The unit of the van der Waals gas equation parameter 'a' in $$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$ is:

We begin with the van der Waals equation for n moles of a real gas

$$\left(P + \frac{a n^{2}}{V^{2}}\right)\,(V - nb) \;=\; nRT$$

In this expression the pressure-correction term $$\frac{a n^{2}}{V^{2}}$$ is added directly to the pressure $$P$$. Because only like quantities can be added, the whole fraction $$\frac{a n^{2}}{V^{2}}$$ must possess exactly the same unit as $$P$$.

Let us denote the unit of pressure by $$[P]$$, the unit of volume by $$[V]$$ and the unit of amount of substance by $$[n]$$. Correspondingly, let $$[a]$$ be the unknown unit of the parameter $$a$$ which we have to determine.

Since the dimensions of the two addends are equal, we can write

$$\Bigl[\frac{a n^{2}}{V^{2}}\Bigr] \;=\; [P]$$

Substituting the separate units we get

$$[a] \,[n]^{2}\,[V]^{-2} \;=\; [P]$$

Rearranging for $$[a]$$ gives

$$[a] \;=\; [P]\,[V]^{2}\,[n]^{-2}$$

Now we replace each symbol by the commonly used practical units in which the van der Waals constants are usually expressed:

• The convenient laboratory unit for pressure is the atmosphere, so $$[P] = \text{atm}$$.
• The convenient laboratory unit for volume is the cubic decimetre, so $$[V] = \text{dm}^3$$.
• The SI-derived unit for the amount of substance is the mole, so $$[n] = \text{mol}$$.

Substituting these explicit units we arrive at

$$[a] \;=\; (\text{atm})\,(\text{dm}^{3})^{2}\,(\text{mol})^{-2}$$

Simplifying the exponent on the volume unit, we finally have

$$[a] = \text{atm}\,\text{dm}^{6}\,\text{mol}^{-2}$$

This exactly matches Option C in the given list.

Hence, the correct answer is Option C.