What are the effects of the presence of air when measuring the vapor pressure curve for water?

Question

How does the slope of $p_{\textrm{total}}$-vs-$T$ change as a function of the amount of air in an air/saturated-water-vapor mix? There are a few possible effects that I know of: (1) The gases are not perfectly ideal, and so intermolecular forces might alter the amounts of air and water vapor; (2) The amount $n_{\textrm{a}}$ of air can change as increasing pressure can drive dissolution of the air in the liquid water (I think this is the Raoult Effect); (3) the extra pressure above the liquid alters its thermodynamic properties, leading to a change in the amount of water vapor in equilibrium with it (this it the Poynting effect I think). How do the sizes of these effects compare?

Knowing the quantitative details for water would be helpful as well, so that we can either explain the experimental results below or see what kind of errors in the experiment led to the results.

Necessary background details of the experiment

We run the following experiment: Place 100-ish mL of water in a small flask, boil it for awhile to expel as much air from the flask as we can, cap it, and then measure the pressure and temperature of the gas inside the flask as it cools. If we have succeeded in expelling all of the air from the flask, then by measuring the pressure and temperature of the gas we get a measurement of the saturated vapor pressure curve for water.

However, if we aren't able to expel all the air, then we only get an estimate of the vapor pressure curve for water due to the fact that we are measuring the total pressure of the gas, which is the sum of the partial pressures of air and water. If we can treat the two components as ideal gases, then the pressure that we measure is given by $ p = p_{\textrm{a}} + p_{\textrm{w}} =\frac{n_{\textrm{a}}R}{V}T + \frac{n_{\textrm{w}}R}{V}T\,, $ where the a and w subscripts are for air and water, respectively, $n$ is the number of moles of the components of the gas, and $V$ and $T$ are the shared volume and temperature of the gases. ($V$ is effectively constant since the volume of the liquid hardly changes given the small-enough amount of condensation that occurs during the cooling.) This shows then that our measured "vapor-pressure" curve for water is going to be shifted up from what it should be. Indeed, we see this in the graph below, where we have done the experiment twice, one in which we didn't do any boiling of the water (and hence didn't expel any air) and one in which we boiled the water for some amount of time. The dashed curve is the "actual" vapor pressure curve for water grabbed from some steam tables somewhere.

Now, if we naively assume that (1) the gases are both ideal gases, and (2) that there is no change in $n_{\textrm{a}}$ (i.e., we are assuming that there is no change in the amount of air dissolved in the liquid water), then the equation above tells us that not only are we overestimating the pressure in the vapor-pressure curve, but we are also overestimating the slope of the vapor-pressure curve, because the local slope is given by $ \frac{dp}{dT} = \frac{n_{\textrm{a}}R}{V} + \frac{n_{\textrm{w}}R}{V} + \frac{R}{V}T\frac{dn_{\textrm{w}}}{dT}\,. $ (All terms on the right-hand side are positive, including the third since the amount of water vapor increases with temperature.)

However! - if I fit the curves above and compute the derivatives, we get the following graph, which I think is inconclusive about how the slope changes. The case of maximal amount of air ("not boiled") has the same slope as the actual vapor-pressure curve, and the case of some air expelled ("boiled") has a smaller slope. This doesn't really make sense, as I would expect a monotonic transition of the slope as the amount of air decreases. So there's clearly some experimental error here.

Notes

1. Related post: What is the effect of atmospheric pressure on the vapour pressure of water?