3
$\begingroup$

From the US National Chemistry Olympiad:

A sample of a pure substance is placed in a sealed, rigid container and the pressure is measured as a function of temperature. Which is the best explanation for the result shown? enter image description here

(A) At lower temperatures, the substance is a mixture of solid and vapor, while at 60 °C the solid melts to give a mixture of liquid and vapor.
(B) At lower temperatures, the substance is a mixture of liquid and vapor, while at 60 °C only liquid is present.
(C) At lower temperatures, the substance is a mixture of liquid and vapor, while at 60 °C only a supercritical fluid is present.
(D) At lower temperatures, the substance consists of vapor only, while at 60 °C only a supercritical fluid is present.

The correct answer here is B, which I don't exactly understand. I understand that the system being a mixture of vapor and liquid at low temperature makes sense, since as the temperature is raised, the vapor pressure of the liquid increases, and the vapor (which there is now more of) exerts more pressure in accordance with the ideal gas law. I also kind of understand why the mixture would, counterintuitively, become only liquid as the temperature is raised: despite the high temperature, the pressure is too high for gas to exist at 60 C. What I don't understand is why the pressure sharply increases at 60 C. How does the liquid suddenly exert so much pressure? And why couldn't a supercritical fluid explain the behavior at 60 C?

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Oh, it's a scenario a bit like chemistry.stackexchange.com/questions/34929/… See, there was only a tiny amount of vapor in your box and liquid expanded on heating to fill it. Now it's trying to break your box apart. $\endgroup$
    – Mithoron
    Commented Oct 2 at 18:37
  • $\begingroup$ In the curved part of the graph, is the liquid contributing to the pressure significantly? I assumed that pressure was only due to the gas. $\endgroup$
    – unstable
    Commented Oct 2 at 19:24
  • $\begingroup$ Yes, it was just vapor pressure, but then there's only incompressible liquid. $\endgroup$
    – Mithoron
    Commented Oct 2 at 19:44
  • 1
    $\begingroup$ The heating takes place at constant average specific volume. $\endgroup$
    – Chet Miller
    Commented Oct 2 at 20:20
  • 1
    $\begingroup$ When you fully fill a glass bottle with water, close it and start to heat it in water bath, what will happen? The glass will shatter, unless the cork, plug or screw cap shoots away first. If you check water coefficients of thermal dilation(bigger than for glass) and compressibility, you will not wonder. $\endgroup$
    – Poutnik
    Commented Oct 3 at 10:33

2 Answers 2

Reset to default
4
$\begingroup$

Ideal gas law applies to the vapor phase and also to supercritical fluids. Now, as the temperature is increased, number of particles per volume unit of vapor increases in addition to kinetic energy per particle. That's why the graph segment is curved rather than a straight line.

At $\pu{60^oC}$ , there is no longer vapor. The liquid has higher number of particles per volume unit. Extra heat provided still increases the average kinetic energy of particles by the same amount with every added temperature unit. Since the liquid cannot expand any more, the pressure goes up more rapidly with temperature. The more the ratio between the density of liquid and gas was, the larger the ratio between the rate of increase of pressure above vs below $\pu{60^oC}$. This effect would diminish if the fluid was near critical.

In fact, the essence of critical points is that approaching it from below causes the densities of liquid and vapor to get more and more similar until they are identical.

Improve this answer
$\endgroup$
3
  • $\begingroup$ So if it became a supercritical fluid, the graph would increase linearly at high temperatures? $\endgroup$
    – unstable
    Commented Oct 2 at 21:47
  • $\begingroup$ @unstable Yes. If you want to learn more, here is a good place to start: aerogel.org/?p=4 $\endgroup$
    – Paul Kolk
    Commented Oct 3 at 19:42
  • $\begingroup$ I edited my answer to include the basics about criticality. In case the readers didn't know. $\endgroup$
    – Paul Kolk
    Commented Oct 3 at 19:54
2
$\begingroup$

Gases are very compressible, liquids are not

Without getting into too much detail or edge cases (like those that arise with supercritical fluids) the issue is simple. Gases are very compressible, liquids are not.

In this case it seems likely that the situation is one where there is not much vapour to start with. So little that the (small) expansion of the volume of the liquid as temperature increases can eliminate the space occupied by the liquid. Once that point is hit there is no remaining space occupied by the most compressible component, the vapour. So there is no option to absorb extra pressure by shrinking the vapour volume. At this point, there is a sharp kink in the curve as the liquid is vastly less compressible than the vapour.

So the curve to the left of the kink reflects the relationship between pressure and volume of the vapour (and, vapour being very compressible, it is a gentle curve). But to the right, the curve reflects the far, far steeper relationship driven by the very low compressibility of the liquid.

More complex edge-cases to do with phase transitions are not relevant.

Improve this answer
$\endgroup$
7
  • $\begingroup$ By "volume of the vapour " is meant molar volume? or what? Total volume decreases with temperature not only by compression. $\endgroup$
    – Paul Kolk
    Commented Oct 5 at 18:39
  • $\begingroup$ Also, note that supercritical fluids are as compressible as gases. Furthermore, liquid completely filling the container does not need to be incompressible as long as its number density is much higher than vapor's at $60^o C$. $\endgroup$
    – Paul Kolk
    Commented Oct 5 at 19:01
  • $\begingroup$ @PaulKolk byt volume of vapour, I mean volume. Physical volume occupied by vapor. Volume not occupied by liquid. The key point is that small expansion of the volume of the liquid eliminates any space for vapour and the resulting pressure/temperature curve is purely about the liquid compressibility. $\endgroup$
    – matt_black
    Commented Oct 5 at 22:46
  • $\begingroup$ Compressibility is a property of vapor at a constant temperature. You wrote: "the curve to the left of the kink reflects the relationship between pressure and volume of the vapour". This is misleading and incomplete. The equilibrium vapor pressure varies with temperature because of two factors - 1)kinetic energy of particles and 2) number density. Both increase non-negligibly with temperature. The surface of the liquid is not something that simply compresses a gas. It also evaporates. Without understanding this, one cannot distinguish between options B and C. $\endgroup$
    – Paul Kolk
    Commented Oct 6 at 21:06
  • $\begingroup$ @PaulKolk Compressibility is a property of all gasses (including vapour). Not just at constant temperature. The simple obvious answer to this question involves the elimination of the vapour by expansion of the liquid. Yes, there are circumstances (but probably at higher temperatures) where supercritical fluids might matter but this question gives no hint that that is relevant. Don't overcomplicate what is intended to be simple question. $\endgroup$
    – matt_black
    Commented Oct 7 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.