To find the dimensional formula of $$a b^{-1}$$ in the equation given by $$\left( P + \frac{a}{V^2}\right)(V-b) = RT$$, where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature, we will first understand the dimensions of each term in the equation. The variables mentioned have their usual meanings in the context of physics and chemistry, associated with the Ideal gas laws and the Van der Waals equation.

The dimensional formula for pressure ($P$) is $[M^1 L^{-1} T^{-2}]$ assuming $P = \frac{Force}{Area}$ and Force = $Mass \times Acceleration$.

The volume ($V$) has a dimensional formula of $[L^3]$.

Temperature ($T$) typically does not factor into the dimensional analysis directly in this context as we are considering the units it would be measured in (e.g., Kelvin), which doesn't directly convert into mass, length, and time. However, $RT$ suggests energy, and since the gas constant $R$ has dimensions including time, we consider energy's dimensions, $[M^1 L^2 T^{-2}]$, but this will not directly affect the dimension we are solving for.

Given these, let's analyze the term $\frac{a}{V^2}$ to deduce $a$'s dimensions:

• $\frac{a}{V^2}$ must have the same dimensions as pressure ($P$) for the equation to be dimensionally consistent, so

$ [a] [L^{-6}] = [M^1 L^{-1} T^{-2}] $

Thus, $a$ has the dimensional formula $[M^1 L^5 T^{-2}]$.

The term we're interested in is $a b^{-1}$, which requires finding the dimension of $b$ as it's being subtracted from $V$, implying it shares dimensions with $V$:

• Hence, $b$ has the dimensional formula $[L^3]$.

Putting it all together for $a b^{-1}$:

$ [a b^{-1}] = [M^1 L^5 T^{-2}] [L^{-3}] = [M^1 L^2 T^{-2}] $

Thus, the correct answer is:

Option D $$[\mathrm{ML}^2 \mathrm{~T}^{-2}]$$.