What are the dimensions of a and b in van der Waal's real gas formula?

In the van der Waal's formula for real gases $\left( p+\frac{a}{V^2}\right)(V-b)=nRT$, what are the dimensions of $a$ and $b$? 


I learnt that $a$ is related to the average force of attraction between the molecules and $b$ is related to the total volume of the molecules, but I fail to understand their dimensions.




Part of your confusion is in the formula you have written. According to the Tanner's General Chemistry webpage [The van der Waals Equation](http://www.tannerm.com/vander.htm), if you are considering 1 mole of a substance, then the formula is


$$(P+a/V^2)(V-b) = RT$$


For $n$ moles


$$(P+an^2/V^2)(V-nb) = nRT$$


The Science HQ website [Van der waal’s equation](http://www.sciencehq.com/chemistry/van-der-waals-equation.html), the constants $a$ and $b$ represent pressure and volume corrections respectively. The units are based on the formula for $n$ moles of substance (the 2nd equation above).


Taking each term on the left hand side individually:


$(P+an^2/V^2)$ is the van der Waals equation equivalent for $P$ in the ideal gas equation, thus this therm must have the same unit, so using $Pa$ as the unit for pressure would require that the units of $a$ to be $Pa L^2 mol^{-2}$ (the website uses the unit $atm$, but the same idea still applies.


In a similar manner, the term $(V-nb)$ is the van der Waals equivalent of $V$ in the ideal gas equation. So to keep the unit as for volume in the second term, the unit for $b$ must be $Lmol^{-1}$.


One thing to always be careful with, as there are a number of units used for both pressure and volume, it is imperative to keep these constant throughout your calculations.


For example, volume can be expressed as $m^3$ or $L$ and several others (hence the $L^2$ term above would be the same as $(m^3)^2$ or simply $m^6$).


One final document is a master list of $a$ and $b$ constants for several chemicals (once again be careful with the units - unit conversion are provided in the document), [van der Waals constants for gases.](http://courses.chem.indiana.edu/c360/documents/vanderwaalsconstants.pdf). Also further explanations of the above can be found on this [Purdue University page entitled Deviations from Ideal Gas Law Behavior](http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/deviation5.html)


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Since the van der Waals equation is derived from the ideal gas equation, the empirical constants $a$ and $b$ need to have dimensions so that the units work out.


The ideal gas constant $R$ has dimensions of $\dfrac{\text{pressure}\cdot\text{volume}}{\text{moles}\cdot\text{temperature}}$ since $R=\frac{pV}{nT}$ for an ideal gas. This is one of the definitions of $R$.


For the van der Waals equation, now $R=\dfrac{(p-\frac{a}{V^2})(V-b)}{nT}$


Since we have $(p-\frac{a}{V^2})$, the term $\frac{a}{V^2}$ needs to have units of pressure for this subtraction to not screw up the units.


$$\therefore \dfrac{a}{V^2}\equiv\text{pressure}\implies a\equiv \text{pressure}\cdot(\text{volume})^2$$


Now, we guarantee that $(p-\frac{a}{V^2})$ has units of pressure:


$$(p-\dfrac{a}{V^2}\equiv (\text{pressure}-\dfrac{\text{pressure}\cdot\text{volume}^2}{\text{volume}^2} \equiv \left(\text{pressure}-\text{pressure}\cdot \dfrac{\text{volume}^2}{\text{volume}^2} \right)\equiv (\text{pressure}-\text{pressure}) \equiv \text{pressure}$$


The constant $b$ is easier. $b$ needs to have dimensions of volume so that $V-b$ still has dimensions of volume.


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