey the ideal gas equation $(PV = nRT)$ only if pressure is low or temperature is high.

Van der waals equation for one mole of Real gas is given by

$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - b} \right] = RT$ . . . . . (1)

And for $n$ moles of gas

$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - nb} \right] = RT$

Where ‘a’ and ‘b’ are constants and called Van der Waals constants. Their values depend upon the nature of gas.

At high pressure $\dfrac{a}{{{V^2}}}$ can be neglected.

$\therefore $equation (1) becomes

$\Rightarrow P[v - b] = RT$

$\Rightarrow PV - Pb = RT$. . . . . (2)

Dividing equation (2) by $RT$ we get

$\Rightarrow \dfrac{{PV}}{{RT}} - \dfrac{{Pb}}{{RT}} = 1$

$\Rightarrow \dfrac{{PV}}{{RT}} = 1 + \dfrac{{Pb}}{{RT}}$ . . . . . (3)

$\Rightarrow Z = \dfrac{{PV}}{{RT}}$ . . . . . (4)

$\Rightarrow Z = 1 + \dfrac{{Pb}}{{RT}}$

Where $Z$ = compressibility factor

Therefore, at high pressure compressibility factor is $1 + \dfrac{{Pb}}{{RT}}$

**Therefore, by the above explanation, the correct option is [A] $1 + \dfrac{{Pb}}{{RT}}$**

**Additional information:**

At high pressure, compressibility factor is greater than $1$.

As $P$ is increased (at constant T). The factor $\dfrac{{Pb}}{{RT}}$ increases. This explains how compressibility increases continuously with pressure.

Van der Waals constants $a$ and $b$ have significance of attractive force among molecules of gas and value of $b$ is a measure of the effective size of gas molecules.

**Note:**A gas which obeys the ideal gas equation $PV = nRT$ under all conditions of temperature and pressure is called ideal gas. Concept of ideal gas is only theoretical when pressure is low and temperature is high. They obey the gas law. Such gases are known as Real gases. All gases are Real gases.

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