When a gas condenses, how do you quantify the amount of the gas which turns to liquid? I'm going through a question in Finn's thermal physics, 

I can't figure out part (c). I understand there will be some pressure change which will cause a certain portion of the gas to condense to water. I can't find a way to calculate that mass. I've tried long-winded methods by equating pressures of the gas to the mercury and calculating a mass through the force of gravitation. This doesn't seem to work however and requires a fair few assumptions about the state of the system. The answer is supposed to be $1.1x10^{-7} g$ but I can't get near that answer. Is anyone willing to guide me in the right direction?
 A: It might be helpful to look at this problem in reverse.  Let's say you have a sealed container attached to a piston which is initially completely filled by a liquid - let's say H$_2$O to be concrete.

If we pull the piston upward, we create a vacuum space above the water.  The more energetic H$_2$O molecules at the surface of the liquid will escape into the vacuum, filling it with H$_2$O vapor.

This net flow of molecules into the vapor phase will continue until the reverse process - free molecules being recaptured by the liquid - occurs at the same rate, at which point the liquid and vapor will be in equilibrium with one another.  The pressure of the gas in this equilibrium state is called the vapor pressure, and is a function of the temperature.
If I continue to pull the piston upward while keeping the system at constant temperature, then more H$_2$O molecules would flow into the vapor phase because the vapor pressure remains the same.  You can calculate how much will evaporate by assuming that the water vapor is an ideal gas; given the gas volume, the vapor pressure, and the temperature, you can calculate the number of moles of H$_2$O vapor which have escaped from the liquid phase.
This will continue until all the liquid is gone. Once the H$_2$O is entirely in the vapor phase, the pressure will decrease as the volume increases.  If you assume ideal gas behavior, then $P\propto 1/V$ in accordance with the ideal gas law.

Imagine now that you start with a container which is filled with vapor and no liquid at all.  As you (isothermally) compress the gas, its pressure rises until it reaches the vapor pressure.  At this point, a small decrease in volume would cause some vapor to condense into the liquid phase while the pressure remains constant.  Continuing to decrease the volume causes more and more gas to condense, until you are finally left with a cylinder filled with pure liquid H$_2$O.
Here's a plot illustrating the pressure of the gas in the cylinder as you isothermally vary the volume.  The green dashed curve is the ideal gas behavior.

The answer provided to you by your text is correct.  At the end of the process, you know the volume of the gas, the pressure of the gas, and the temperature of the gas, so you can find out how many moles of H$_2$O are in the vapor phase.  Comparing this to the amount of gas you started with, you can find how many moles of H$_2$O condensed during the compression.
A: If you have a vapor and a liquid in a container at equilibrium, the equation that applies is $$m_LV_L+m_VV_V=V$$where $m_L$ and $m_V$ are the mass of liquid and vapor respectively, $V_L$ and $V_V$ are the specific volumes of the liquid and vapor respectively, and V is the container volume.  In addition, if m is the total mass of vapor and liquid, then $$m_L+m_V=m$$Another version of the first equation is $$(1-x)V_L+xV_V=
\frac{V}{m}=\bar{V}$$where x is the mass fraction of vapor (aka, the quality for the case of steam and liquid water), and $\bar{V}$ is the average specific volume of the mixture of vapor and liquid.
