Pressure: Sealed bottle physics

[SteveA001]

TL;DR Summary

Sealed bottle physics

If a bottle is partially filled with hot water and sealed, together what some air, what happens to the pressure inside the bottle as it cools to room temperature? Assumptions: The temperature of the air in the bottle will initially be about the hot water temperature as the bottle is first sealed, and the pressure inside will also equal the outside air pressure as the bottle is sealed. I notice my vacuum flask does not appear to build up a noticeable vacuum as I open it after my coffee has cooled, and the same lack of vacuum appears with a zip bag filled with water as it cools (no balloons handy unfortunately). So on the one hand experiment appears to indicate that a vacuum is not created, but on the other hand, would the pressure inside the bottle not be at water vapour pressure, which lowers as the temperature drops, and which starts at ambient pressure?

 

[Arjan82]

It is my experience that when pouring hot liquid in a thermal flask and wait until it eventually cooled, there is a noticeably lower pressure when I open it. So maybe your vacuum flask wasn't as sealed as you hoped? In theory you can also have some outgassing of the liquid due to the lower pressure which undoes the lower pressure, but I don't know how strong that effect can be since boiling usually removes a lot of dissolved gasses.

There is not an actual vacuum of course after cooling. The lower temperature makes the density of air to want to become larger again, but since that's not possible you get a lower pressure instead. This does explain why the zip-bag doesn't work, the zip-bag can change shape and thus accommodate the lower density by decreasing the volume of the air.

The water vapor pressure has nothing to do with this. It is the temperature and pressure at which water vapor condenses again. Unless, of course, a significant part of air was replaced by water vapor which then condenses. That actually would generate quite a strong 'vacuum'.

(For me a 'vacuum' is still equivalent to 'no air at all' but apparently it is often used to mean 'pressure lower than atmospheric')

 

[Chestermiller]

It is my experience that when pouring hot liquid in a thermal flask and wait until it eventually cooled, there is a noticeably lower pressure when I open it. So maybe your vacuum flask wasn't as sealed as you hoped? In theory you can also have some outgassing of the liquid due to the lower pressure which undoes the lower pressure, but I don't know how strong that effect can be since boiling usually removes a lot of dissolved gasses.

There is not an actual vacuum of course after cooling. The lower temperature makes the density of air to want to become larger again, but since that's not possible you get a lower pressure instead. This does explain why the zip-bag doesn't work, the zip-bag can change shape and thus accommodate the lower density by decreasing the volume of the air.

The water vapor pressure has nothing to do with this.

This is not correct. When the lid is fastened, the total pressure in the container is 1 atm., and it is comprised of the water vapor in the head space at a partial pressure equal to its equilibrium vapor pressure at the temperature of the hot liquid, and air in the head space at a partial pressure of 1 atm. minus the partial pressure of the water vapor. When the system cools, most of the water vapor condenses, leaving water vapor in the head space at a much lower partial pressure, equal to the equilibrium vapor pressure of the much cooler liquid in the container. The pressure in the head space also decreases a little because the partial pressure of the air decreases a little. But, the dominant effect is water vapor condensing from the gas phase. This is the basic principle behind vacuum sealing of items like soup cans and preservatives.

 

[SteveA001]

Ah, yes. I have just tried it with a plastic bottle and it did indeed deform as it cooled, indicating a pressure drop, as you say. It appears I was distracted by my vacuum flask, strange that it doesn't appear to pop or offer resistance when opened when cooled. My original interest was in the magnitude of the vacuum created in such a set up? How would it be calculated? I hit upon differences in vapour pressure, before being distracted by my flask, but you say that is not the way to go anyway?

 

[Chestermiller]

Ah, yes. I have just tried it with a plastic bottle and it did indeed deform as it cooled, indicating a pressure drop, as you say. It appears I was distracted by my vacuum flask, strange that it doesn't appear to pop or offer resistance when opened when cooled. My original interest was in the magnitude of the vacuum created in such a set up? How would it be calculated? I hit upon differences in vapour pressure, before being distracted by my flask, but you say that is not the way to go anyway?

I do say that the vapor pressure change is key to this. Why don't I help you get started in modeling this. Suppose I have a rigid bottle of total volume V an a head space of v. Assume that air does not significantly dissolve in the liquid water. Let the initial temperature be 160 F. What is the specific volume of liquid water at this temperature? What is the equilibrium vapor pressure of water at this temperature? What is the specific volume of water vapor at this temperature? In terms of V, what is the mass of water in the container? What is the partial pressure of air in the head space?

 

[JT Smith]

When the lid is fastened, the total pressure in the container is 1 atm., and it is comprised of the water vapor in the head space at a partial pressure equal to its equilibrium vapor pressure at the temperature of the hot liquid, and air in the head space at a partial pressure of 1 atm. minus the partial pressure of the water vapor.

I think the assumption that the water vapor partial pressure is at equilibrium isn't necessarily the case. It likely depends on the geometry of the bottle and how long between when you pour in the liquid and seal it but in general I think it would be something less.

 

[A.T.]

It is my experience that when pouring hot liquid in a thermal flask and wait until it eventually cooled, there is a noticeably lower pressure when I open it.

This is also how some sealing methods in home canning work.
https://en.wikipedia.org/wiki/Weck_jar

 

[Chestermiller]

I think the assumption that the water vapor partial pressure is at equilibrium isn't necessarily the case. It likely depends on the geometry of the bottle and how long between when you pour in the liquid and seal it but in general I think it would be something less.

You think that? Have you done any calculations to quantify this?

 

[gmax137]

to @SteveA001 , I really encourage you to work with @Chestermiller as offered in Post#5

 

[SteveA001]

Ah, yes. I have just tried it with a plastic bottle and it did indeed deform as it cooled, indicating a pressure drop as you say. It appears I was distracted by my vacuum flask performance. Strange that it doesn't appear to pop or offer resistance when opened. My original inquiry, before was the magnitude of the vacuum How would it be calculated? I hit upon differences in vapour pressure, before being distracted by the flask, but you say that is not the way to go anyway?

I do say that the vapor pressure change is key to this. Why don't I help you get started in modeling this. Suppose I have a rigid bottle of total volume V an a head space of v. Assume that air does not significantly dissolve in the liquid water. Let the initial temperature be 160 F. What is the specific volume of liquid water at this temperature? What is the equilibrium vapor pressure of water at this temperature? What is the specific volume of water vapor at this temperature? In terms of V, what is the mass of water in the container? What is the partial pressure of air in the head space?

What is the specific volume of liquid water at this temperature? 1000 kg/m^3
What is the equilibrium vapor pressure of water at this [114 C] temperature? 166,000 Pa
What is the specific volume of water vapor at this temperature? 1 kg/m^3
In terms of V, what is the mass of water in the container? 1000(V-v) kg
What is the partial pressure of air in the head space? Presumably it rises with temperature, as does the proportion of water vapour?

 

[Chestermiller]

Ah, yes. I have just tried it with a plastic bottle and it did indeed deform as it cooled, indicating a pressure drop as you say. It appears I was distracted by my vacuum flask performance. Strange that it doesn't appear to pop or offer resistance when opened. My original inquiry, before was the magnitude of the vacuum How would it be calculated? I hit upon differences in vapour pressure, before being distracted by the flask, but you say that is not the way to go anyway?What is the specific volume of liquid water at this temperature? 1000 kg/m^3
What is the equilibrium vapor pressure of water at this [114 C] temperature? 166,000 Pa

160 F is 71.1 C. The equilibrium vapor pressure of water at this temperature is 32700 Pa

What is the specific volume of water vapor at this temperature? 1 kg/m^3

From the steam tables, 4.84 kg/m^3

In terms of V, what is the mass of water in the container? 1000(V-v) kg

$$m=\frac{(V-v)}{0.001}+\frac{v }{4.84}$$

What is the partial pressure of air in the head space? Presumably it rises with temperature, as does the proportion of water vapour?

p=100000-32700=67300 Pa

OK so far?

 

[SteveA001]

160 F is 71.1 C. The equilibrium vapor pressure of water at this temperature is 32700 Pa

D'oh, I googled rather than calculating.

From the steam tables, 4.84 kg/m^3

I found a table online(https://www.thermexcel.com/english/tables/vap_eau.htm), added an exponential trend line in Excel and got this formula: Specific volume (steam) = 1.6888*(AbsolutePressureInBar)^-0.942. This agrees with your value for 0.327 bar.

$$m=\frac{(V-v)}{0.001}+ \frac{v}{4.84}$$

I see.

p=100000-32700=67300 Pa

OK so far?

Well, I guess given that you calculated masses in the previous paragraph the answer isn't as simple as subtracting the above from atmospheric pressure, leaving a third of a bar. So my puzzle now is to relate mass to pressure.

 

[Chestermiller]

D'oh, I googled rather than calculating.

I found a table online(https://www.thermexcel.com/english/tables/vap_eau.htm), added an exponential trend line in Excel and got this formula: Specific volume (steam) = 1.6888*(AbsolutePressureInBar)^-0.942. This agrees with your value for 0.327 bar.

I see.

Well, I guess given that you calculated masses in the previous paragraph the answer isn't as simple as subtracting the above from atmospheric pressure, leaving a third of a bar. So my puzzle now is to relate mass to pressure.

If the total pressure is 1 bar = 100000 Pa, and the water vapor partial pressure is 32700 Pa, the partial pressure of the air is 67300 Pa.

You now cool the container and its contents to 20 C. What is the equilibrium vapor pressure of water at 20 C? Assuming the changes in the volume of liquid water is negligible when part of the water condenses, what is the partial pressure of the air when it is cooled from 71.1 C to 20 C? What is the total pressure at 20 C?

 

[JT Smith]

You think that? Have you done any calculations to quantify this?

No, wouldn't know how to do that.

But I did a couple of kitchen experiments to test this a few years ago. I poured near-boiling water into a small jar, placed the lid loosely on top, and then after 10-15 seconds I screwed the lid tight. When cooled to room temperature I inverted the jar in a basin of water and unscrewed the lid, allowing water to enter because of the vacuum created. Then I sealed it again, dried it off, and weighed it. I assumed that the initial air temperature was initially the same as the water temperature, which is probably not quite true. And I figured that the water vapor partial pressure was effectively zero at room temperature.

The water temperature in the jar was initially about 90°C. At that temperature equilibrium vapor pressure is 0.7bar. But the amount of water that entered the inverted jar suggested that the water vapor partial pressure was actually about 0.5bar. In another test where I screwed the lid down immediately it was only 0.4bar.

 

[Chestermiller]

No, wouldn't know how to do that.

But I did a couple of kitchen experiments to test this a few years ago. I poured near-boiling water into a small jar, placed the lid loosely on top, and then after 10-15 seconds I screwed the lid tight. When cooled to room temperature I inverted the jar in a basin of water and unscrewed the lid, allowing water to enter because of the vacuum created. Then I sealed it again, dried it off, and weighed it. I assumed that the initial air temperature was initially the same as the water temperature, which is probably not quite true. And I figured that the water vapor partial pressure was effectively zero at room temperature.

The water temperature in the jar was initially about 90°C. At that temperature equilibrium vapor pressure is 0.7bar. But the amount of water that entered the inverted jar suggested that the water vapor partial pressure was actually about 0.5bar. In another test where I screwed the lid down immediately it was only 0.4bar.

The diffusivity of water vapor in air is about 0.25 cm^2/sec. If the head space is about 1 cm thick, it would take about 4-10 seconds for the water vapor to equilibrate (and purge the corresponding amount of air from the head space). In your experiment, while the lid was being set in place, maybe the lid was put in place very rapidly and the air partial pressure was closer to 1 bar to start, and when the water vapor equilibrated, the total pressure in the bottle was higher than 1 bar. But, putting the lid on gradually, as is usually the situation in canning operations, would virtually guarantee that the partial pressure of the air in the head space is 1 bar minus the equilibrium vapor pressure of the water.

 

[SteveA001]

You now cool the container and its contents to 20 C. What is the equilibrium vapor pressure of water at 20 C?

~2350 Pa

Assuming the changes in the volume of liquid water is negligible when part of the water condenses, what is the partial pressure of the air when it is cooled from 71.1 C to 20 C?

I'm still coming to grips with the concept of partial pressures, having only seen it in relation to parts of air rather than to the entire air. Based on the earlier example partialPressureOfAirInHeadspace = totalPressure - 2350? Is it something that can be found from lookup tables?

What is the total pressure at 20 C?

partialPressureOfAirInHeadspace + 2350?

 

[Chestermiller]

~2350 Pa

I'm still coming to grips with the concept of partial pressures, having only seen it in relation to parts of air rather than to the entire air. Based on the earlier example partialPressureOfAirInHeadspace = totalPressure - 2350? Is it something that can be found from lookup tables?

partialPressureOfAirInHeadspace + 2350?

No. Based on the ideal gas law, partial pressure of air in cooled sealed container = $67300\frac{273+20}{273+71}=57300\ Pa$. So the total pressure in the quenched container is 57300 + 2350 = 59700 Pa. This is 60% of the 1 bar pressure that was present when the lid went on at 71 C. Quite a vacuum!!

 

[SteveA001]

Sneaky, p1/t1= p2/t2. So if the initial temperature were 95C the vacuum would be 15% of atmospheric pressure. And at 100 C it would be 1 atmosphere of vacuum.

But did we use the specific volume of water vapor (4.84 kg/m^3) and the mass calculated earlier?

 

[Chestermiller]

Sneaky, p1/t1= p2/t2. So if the initial temperature were 95C the vacuum would be 15% of atmospheric pressure. And at 100 C it would be 1 atmosphere of vacuum.

But did we use the specific volume of water vapor (4.84 kg/m^3) and the mass calculated earlier?

We would take those into account if we wanted to include the effect on the air partial pressure of the change in liquid volume as a result of condensation.

 

[SteveA001]

We would take those into account if we wanted to include the effect on the air partial pressure of the change in liquid volume as a result of condensation.

Oh, that would be a small effect then.

Thanks for your help.

Now that the answer is quantified, lots of everyday observations can be explained. For example why vacuum flasks and hot water bottles have hard screw tops. I am also testing a preheated vacuum flask with large airspace and little liquid to see if a near 1-atmosphere vacuum in a 0.5-litre flask can make a pop when cooled and opened. I wonder also whether these equations could be tested by seeing how much water is sucked into the flask submerged in water as it is opened.

 

[JT Smith]

The diffusivity of water vapor in air is about 0.25 cm^2/sec. If the head space is about 1 cm thick, it would take about 4-10 seconds for the water vapor to equilibrate (and purge the corresponding amount of air from the head space). In your experiment, while the lid was being set in place, maybe the lid was put in place very rapidly and the air partial pressure was closer to 1 bar to start, and when the water vapor equilibrated, the total pressure in the bottle was higher than 1 bar. But, putting the lid on gradually, as is usually the situation in canning operations, would virtually guarantee that the partial pressure of the air in the head space is 1 bar minus the equilibrium vapor pressure of the water.

No, the lid was placed gently on the jar.

Do the experiment yourself at home. All you need is a jar and a scale and a sink. Plus the willingness to waste thirty minutes or so which (like me) you clearly have since you post here :-)

 

[Chestermiller]

OK. Here is the detailed model and how it plays out.

Let $v_L(T)$ = specific volume of saturated liquid water at temperature T

$v_V(T)$ = specific volume of saturated water vapor at temperature T

$P^*(T)$ = equilibrium vapor pressure of water at temperature T

V = volume of container

v = volume of head space at initial temperature $T_H$

$T_H$ = Initial high temperature when lid is put in place

$T_C$ = final cold temperature

$p_{atm}$ = atmospheric pressure

INITIAL HIGH TEMPERATURE

partial pressure of water vapor = $P^*(T_H)$

partial pressure of air in heat space = p_{atm}-$P^*(T_H)$

mass of water in can = m = $\frac{V-v}{v_L(T_H)}+\frac{v}{v_V(T_H)}$

FINAL LOW TEMPERATURE
$$m(1-x)v_L(T_C)+mxv_V(T_C)=V$$where x = mass fraction vapor in final state. Solving for x gives: $$x=\frac{\frac{V}{m}-v_L(T_C)}{v_V(T_C)-v_L(T_C)}$$

Volume of gas phase in final cold state = $mxv_V(T_C)=mv_V(T_C)\frac{\frac{V}{m}-v_L(T_C)}{v_V(T_C)-v_L(T_C)}$

Partial pressure of air in final state = $[p_{atm}-P^*(T_H)]\frac{T_C}{T_H}\frac{v}{mv_V(T_C)\frac{\frac{V}{m}-v_L(T_C)}{v_V(T_C)-v_L(T_C)}}$

Partial pressure of water vapor in final state = $P^*(T_C)$

 

[SteveA001]

Nice work. I've run some numbers through it, in Excel. Using the same temperatures as before (71C to 20C) for a flask 98% full of water and then 2% full, producing answers of 30% of atmospheric pressure and 60% (the second one being same as before, when not taking condensation into account). Basically condensation can't be ignored when the flask is full, which is counter-intuitive?

So how to maximise the pop when opening a container for a given temperature. One extreme gives high vacuum with little volume, and the other a high volume of little vacuum. Hmm.

 

[Chestermiller]

Nice work. I've run some numbers through it, in Excel. Using the same temperatures as before (71C to 20C) for a flask 98% full of water and then 2% full, producing answers of 30% of atmospheric pressure and 60% (the second one being same as before, when not taking condensation into account). Basically condensation can't be ignored when the flask is full, which is counter-intuitive?

Not really. Condensation compresses the air.

So how to maximise the pop when opening a container for a given temperature. One extreme gives high vacuum with little volume, and the other a high volume of little vacuum. Hmm.

 

[JT Smith]

v = volume of head space at initial temperature $T_H$

Did you mean $v$ here instead of v?

 

[Chestermiller]

Did you $v$ here instead of v?

They're the same.

 

[JT Smith]

I redid my little kitchen experiment this morning, being more careful to allow enough time for the vapor to equilibrate. I left the lid off for about 15 seconds and then placed it on the jar with just a crack open for another 15 seconds. Then I sealed it.

When I originally did this I wasn't being rigorous with the calculations as I was mostly interested in the final result: how much pressure was in the container after it cooled. One thing I didn't include was the fact that the liquid would contract by a small but significant amount.

So using your detailed model I crunched the numbers. Can you check to see if I did it right? I'd appreciate it.

I got a total pressure of 0.23 bar using the following values.

V = 237.9 ml
$v$ = 39.5 ml
$T_H$ = 90°C
$T_C$ = 15°C
$p_{atm}$ = 1.004 bar

$v_L(T_H)$ = 1.036 ml/g
$v_V(T_H)$ = 2359.1 ml/g
$P^*(T_H)$ = 0.702 bar

$v_L(T_C)$ = 1.000 ml/g
$v_V(T_C)$ = 77875 ml/g
$P^*(T_C)$ = 0.017 bar

 

[SteveA001]

I redid my little kitchen experiment this morning, being more careful to allow enough time for the vapor to equilibrate. I left the lid off for about 15 seconds and then placed it on the jar with just a crack open for another 15 seconds. Then I sealed it.

When I originally did this I wasn't being rigorous with the calculations as I was mostly interested in the final result: how much pressure was in the container after it cooled. One thing I didn't include was the fact that the liquid would contract by a small but significant amount.

So using your detailed model I crunched the numbers. Can you check to see if I did it right? I'd appreciate it.

I got a total pressure of 0.23 bar using the following values.

V = 237.9 ml
$v$ = 39.5 ml
$T_H$ = 90°C
$T_C$ = 15°C
$p_{atm}$ = 1.004 bar

$v_L(T_H)$ = 1.036 ml/g
$v_V(T_H)$ = 2359.1 ml/g
$P^*(T_H)$ = 0.702 bar

$v_L(T_C)$ = 1.000 ml/g
$v_V(T_C)$ = 77875 ml/g
$P^*(T_C)$ = 0.017 bar

That's roughly what I get using your numbers. Then I think you add 'Partial pressure of water vapor in final state' to 'Partial pressure of air in final state' to get 'Total pressure in the quenched container', making 22800 Pa, or 22.7 % of atmospheric pressure.

An Excel sheet is attached, if this site permits.

 

[JT Smith]

Thank you. I'm glad I didn't screw up the math.

My experimental results are still at odds with the model however. I calculate an initial water vapor partial pressure of about 0.5 bar instead 0.7 bar, essentially the same as what I'd determined with a much rougher model. So either I'm making some poor assumption, screwing up the actual work, or your model is an idealization that doesn't always hold in the real world.

You are an esteemed scientist and I'm just some guy on the internet so the good money would be placed on you being right. I'm often wrong. I just can't see what I'm screwing up.

 

[JT Smith]

One other thing: I noticed that the headspace temperature wasn't the same as the water. Even when I sealed up the opening with plastic and put an oven mitt over it to insulate it was still about 10°C lower. When the vapor and liquid are not in thermal equilibrium how do you determine the vapor pressure?

I recall reading a paper years ago where the author couldn't fit the data in his system using the water temperature in his model. But using the vapor temperature gave a great fit. He never justified that choice via first principles, he just kind of went with it. If I do the same it seems to work for me too. That is, if I use the saturated pressure for the headspace temperature instead of the water temperature the model is a pretty close match to what I measured. That is, the saturated vapor pressure at 80°C is roughly 0.5 bar.