Saturated vapor pressure
This is purely an equilibrium phenomenon. The vapor pressure of a fluid is generally just a function of the temperature. This pressure is just the partial pressure of the fluid, not anything else. To rationalize why this is, consider what the reaction looks like:

$$\ce{A(l) &lt;=&gt; A(g)}$$

$$\mathrm{K}_{eq} = \frac{P_{\ce{A(g)}}}{\ce{[A(l)]}}$$

The partial pressure is only dependent on factors that affect the activity of the pure substance. Since the liquid is essentially incompressible, the total external pressure does not have an effect.