Relationship between pressure and temperature for liquids

[sss1]

Homework Statement

Is there a relationship between pressure and temperature for liquids? I know for most gassed they obey the ideal gas law, PV=nRT, which means that pressure is directly proportional to temperature. Is there also such a proportionality for liquids ( I know the gas law doesn’t apply for liquids but is there some other relationship?). And with some googling I saw the experiment where you place some liquid into a bell jar and remove the air and some time after the liquid starts to boil. Is that because the temperature of the liquid was increased or it just became easier for the surface liquid to escape because now there’s less force pushing down on it and temperature hasn’t actually changed?

Relevant Equations

NA

^^

 

[DrClaude]

Homework Statement: Is there a relationship between pressure and temperature for liquids? I know for most gassed they obey the ideal gas law, PV=nRT, which means that pressure is directly proportional to temperature.

at constant volume and number of particles (or density)! You have to remember that you are looking a the relation between more than two quantities.

Is there also such a proportionality for liquids ( I know the gas law doesn’t apply for liquids but is there some other relationship?).

The ideal gas law works well because the interaction between gas particles is most often negligible, so all gases behave the same. Interactions are very important in liquids, so there is no such simple, universal law.

And with some googling I saw the experiment where you place some liquid into a bell jar and remove the air and some time after the liquid starts to boil. Is that because the temperature of the liquid was increased or it just became easier for the surface liquid to escape because now there’s less force pushing down on it and temperature hasn’t actually changed?

The lower the atmospheric pressure, the lower the temperature at which a liquid will boil. Your explanation as to why that is basically correct. See
https://chem.libretexts.org/Bookshe...er/States_of_Matter/Phase_Transitions/Boiling

 

[sss1]

at constant volume and number of particles (or density)! You have to remember that you are looking a the relation between more than two quantities.

Yes, that's true, I wasn't being specific enough.

The ideal gas law works well because the interaction between gas particles is most often negligible, so all gases behave the same. Interactions are very important in liquids, so there is no such simple, universal law.

That makes sense!

Your explanation as to why that is basically correct.

Explanation as in it just became easier for the surface liquid to escape because now there’s less force pushing down on it and temperature hasn’t actually changed?

 

[Chestermiller]

For a single phase liquid, $$\frac{1}{\rho}d\rho=-\alpha dT+\frac{dP}{B}$$ahere $\rho$ is the density, $\alpha$ ia the volumetric coefficient of thermal expansion, T is the temperature, P is the pressure, and B is the bulk modulus.

 

[jbriggs444]

Explanation as in it just became easier for the surface liquid to escape because now there’s less force pushing down on it and temperature hasn’t actually changed?

"Boiling" is not so much about surface liquid escaping as it is about the behavior underneath the surface.

If you remove enough pressure, liquid beneath the surface can evaporate into newly formed bubbles. Instead of evaporation taking place only at the surface, you have the potential for "evaporation" everywhere.
Similar to a shaken bottle of soda when the pressure is released.

What I would expect for a beaker of water in a bell jar that is being evacuated would be for the surface water to begin evaporating first. Since the partial pressure of water vapor in the chamber is less than the vapor pressure of the water in the beaker, evaporation will take place.

One effect of the evaporation would be to cool the surface of the water.

As the total pressure [not just water vapor pressure] in the bell jar decreases below the vapor pressure of the submerged water, boiling becomes possible beneath the surface. This would take place in preference to plain old evaporation at the surface because the surface had been chilled by the prior evaporation.

If the beaker is well insulated, the result should be boiling water with bits of ice floating in it and then just the remaining ice sublimating into the vacuum.

 

[sss1]

As the total pressure [not just water vapor pressure] in the bell jar decreases below the vapor pressure of the submerged water, boiling becomes possible beneath the surface. This would take place in preference to plain old evaporation at the surface because the surface had been chilled by the prior evaporation.

That's interesting, I've never heard of the term vapour pressure before.
How does that come into play here?
When you remove air inside the bell jar, it becomes easier for the water molecules to evaporate? And thus more vapour pressure? With some googling I found that liquid starts to boil when the vapour pressure is equal to the atmospheric pressure? Why is that?

 

[haruspex]

That's interesting, I've never heard of the term vapour pressure before.
How does that come into play here?
When you remove air inside the bell jar, it becomes easier for the water molecules to evaporate? And thus more vapour pressure? With some googling I found that liquid starts to boil when the vapour pressure is equal to the atmospheric pressure? Why is that?

The pressure of a gas mixture is the sum of the pressures ("partial pressures ") of the constituents. Each constituent behaves pretty much as though the others are not there.
For water vapour, its partial pressure is often called its vapour pressure. It is limited by the temperature. The higher the temperature the higher the vapour pressure can go (up to saturation vapour pressure, SVP).
As long as the VP is below the SVP, evaporation will take place from the surface, but its rate is limited by the exposure area. As @jbriggs444 notes, the molecules which evaporate are the most energetic, so their evaporation lowers the temperature at the surface. A flow of energy from further down (mostly convection) maintains it.

Boiling point is when SVP equals ambient total pressure. As @jbriggs444 wrote, at this point bubbles can form within the water. That greatly increases surface area and evaporation proceeds much faster, so long as more heat is supplied to maintain the temperature.

This LibreText description is misleading. It fails to distinguish between boiling and evaporation. Defining boiling as what happens at boiling point is particularly unenlightening. Couldn't find a way to give feedback.

 

[sss1]

Cool, but when the beaker of water is placed into the bell jar and air is being removed, the water begins to boil very quickly, which means that the bell jar speeds up the evaporation rate so that the VP reaches SVP? is that because now due to reduced air pressure, the temperature of water is essentially increased? This sounds counterintuitive but I googled the definition of temperature and it said the average kinetic energy, which I guess makes sense because with reduced suppression from air pressure then the molecules can move faster and as a result more evaporation occurs?

 

[jbriggs444]

Cool, but when the beaker of water is placed into the bell jar and air is being removed, the water begins to boil very quickly, which means that the bell jar speeds up the evaporation rate so that the VP reaches SVP?

It is difficult to follow what you are saying here.

If the vapor pressure reaches the saturated vapor pressure then evaporation stops. An equilibrium is reached where vapor molecules are condensing into the liquid at the same rate that liquid molecules are evaporating into the surrounding volume.

The vacuum pump attached to the bell jar continuously removes air and vapor so that saturated vapor pressure is not reached. The fact that boiling is occurring is evidence that total pressure must be less than the saturated vapor pressure (for the liquid at its current temperature).

is that because now due to reduced air pressure, the temperature of water is essentially increased?

No. The temperature does not increase. It decreases.

It takes energy to free a water molecule from the liquid water so that it escapes into the surrounding volume. This energy is supplied by the water molecule's kinetic energy. If it has enough energy and it located at the surface of the water, is escapes and has its energy reduced because of the escape. This energy loss is the "latent heat of vaporization".

The liquid water has lost a high energy molecule. So its average kinetic energy decreases.

The surrounding volume has picked up a molecule with reduced energy. So its average kinetic energy usually decreases as well.

This sounds counterintuitive but I googled the definition of temperature and it said the average kinetic energy, which I guess makes sense because with reduced suppression from air pressure then the molecules can move faster and as a result more evaporation occurs?

No. The reduced pressure does essentially nothing to alter the energy of the water molecules.

The increased evaporation is, as @haruspex points out, due to the dramatically increased surface area made available due to spontaneously forming bubbles beneath the surface.

Bubble formation is made possible since air pressure (and, hence, water pressure) is not pushing back on tiny bubbles to prevent them from expanding.

 

[haruspex]

Bubble formation is made possible since air pressure (and, hence, water pressure) is not pushing back on tiny bubbles to prevent them from expanding.

Small clarification…

air pressure (and, hence, water pressure) is not pushing back sufficiently on tiny bubbles to prevent them from expanding.

 

[Lnewqban]

Homework Statement: Is there a relationship between pressure and temperature for liquids?

Inducing boiling by gradually reducing the pressure of any gas on top of a liquid is difficult.
More and more vapor jumps out to work against any vacuum that a machine could create.

Boiling can appear and disappear almost instantaneously for cases in which the liquid is in direct contact with a solid that suddenly creates very low pressure within the mass of liquid by moving away from the (non-compressible/extensible) interphase.

Please, see:
https://en.wikipedia.org/wiki/Phase_diagram#3-dimensional_diagrams

https://en.wikipedia.org/wiki/Cavitation

 

[sss1]

It is difficult to follow what you are saying here.

If the vapor pressure reaches the saturated vapor pressure then evaporation stops. An equilibrium is reached where vapor molecules are condensing into the liquid at the same rate that liquid molecules are evaporating into the surrounding volume.

The vacuum pump attached to the bell jar continuously removes air and vapor so that saturated vapor pressure is not reached. The fact that boiling is occurring is evidence that total pressure must be less than the saturated vapor pressure (for the liquid at its current temperature).No. The temperature does not increase. It decreases.

It takes energy to free a water molecule from the liquid water so that it escapes into the surrounding volume. This energy is supplied by the water molecule's kinetic energy. If it has enough energy and it located at the surface of the water, is escapes and has its energy reduced because of the escape. This energy loss is the "latent heat of vaporization".

The liquid water has lost a high energy molecule. So its average kinetic energy decreases.

The surrounding volume has picked up a molecule with reduced energy. So its average kinetic energy usually decreases as well.No. The reduced pressure does essentially nothing to alter the energy of the water molecules.

The increased evaporation is, as @haruspex points out, due to the dramatically increased surface area made available due to spontaneously forming bubbles beneath the surface.

Bubble formation is made possible since air pressure (and, hence, water pressure) is not pushing back on tiny bubbles to prevent them from expanding.

Alright, so evaporation occurs at surface where particular water molecules at the surface has enough energy to escape, lowers the temperature of the water. Air and vapour is removed by the bell jar so SVP( a point where the rate of evaporation and condensation is equal) is never reached. Which makes bubble formation possible since now the air pressure is not enough to stop them from forming. And so evaporation rate is increased.
Then since SVP is never reached, does that mean the water in the breaker is not boiling?

 

[jbriggs444]

Then since SVP is never reached, does that mean the water in the breaker is not boiling?

Boiling occurs when the saturated vapor pressure that the liquid would have exceeds the total pressure that the surrounding volume does have.

The water in beaker is boiling.

Consider the situation at the wall of a small, stable vapor bubble in the water exactly at the boiling point:

The bubble is filled with nothing but vapor. That vapor is at the saturated vapor pressure for the fluid at the current temperature. Because the bubble contains nothing but vapor, this is also the total pressure within the bubble. The surrounding fluid is also under this same pressure. The bubble is in equilibrium between the water encroaching and shrinking it or the gas expanding and growing it. So the bubble neither grows nor shrinks.

Because the vapor in the bubble is exactly at SVP, the rate of evaporation and of condensation are equal. The bubble content is in equilibrium. There is no net evaporation into the bubble or condensation from the bubble. So the content of the bubble does not change over time.

If the ambient pressure is reduced below this equilibrium point ("the boiling point"), then the equilibrium will be broken. The bubble (at SVP) will expand within the liquid (now below SVP). This will result in a reduced pressure within the bubble. The rate of evaporation will then exceed the rate of condensation. There will be a net flow of vapor into the bubble. This process will continus with the bubble expanding and filling with newly evaporated vapor.

Convection will cause the bubble to float to the surface and pop while new bubbles are forming below. This is the familiar "rolling boil" that one sees in a pot on the stove.

 

[mtcmech1]

For a single phase liquid, $$\frac{1}{\rho}d\rho=-\alpha dT+\frac{dP}{B}$$ahere $\rho$ is the density, $\alpha$ ia the volumetric coefficient of thermal expansion, T is the temperature, P is the pressure, and B is the bulk modulus.

I was doing some research on a general approximation for the relationship between a liquid's temperature, pressure and density and came across the attached response of yours at some point. This is an absolute godsend of an equation, but I was hoping to get some background as to where it came from (most importantly) and the context in which it was used.

Cheers!
Mech1

 

[jbriggs444]

For a single phase liquid, $$\frac{1}{\rho}d\rho=-\alpha dT+\frac{dP}{B}$$ahere $\rho$ is the density, $\alpha$ ia the volumetric coefficient of thermal expansion, T is the temperature, P is the pressure, and B is the bulk modulus.

I was doing some research on a general approximation for the relationship between a liquid's temperature, pressure and density and came across the attached response of yours at some point. This is an absolute godsend of an equation, but I was hoping to get some background as to where it came from (most importantly) and the context in which it was used.

With some algebra, it is pretty much true by definition. It is a consequence of the fact that the volume changes from temperature and from pressure are independent. The effects superimpose so that the combined [infinitesimal] effect is the sum of the individual [infinitesimal] effects.

The bulk modulus $B$ is, by definition, the ratio of infinitesimal pressure change to the resulting infinitesimal volume change per unit volume.$$B = -V \frac{dP}{dV}$$The volumetric coefficient of thermal expansion $\alpha$ is pretty much the same thing, except that it is for temperature rather than pressure and the scale is inverted.$$\frac{1}{\alpha} = V \frac{dT}{dV}$$We can do a little bit of algebra, combine the two effects together and write that$$dV = -\frac{V dP}{B} +V\alpha dT$$As we know, density $\rho$ is given by mass over volume:$$\rho = \frac{m}{V}$$If we evaluate the first derivative of density with respect to volume, we will be using the rule for $x^{-1}$: $d(kx^n) = kn x^{n-1} dx$. So we get $$d \rho = d(\frac{m}{V}) = -\frac{m}{V^2}dV = -\frac{\rho}{V} dV$$Dividing by $\rho$ and substituting our previous expression for $dV$ we get:$$\frac{d\rho}{\rho} = - \frac{-V\frac{dP}{B} + \alpha VdT}{V} = \frac{dP}{B} - \alpha dT$$

 

[mtcmech1]

Thank you very much, that is very helpful. I was unsure about the ability to superimpose the components and out of precaution did not.

Thanks!
Mech1

 

[jbriggs444]

I was unsure about the ability to superimpose the components and out of precaution did not.

One note of caution that you probably do not need...

For infinitesimal changes in temperature and pressure, the resulting infinitesimal changes in volume add.

For larger changes in temperature and pressure life is not as simple. It is conceivable that the bulk modulus and the coefficient of thermal expansion change. Even if they do not change, One would want to treat the two sorts of expansion as percentage increases and multiply the scale factors rather than add the deltas. [That bit is the same thing as knowing that a 10% increase and a 10% increase delivers a 21% result]

 

[Chestermiller]

Thank you very much, that is very helpful. I was unsure about the ability to superimpose the components and out of precaution did not.

Thanks!
Mech1

It all follows from the laws of partial differentiation. The specific volume v is a function of temperature and pressure: $v=v(T,P)$. So $$d\ln{v}=\frac{\partial \ln{v}}{\partial T}dT+\frac{\partial \ln{v}}{\partial P}dP$$The specific volume is the inverse of the density $\rho$: $$d\ln{v}=-d\ln{\rho}$$