How do I calculate van der Waals constants with 2 unknowns (a and b)?

[Saado]

How would I go about calculating the units for the van der waals constants if I have 2 unknowns? (a and b)
(P + a (n/V)^2) (V - n b) = n R T

p=Pressure
v=Volume
n=moles
R=gas constant
T=temperature

 

[DrClaude]

How would I go about calculating the units for the van der waals constants if I have 2 unknowns? (a and b)
(P + a (n/V)^2) (V - n b) = n R T

p=Pressure
v=Volume
n=moles
R=gas constant
T=temperature

Units have to be consistent: if you add/substract quantities with units, they must have the same units.

 

[Saado]

Yeah I know that. How would I go about getting the units of a and b from the units above? It seems to me that when I try to re-arrange, I'll always have 2 unknowns (a and b) so I can't work out the individual units for those constants

 

[DrClaude]

Yeah I know that. How would I go about getting the units of a and b from the units above? It seems to me that when I try to re-arrange, I'll always have 2 unknowns (a and b) so I can't work out the individual units for those constants

I don't see how you get that. Take $b$ for instance. It has units such that $nb$ and $V$ have the same units. The common notation for this is
$$
[n] [ b ] = [V]
$$
Can you use that to figurs the units of $b$?

 

[Saado]

Would that not simply be the units for volume? meter cubed per mol?

Could you take me through the steps of why that is the case using algebra? It isn't intuitive to me at the moment.

 

[DrClaude]

Would that not simply be the units for volume? meter cubed per mol?

Could you take me through the steps of why that is the case using algebra? It isn't intuitive to me at the moment.

In SI units, volume is expressed in cubic meters. So you have
\begin{align}
[V] &=\textrm{m}^3 \\
[n] &= \textrm{mol}
\end{align}
Therefore
$$
[ b ] = \frac{[V]}{[n]} = \textrm{m}^3 \textrm{mol}^{-1}
$$
That way, you have that $n b$ will have units of volume (m³ in this case), so that you can indeed calculate $V - n b$ and recover something with the units of a volume.

You now need to figure out the units of $a$ to recover a pressure (Pa = N m^-2 = kg m^-1 s^-2 in SI units).

 

[Saado]

Right, I understood that much. Maybe I should have expressed my query better.
As a test of algebra, I was given the above equation: (P + a (n/V)^2) (V - n b) = n R T

With absolutely no context and told that a and b are constants.
I recognized the other terms from eqns like PV=nRT and pv=1/3 nmv^2
How would I get from (P + a (n/V)^2) (V - n b) = n R T to B=V/n? I don't see how you could deduce with algebra how you are supposed to isolate the terms a and b when you know neither of their units.

 

[DrClaude]

How would I get from (P + a (n/V)^2) (V - n b) = n R T to B=V/n?

I guess you mean here $[ b ] = [V] / [n]$? (The brakets around b are causing problems since it also means bold.) This comes simply from the fact that you have the term $(V - n b)$. The only way for this to make sense is if both $V$ and $nb$ have the same units. An expression like $3 \mathrm{m} - 10 \mathrm{s}^{-1}$ is completely meaningless. You therefore set $[V] = [n][ b ]$ and find out the units for $b$.

I don't see how you could deduce with algebra how you are supposed to isolate the terms a and b when you know neither of their units.

I'm not sure what you are asking here. To work with an equation like (P + a (n/V)²) (V - n b) = n R T requires no knowledge of the units of the different variables. It's only when you need to input numerical values that units are necessary.

 

[Saado]

THANK YOU! I don't know why I didn't think of that before. Yes, this makes perfect sense.