The equation of state for a real gas is given by (P+a/V²)(V-b)= RT. The dimensions of the constant a are

To find the dimensions of the constant $a$ in the given equation of state for a real gas:
$(P+\frac{a}{V^2}$$\left)\right(V-b)=RT$

We apply the principle of homogeneity of dimensions. According to this principle, physical quantities can be added or subtracted only if they have the same dimensions. This means that each term in a sum or difference must have the same dimensional formula.

In the expression $(P+\frac{a}{V^2}$$)$, the constant term $\frac{a}{V^2}$ is added to pressure $P$. Therefore, the dimensions of $\frac{a}{V^2}$ must be identical to the dimensions of pressure $P$:
$[\frac{a}{V^2}$$]=[P]$

Rearranging the equation to solve for the dimensions of $a$, we get:
$\left[a\right]=\left[P\right]\times \left[V^2\right]$

Now, let's write down the dimensional formulas for pressure $P$ and volume $V$:
1. Pressure ($P$) is force per unit area. Since force has dimensions $\left[ML{T}^{-2}\right]$ and area has dimensions $\left[L^2\right]$:
$\left[P\right]=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$
2. Volume ($V$) has dimensions $\left[L^3\right]$. Therefore, volume squared ($V^2$) has dimensions:
$\left[V^2\right]={\left[L^3\right]}^2=\left[L^6\right]$

Substituting these dimensional formulas back into the equation for $\left[a\right]$:
$\left[a\right]=\left[M{L}^{-1}{T}^{-2}\right]\times \left[L^6\right]$
$\left[a\right]=\left[M{L}^{-1+6}{T}^{-2}\right]$
$\left[a\right]=\left[M{L}^5{T}^{-2}\right]$

Thus, the dimensional formula for the constant $a$ is $\left[M{L}^5{T}^{-2}\right]$.