Dimensionless consistency and quantities

Question

I am a chemical engineering student learning about dimensionless quantities. This is a practice question that I am trying.

The Van der Waals equation of state can be used to predict the behaviour of non-ideal gas, and it reads $$ \left( P + \frac{n^2a}{V^2}\right) (V - nb)= nRT, $$ where

• $P$ is the pressure, in $\rm N/m^2$,
• $n$ is number of moles, in $gmol$,
• $V$ is the volume, in $\rm m^3$,
• $T$ is the temperature, in $\rm K$,
• $R$ is the molar gas constant, in $atm.L / gmol.K$, and
• $a$ and $b$ are constants.

I am trying to show that this equation is dimensionally consistent. But first I must find the units of $a$ and $b$.

I learned that units on the left must be equals to units on the right if it is dimensionally consistent. Knowing that, I found that $nRT= atm.L$

However, now I get stuck because on the left hand side of the equation, it’s something multiply by something. Meaning the whole thing has a unit of $atm.L$ . But how do I find them individually?

In short, It’s wrong to say that $( P + \frac{n^2a}{V^2})$ has a unit of $atm.L$ so how do I find units of a and b ? How do I know what is the unit of $( P + \frac{n^2a}{V^2})$ and also the other part of the equation?

Answer 1

In the term $ (p+ \frac {n^2a}{v^2}) $ your adding pressure to something. That means in order for dimension consistency to hold $\frac {n^2a}{v^2}$ must also have units of pressure i.e. $\frac N{m^2}$. Do a little cross multiplying you get the units of a to be $ \frac {NL^2}{m^2(mol)^2} $
If you want you could break down the units of newton further.
Similarly units of b will be $ \frac L{(mol)} $

The whole of $ (p+ \frac {n^2a}{v^2}) $ has units of pressure.
The whole of $ (v - nb) $ has units of L.
Which equals the right side's $nrt$.

Answer 2

For a formula to be dimensionally consistent, the dimensions must match for any two terms that are (i) equal to each other, or (ii) added together. This means that:

• In the first factor, $[P]=[n^2a/V^2]$ (i.e. the second term must have the units of pressure), so that $[a]=[PV^2/n^2]$ and $a$ must have units of $\rm J\:m^3 /mol^2$.
• In the second factor, $[V]=[nb]$, so that $[b]=[V/n]$ is a molar volume, i.e. it must have units of $\rm m^3/mol$.

This then means that the left-hand side must have dimensions of $[PV]=[E]$ i.e. a pressure-volume product, which has dimensions of energy. This is also true of the right-hand side: since the molar gas constant has dimensions $[R]=[E/nT]$ (or in other words units of $\rm J\:mol^{-1}\: K^{-1}$) the product $nRT$ also has dimensions of energy.

Answer 3

The other answers to find the constants $a$ and $b$ on the left hand side of the equation seem to be correct. But you have to be careful. The units used in any equation have to be taken consistently from one unit system. This seems not to be the case here. The LHS uses for pressure and volume the SI units $N/m^2$ and $m^3$. The RHS gives the molar gas constant $R$ in incoherent units of $$\frac{atm\cdot L}{gmol\cdot K}$$ where pressure is in atm (1 atm =$1.01325×10^5 N/m^2$, volume is in L (1 liter=$1\cdot 10^{-3}m^3$). Also gmol is an outdated name for mol. The equation has not only to be dimensionally correct. It also has use the same units everywhere. To be consistent with the LHS, the RHS of the equation must also use SI units!