At very high pressures, the compressibility factor of one mole of a gas is given by:

The quantity that is asked for is the compressibility factor, defined for one mole as

$$Z=\dfrac{P\,V}{R\,T}.$$

To connect $$P$$, $$V$$ and $$T$$ for a real gas we start from the van der Waals equation

$$\left(P+\dfrac{a}{V^{2}}\right)\,(V-b)=R\,T.$$

We first expand the product on the left hand side:

$$P(V-b)+\dfrac{a}{V^{2}}\,(V-b)=P\,V-P\,b+\dfrac{a}{V}-\dfrac{a\,b}{V^{2}}.$$

Equating this expression to $$R\,T$$ we obtain

$$P\,V-P\,b+\dfrac{a}{V}-\dfrac{a\,b}{V^{2}}=R\,T.$$

Now we isolate the term $$P\,V$$ because it appears directly in $$Z$$:

$$P\,V=R\,T+P\,b-\dfrac{a}{V}+\dfrac{a\,b}{V^{2}}.$$

Dividing every term by $$R\,T$$ converts this into the desired compressibility factor:

$$Z=\dfrac{P\,V}{R\,T}=1+\dfrac{P\,b}{R\,T}-\dfrac{a}{R\,T\,V}+\dfrac{a\,b}{R\,T\,V^{2}}.$$

We are interested in the behaviour at very high pressures. At such large pressures

• the volume $$V$$ becomes extremely small, which means the dominant effect is the finite size of the molecules (parameter $$b$$),
• the term proportional to $$a$$ (intermolecular attractions) becomes insignificant compared with the huge repulsive contribution represented by $$P\,b$$.

Consequently we neglect the last two attraction-related terms and keep only the first correction term. Hence

$$Z\;\approx\;1+\dfrac{P\,b}{R\,T}.$$

This matches option A.

Hence, the correct answer is Option A.