Plotting a Langmuir adsorption isotherm We have,

$$\theta = \frac{Kp_\mathrm a}{1+Kp_\mathrm a}$$ 

where, the fractional coverage $\theta$ is defined as the number of sites occupied by adsorbate $A$ over the total number of sites. $p_\mathrm a$ is the partial pressure of the adsorbate and $K$ is the equilibrium constant.
$$\theta = \frac{[A_{\text{ad}}]}{[S_{\text{total}}]}$$

I need to plot the following data to verify the monolayer adsorption of $\ce{CO}$ on 1 g of $\ce{Pt}$ powder.

$$\begin{array}{cc}
\hline
\text{Equilibrium Pressure (P)/mmHg} &amp; \text{Adsoprtion Amount(V)/ mmHg} \\ \hline
 2.50 &amp; 0.507\\
5.00 &amp; 0.92\\
10.0 &amp; 1.51\\
15.0 &amp; 1.97\\
20.00 &amp; 2.32\\
30.00 &amp; 2.75\\ 
40.00 &amp; 3.13\\ \hline
\end{array}$$

The question gives me the following hint


  Because adsorption amount $V$ is proportional to $\theta$, $\theta$ can be expressed as $\frac{V}{V_\mathrm m}$ where $V_\mathrm m$ is the adsorption amount at $\theta = 1$ 


I generated the following graph in Mathematica, simply by plotting $V$ vs $P$



I don't quite understand what the hint wants me to do? Does that require that I normalise the data fashion, i.e by dividing by a particular $V_\mathrm m$, so instead of having $V$ on the y-axis, I have $\theta$ varying from 0 to 1.

Is $V_\mathrm m$ something I obtain from the data/graph by calculating, or is it a tabulated value that I can look up in table somewhere.
Self Answer

Additionally, I plotted $\frac{1}{V}$ and $\frac{1}{p}$, as suggested by @Chester Miller in the comments below, and indeed I get a straight line. 



A linear model fit was performed and the equation $4.39274 x+0.214246$ fits the data with an $R^2 = 0.999929 $. From my limited knowledge on this topic (I haven't formally studied this subject), the intercept should give me $\frac{1}{\text{adsorption capacity}}$, and the slope should give some indication of the affinity of the adsorption sites?

So going back to the initial equation, $$\theta = \frac{V}{V_\mathrm m} = \frac{Kp_\mathrm a}{1+Kp_\mathrm a} $$

And now considering $$\frac{V_\mathrm m}{V} = \frac{1+Kp_\mathrm a}{Kp_\mathrm a} = \frac{1}{Kp_\mathrm a} + 1$$

and dividing throughout by $V_\mathrm m$ we get,

$$\frac{1}{V} = \frac{1}{V_\mathrm m} + \frac{1}{V_\mathrm mKp_\mathrm a}$$

And we get the appropriate values for $V_\mathrm m$ and $K$ from the slope and intercept of the $1/V$ vs $1/p$ plot, and then obtaining $\theta$ and plotting vs. $p$ is a trivial exercise.

Anyway, using the calculated parameters, I once again plotted $V$ vs $p$ and fit a curve of the form $\frac{V_\mathrm mKp_\mathrm a}{1+Kp_\mathrm a}$