1
$\begingroup$

In this link

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/phase.php

It says (in the latter part ) that for all combinations of Pressure and Temperature along line BC

the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

And C is the Critical Point. Similar thing has been said for other lines also

Now 100℃ (or boiling point of the liquid) will lie somewhere on BC

So according to the link rate of boiling of water to form gas is the same as the rate of condensation of gas to form water.

But I have read that at boiling point all the liquid changes to gas or there is no equilibrium at that point.

So why then it says- the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

I am not good at chemistry so this might not be a good question. But I was trying to understand phase Diagrams for the first time !

$\endgroup$
  • $\begingroup$ The phase diagram shown is an imaginary diagram. It is not for water, and there are no temperatures or pressures given on the axises so you can't pull 100 ℃ out of thin air. $\endgroup$ – MaxW Mar 4 '17 at 17:53
  • $\begingroup$ @MaxW No I don't refer to 100℃. I meant that he says that along line BC rate of boiling = rate of condensation. But I have read it can be rate of evaporation which can be equal to rate of condensation. At boiling point there is no equilibrium . Is there ? $\endgroup$ – Shashaank Mar 4 '17 at 18:42
  • $\begingroup$ The boiling point of a liquid is the temperature at which 1 atmosphere of vapor pressure is created. If you look at the lower diagram on the weblink you can see that there is noting particularity unique about the boiling point. The lower diagram shows that the substance has a vapor pressure in equilibrium with the liquid phase below the boiling point and above the boiling point (albeit at a higher pressure than 1 atmosphere). $\endgroup$ – MaxW Mar 4 '17 at 19:05
  • $\begingroup$ I got persnickety because carbon dioxide, for example, doesn't have a "boiling point." The critical pressure is above one atmosphere. $\endgroup$ – MaxW Mar 4 '17 at 19:09
1
$\begingroup$

In the phase diagram of a pure substance there are regions of a single phase e.g. solid, liquid or vapour. The boundary lines you refer to are where the substance exits as $2$ two phases, say solid and vapour in equilibrium with one another.

The Gibbs phase rule states that the number of degrees of freedom, F, which is the number of (intensive) variables that can be varied is given by $$F = C-P+2$$ where C is the number of components, $1$ for a pure substance, and P the number of phases, which is $1$ anywhere in the diagram except on a boundary line. Therefore the number of degrees of freedom is $2$, (except on a boundary line) which means that pressure and temperature can both be varied independently.

The boundary lines indicates the condition of P and T where there are two phases in equilibrium, say solid and vapour. As there are $2$ phases there is now only $1$ degree of freedom so that temperature and pressure are no longer able to be varied independently. Thus on the liquid /vapour curve, if the pressure is increased some vapour is condensed (heat is given out) and the temperature increases to a new value but one that that is still on the boundary line, i.e. vapour is still in equilibrium with the liquid. Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing. (Any point along the liquid/vapour boundary line can be considered to be boiling, but we define the normal boiling point to be the boiling temperature at $1$ atm pressure.)

At the triple point, where the three boundary curves meet, there are no degrees of freedom and this point is fixed. At and above the critical point the density of the liquid and vapour are the same and the liquid's meniscus disappears. The super-critical fluid looks very much like the swirling foam that one sees for a short while on pouring a bottle of Guinness into a glass.

The boundary lines can be calculated for a substance using the Clausius and Clausius-Clapeyron equations.

$\endgroup$
  • $\begingroup$ This seema to be better. Just confirm what I have understood is right or not. The boiling I am referring to , the normal one at say 100℃ is not an equilibrium ( IS THIS MUCH RIGHT) and it is not in equilibrium because Pressure is constant but temperature is rising ( but at 100℃ temperature is constant till boiling happens or non equilibrium exist.). But here both P & T are varying and hence equilibrium exist. That means I would have got the ' normal boiling point ( out of equilibrium ) , if I would have searched it inside the region of line BC towards the liquid side , not on the line BCitself $\endgroup$ – Shashaank Mar 4 '17 at 18:40
  • $\begingroup$ RE: "Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing." // Melting-freezing is only true for solid/liquid curve, not any curve on phase diagram. $\endgroup$ – MaxW Mar 4 '17 at 19:12
  • $\begingroup$ Re: "Any point along the liquid/vapour boundary line can be considered to be boiling..." // Not really "boiling." Boiling doesn't really have any meaning in phase diagram other than there is a temperature which is the "boiling point." The phase diagram is about adiabatic equilibrium. You can't have "boiling" without the addition of heat. $\endgroup$ – MaxW Mar 4 '17 at 19:18
  • $\begingroup$ @MaxW So what I said is that wrong ? $\endgroup$ – Shashaank Mar 4 '17 at 19:18
  • $\begingroup$ @MaxW Can you if you wish write an answer as to why the link says that rate of boiling is equal to rate of condensation along BC. I have no problem with lime BC but the usage " equal". Because I have read that at boiling equilibrium cannot exist so rates cannot be equal. That is why it is boiling. If you think there is any fault in the link please point it too. $\endgroup$ – Shashaank Mar 4 '17 at 19:29
-1
$\begingroup$

An excerpt from the link you provided:

The figure below shows an example of a phase diagram, which summarizes the effect of temperature and pressure on a substance in a closed container...

When heating a liquid in a rigid, close container, no boiling* will occur. The pressure (and therefore density) of the vapour will rise as the temperature is rised. If you carry this on for long enough, the temperature on the liquid surface will reach what is called a critical temperature, $T_{c}$. At $T_{c}$ the pressure of the system is called critical pressure, $p_{c}$. At this point the container will be filled with a supercritical fluid.

* Boiling, in this case, refers to free vapourization throughout the liquid.

$\endgroup$
  • $\begingroup$ Ok I understand Boiling here means the normal vapourization. So I was right that at boiling there is no equilibrium. Right ? And one thing more , I was thinking that even the pressure varies. At normal boiling pressure is constant. Even this might be a difference. $\endgroup$ – Shashaank Mar 4 '17 at 9:24
  • $\begingroup$ And one thing more. Throughout the line BC the liquid and gas coexist in equilibrium. Is saying equilibrium right ? But then what's special about the Critical Point . $\endgroup$ – Shashaank Mar 4 '17 at 9:26
  • $\begingroup$ First question: With an open container, the pressure of the vapour equals the external pressure. When two phases are at an equilibrium, the system temperature is not freely variable if the pressure is set, and vice versa. Second: Yes, it is correct. Throughout the BC line the liquid has some vapour pressure, which corresponds to some temperature. $\endgroup$ – Bdrs Mar 4 '17 at 9:40
  • $\begingroup$ Ok I got it. So in the closed box the temperature and pressure are dependent on each other just like you said and hence equilibrium . But what is special about critical temperature. Is it the maximum temperature at which they coexist ? $\endgroup$ – Shashaank Mar 4 '17 at 9:51
  • $\begingroup$ At the critical temperature the interface between the gas and the liquid disappears, and a supercritical liquid is formed. $\endgroup$ – Bdrs Mar 4 '17 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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