When looking at a graph of comparing temperature and specific volume, we see that for increasing pressures, the boiling point increases and that the length of the saturated liquid-vapor line decreases.

So, there's really two questions here, but the following first question will help with answering my main question.

I understand this as because of an increase in pressure, molecules are harder to push apart (as essentially pressure is an external force keeping the molecules together). Is this correct?

However, the main question is why does the saturated liquid-vapor line decrease in length as pressure increases?

The way I understand it is that because we needed a higher boiling point at high pressures, we've already inputted a lot of energy to get to this point, and so the specific volume of the liquid is already quite high for higher pressures, as inputting energy essentially tries to push apart these molecules. Hence, to reach the saturated vapor point, we don't need as much energy to keep forcing particles apart (so the increase in specific volume is not as much).

Refer to graph 2-13 on http://engr.bd.psu.edu/davej/classes/thermo/chapter2.html

enter image description here

  • $\begingroup$ If you can post the one graph that you are questioning, it would be somewhat easier to answer the question that was posed. Your link shows several graphs. $\endgroup$ – David White Aug 5 '18 at 5:17
  • $\begingroup$ @DavidWhite I've attached the graph! $\endgroup$ – 14tim4 Aug 5 '18 at 5:22

The graph represents temperature vs. specific volume for a pure component. Because this is a pure component, several facts are known:

1) In order to gather the given data, some container was partially filled with a given liquid, with the remainder of the container filled with the vapor of the same substance. The container contained a substantial fraction of liquid and a substantial fraction of vapor (e.g., 50% liquid and 50% vapor), to ensure that some liquid and some vapor remained in the container as the liquid temperature was increased.

2) The liquid density goes down (liquid specific volume increases) as the liquid temperature increases.

3) There is a relationship between the temperature of the liquid and the pressure inside the container. This relationship is determined by the vapor pressure of the liquid, which can be calculated by the Antoine equation (see https://en.wikipedia.org/wiki/Antoine_equation). As the temperature inside the container goes up, the vapor pressure of the liquid goes up. As a result, more liquid goes into the vapor phase, and since this is a closed container, the pressure in the container goes up as a result. As the pressure increases, the density of the vapor phase goes up (vapor specific volume decreases).

4) At some point, as the temperature increases, the liquid density decreases enough to equal the vapor density, and the critical point is reached. At this point, there is only one phase in the container, as the liquid and vapor phase densities are equal.

Because increasing temperature causes liquid specific volume to increase, and at the same time causes vapor specific volume to decrease due to the dependencies between temperature, vapor pressure, and specific volume, the width of the saturated liquid-vapor line decreases as temperature increases, until you reach the critical temperature.

  • $\begingroup$ So, what you're saying is, suppose we kept the pressure constant at a low value and we increase temperature, then the density of the liquid decreases, but is still too high to match the density of the vapor, so the width of the saturated liquid-vapor line is long. However, if we kept the pressure constant at a high value, then increase temperature, the density of the liquid goes down, but because we are at a higher pressure, the density of the vapor is already high, so the width is less. $\endgroup$ – 14tim4 Aug 5 '18 at 21:52
  • $\begingroup$ No - what I am saying is that pressure is a function of temperature for a saturated liquid, per the Antoine equation. If you increase temperature, some of the saturated liquid will go into the vapor phase, and the pressure will increase as a result. $\endgroup$ – David White Aug 6 '18 at 17:18

Regarding your first question, as the temperature gets higher (forcing the molecules further apart) higher pressures are needed to liquify the vapor. Above the critical point (top of the chart), where liquid and vapor phases are indistinguishable and the heat of vaporization becomes zero, the temperatures become so high that the vapor cannot be liquefied by pressure alone.

To answer your second (main) question, as pressure and temperature increase and you approach the critical point the difference in the specific properties of the liquid and vapor phases necessarily become smaller an smaller. For example, from the saturated steam table for two pressures we have the following:


$$p_{sat} = 0.10135 MP_a$$ $$v_f = 0.001044$$ $$v_g = 1.6729$$ The specific volume of the vapor is about 1600 times greater than the liquid.

for $$p_{sat} = 360 MP_a$$ $$v_f = 0.0011893$$ $$v_g = 0.006945$$ The specific volume of the vapor is about 6 times greater than the liquid.

Hope this helps.


First, I will give you my interpretation of the graph, because I had to stare at it for a bit to make sense out of it. Here is what I imagine each line represents.

You fill a vessel completely with cold liquid and seal it. At the very top (where vapor would be if there were any) there is a pressure relief valve set to allow fluid to escape if the pressure reaches a desired threshold. Each red line corresponds to a different threshold, increasing in pressure toward the top of the graph. Define "specific volume" as the volume of the container divided by the total mass of the stuff inside, whether it's liquid, vapor, or both. Then you start adding heat, which increases the temperature of the liquid and therefore the pressure. Pretty soon the threshold pressure is reached and the valve starts leaking. Things thereafter proceed in three stages, with the pressure always remaining at the constant determined by the valve setting: 1) the temperature of the liquid increases, and it expands (its specific volume increases) causing some liquid to be ejected through the valve; this is the straight section of each line on the left side of the graph. 2) As you continue adding heat the temperature reaches the boiling point and stops increasing; instead the added heat goes into boiling liquid into vapor, with vapor constantly being pushed through the valve to make room for more; this is the flat straight section of each line. 3) Once all the the liquid is boiled away, leaving only vapor in the container, the temperature begins to increase again, and as it does the vapor expands; this is the upward sloping part of each line on the right hand side.

Now that that's out of the way, your "main question" is why does the middle, horizontal line decrease in length for higher pressures. One answer is that the vapor is more compressible than the liquid: at higher pressures there's less difference in the densities of liquid vs. vapor because the vapor has shrunk more in response to being squeezed. Less difference in densities means a smaller range in "average density" (or specific volume) in going from all liquid to all gas, which is what the length of the horizontal line shows. The horizontal line shrinks to zero when liquid and vapor densities are the same.

I know some of this answer is reiterating stuff already offered in the others. But even after reading them it took me a minute to make sense of the graph, so I figured I'd share my thoughts.


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