1. Given equation of state

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

We need the dimensions of

$\frac{a}{b^2}$

2. Find dimensions of $a$

From the term

$P+\frac{a}{V^2}$

both terms must have the same dimensions. Hence,

$\left[\frac{a}{V^2}\right]=[P]$

So,

$[a]=[P][V^2]$

3. Find dimensions of $b$

From the term

$V-b$

both terms must have same dimensions. Therefore,

$[b]=[V]$

Thus,

$[b^2]=[V^2]$

4. Now compute dimensions of $\dfrac{a}{b^2}$

$\left[\frac{a}{b^2}\right]=\frac{[a]}{[b^2]}=\frac{[P][V^2]}{[V^2]}=[P]$

So $\dfrac{a}{b^2}$ has the dimensions of pressure.

5. Check options

• A: $P$ → correct
• B: $RT$ → not pressure
• C: $PV$ → not pressure
• D: $R$ → not pressure

Therefore, the correct option is:

$\boxed{A}$