The dimensional formula of the constant a in van der Waal's gas equation (P+a/V²)(V-b)=RT is

The correct option is [ML⁵T⁻²].

Step-by-step Explanation:

The given van der Waals gas equation is:
$(P+\frac{a}{V^2})(V-b)=RT$
where:
- $P$ represents pressure,
- $V$ represents volume,
- $T$ represents temperature,
- $R$ is the universal gas constant, and
- $a$ and $b$ are constants.

According to the principle of homogeneity of dimensions, physical quantities with the same dimensions can be added to or subtracted from each other. This means that in the expression $(P+\frac{a}{V^2})$, the term representing pressure ($P$) and the term $\frac{a}{V^2}$ must have the exact same dimensional formula.

Therefore, we can write:
$\left[P\right]=\left[\frac{a}{V^2}\right]$

Rearranging the equation to solve for the constant $a$:
$\left[a\right]=\left[P\right]·\left[V^2\right]$

Now, let's write down the dimensional formulas for pressure ($P$) and volume ($V$):
1. Pressure is defined as force per unit area:
$P=\frac{\text{Force}}{\text{Area}}$
The dimensional formula of Force is $\left[{\text{MLT}}^{-2}\right]$ and Area is $\left[{\text{L}}^2\right]$. Thus:
$\left[P\right]=\frac{\left[{\text{MLT}}^{-2}\right]}{\left[{\text{L}}^2\right]}=\left[{\text{ML}}^{-1}{\text{T}}^{-2}\right]$

2. The dimensional formula of Volume ($V$) is:
$\left[V\right]=\left[{\text{L}}^3\right]$
Thus, the dimensions of $V^2$ are:
$\left[V^2\right]={\left[{\text{L}}^3\right]}^2=\left[{\text{L}}^6\right]$

Substitute the dimensions of $P$ and $V^2$ back into the equation for $\left[a\right]$:
$\left[a\right]=\left[{\text{ML}}^{-1}{\text{T}}^{-2}\right]·\left[{\text{L}}^6\right]$
$\left[a\right]=\left[{\text{ML}}^{(-1+6)}{\text{T}}^{-2}\right]$
$\left[a\right]=\left[{\text{ML}}^5{\text{T}}^{-2}\right]$

Hence, the dimensional formula of the constant $a$ is [ML⁵T⁻²].