What is the pressure in a closed container dependent on?

I think we are comparing apples and oranges. Upon heating, the stretched rubber band... is now less stretched!In summary, the discussion revolved around the pressure inside an elastic container filled with water and its dependence on temperature and other factors. The participants conducted experiments and shared their observations, with some hypothesizing that the pressure increase could be due to the elasticity of the balloon material and thermal expansion, while others suggested it could be due to the vapor pressure of the water. Overall, it was agreed that further experiments were needed to fully understand the factors affecting the pressure inside the container.
  • #1
DarekGTR
Hello!

I have a problem : I have an elastic container (I call it "balloon"). I fill it 100% with water (no air bubbles) and close it. There is a pressure sensor connected to the balloon to measure the pressure inside it.

I would like to ask You : what factors have influence on the pressure inside the balloon, when it is lying on the table and no one is moving this?

I noticed that the pressure sensor says, that the pressure increases, when I heat the balloon a little (just a few C degrees) with a hair dryer. I was expecting that, but there is on thing I don't understand and I would like to ask about: When I left the balloon with water (and of course with sensor) for the night, the next day in the morning the sensor says that the pressure increased during night. But the temperature at night didn't increased - it decreased. I don't fully understand this, that is I would like to ask: what else could the pressure inside could be dependent on?

Thanks for help!

Darek
 
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  • #2
If the balloon material got colder, its elastic properties perhaps increased, i.e. its elasticity/stretchability decreased, raising the pressure inside the balloon. ## \\ ## I anticipate if you cooled it even colder, the trend would continue. The vapor pressure of the water when you cool it has little effect because it is already quite low at room temperature and also you didn't leave any room for air bubbles. You cool the balloon down, and it simply wants to squeeze the water tighter.
 
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  • #3
How much did it change?

Could it be that the pressure sensor is temperature sensitive?

Could it be the weather changed? High pressure zone moved though at night?
 
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  • #4
Charles: I did a little experiment. I took hair dryer and warmed the balloon with water just about 4 C degrees. The pressure inside increased sth about 7.5 hPa in a few minutes. Do you think that if I would take this balloon to the fridge, the pressure would also increase?

CWatters: It changed 50 hPa during night. Actually it is hard to say if the pressure sensor is temperature sensitive, the data sheet says that the sensor can work in the range -40 C to 125 C degrees. The weather could change, but I think there is no way the pressure will change 50 hPa during night.

Thanks!
 
  • #5
DarekGTR said:
Charles: I did a little experiment. I took hair dryer and warmed the balloon with water just about 4 C degrees. The pressure inside increased sth about 7.5 hPa in a few minutes. Do you think that if I would take this balloon to the fridge, the pressure would also increase?

CWatters: It changed 50 hPa during night. Actually it is hard to say if the pressure sensor is temperature sensitive, the data sheet says that the sensor can work in the range -40 C to 125 C degrees. The weather could change, but I think there is no way the pressure will change 50 hPa during night.

Thanks!
It sounds like a good experiment (to put it in the refrigerator). I say, yes, I am expecting a pressure increase in the refrigerator from the previous observations, as well as I think the hypothesis of the temperature dependence of the elasticity might be valid.
 
  • #6
Charles Link said:
It sounds like a good experiment (to put it in the refrigerator). I say, yes, I am expecting a pressure increase in the refrigerator from the previous observations, as well as I think the hypothesis of the temperature dependence of the elasticity might be valid.

I think it's a good experiment too.

Do you really think raising the temp increases pressure AND lowering the temp increases pressure? That seems kind of odd?
 
  • #7
gmax137 said:
I think it's a good experiment too.

Do you really think raising the temp increases pressure AND lowering the temp increases pressure? That seems kind of odd?
Yes, because of the way the water is packaged in the balloon with the large amount of liquid. If it were just mostly water vapor and it was just an experiment on gas pressure and vapor pressure, the result would be the pressure would most likely simply increase with temperature due to the dramatic increase in vapor pressure with temperature. Any temperature dependent elasticity effect from the balloon material could affect the result, but the vapor pressure change would be the most dominant effect, at least until all of the condensate was evaporated. The way the OP has packaged it, with mostly liquid and little or no vapor at ambient temperature, will cause the elasticity of the material to be what gets measured upon cooling below the ambient temperature.
 
  • #8
Charles Link said:
cause the elasticity of the material to be what gets measured upon cooling below the ambient temperature.
Are you not also measuring the thermal expansion coefficient of stretched rubber? Which is negative, by the way.
 
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  • #9
jbriggs444 said:
Are you not also measuring the thermal expansion coefficient of stretched rubber? Which is negative, by the way.
I think the thermal expansion coefficient is positive for stretched rubber, but in any case, when the balloon with the liquid is cooled, that would seem to be the primary factor affecting the pressure.
 
  • #10
Charles Link said:
I think the thermal expansion coefficient is positive for stretched rubber, but in any case, when the balloon with the liquid is cooled, that would seem to be the primary factor affecting the pressure.
http://www.physlink.com/education/askexperts/ae478.cfm
 
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  • #11
What kind of pressure sensor are you using?
 
  • #13
Charles Link said:
I think I disagree with the "experts" here. Rubber and plastic normally become softer and more pliable and more easily stretched upon heating. e.g. Why does very cold rubber become hard like a rock?
I think we are comparing apples and oranges. Upon heating, the stretched rubber band may not shrink in volume, but it will shrink in length.
 
  • #14
jbriggs444 said:
I think we are comparing apples and oranges. Upon heating, the stretched rubber band may not shrink in volume, but it will shrink in length.
Reading this as well as a google of the subject shows that rubber and plastics tend to be a "strange bird" in that they contract upon heating. Additionally though, which isn't mentioned, is they do become more brittle and less flexible upon cooling. As far as the contraction upon heating, they didn't seem to be very quantitative. In addition, they didn't give the result for a simple experiment with the stretched rubber band where a small amount of heat is applied when it is connected to a spring balance. I believe upon heating the force on the spring balance will decrease, and if the rubber band is cooled, the force will increase.
 
  • #15
Charles Link said:
I believe upon heating the force on the spring balance will decrease, and if the rubber band is cooled, the force will increase.
Fair enough. But experiment trumps belief.
 
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  • #16
jbriggs444 said:
Fair enough. But experiment trumps belief.
Yes, I would very much like to see the results of such an experiment. I can see where the result might go either way, and with what we have googled, the outcome is very much a coin-flip in which way it might go. :)
 
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  • #17
Khashishi said:
What kind of pressure sensor are you using?
I am using this sensor:
upload_2017-8-31_9-36-20.png

And this is the change in pressure [hPa] in time [seconds] from 4 PM to 9 AM during last night:
Capture.PNG
 
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  • #18
The Young-Laplace equation https://en.wikipedia.org/wiki/Young–Laplace_equation gives the difference in pressure between the inside and outside due to the surface tension of the balloon.
If we can neglect gravity, the balloon should be spherical, so R and gamma should be constant across the surface and it should be easy to calculate. The change over time could be due to changes in the surface tension due to temperature or age, or due to changes in external pressure. I suspect the external pressure changes during the night.
We probably can't neglect gravity, so the shape is more complicated, but I still guess the external pressure is changing.
 
  • #19
What is the material of the "balloon"? "Elastic" can mean even a metallic container.
 
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  • #20
@DarekGTR : There's no way to understand this problem without knowing the design of the containment vessel , the orientation it is set on the table and the location and arrangement of the pressure sensor connection .

Given that there is actually any real effect to explain then there could be many different causes .

Please post a diagram of the set up with some leading dimensions .
 
  • #21
Without knowing how big the balloon is, what the shape and material is and how you manage to put a small pressure sensor on it it is difficult to answer your question. Also the relative temperatures would help and if possible the barometric readings. Are you at sea level or at altitude?

Also if my old memory serves me rightly, doesn't water's viscosity increase as it cools?
 
  • #22
DarekGTR said:
Hello!

I have a problem : I have an elastic container (I call it "balloon"). I fill it 100% with water (no air bubbles) and close it. There is a pressure sensor connected to the balloon to measure the pressure inside it.

I would like to ask You : what factors have influence on the pressure inside the balloon, when it is lying on the table and no one is moving this?

I noticed that the pressure sensor says, that the pressure increases, when I heat the balloon a little (just a few C degrees) with a hair dryer. I was expecting that, but there is on thing I don't understand and I would like to ask about: When I left the balloon with water (and of course with sensor) for the night, the next day in the morning the sensor says that the pressure increased during night. But the temperature at night didn't increased - it decreased. I don't fully understand this, that is I would like to ask: what else could the pressure inside could be dependent on?

Thanks for help!

Darek
I will suggest that you put the container/balloon in ice to decrease the temperature and see what will happen.. I believe at night there is fast moving cold air that is why your pressure readings increaases inside compared to outside.. "fast moving air" might your possible factor.
 

1. What factors can affect the pressure in a closed container?

The pressure in a closed container can be affected by several factors, including the volume of the container, the temperature of the gas inside, and the number of gas molecules present.

2. How does the volume of a closed container affect the pressure?

According to Boyle's Law, the volume of a gas is inversely proportional to its pressure. This means that as the volume of a closed container decreases, the pressure inside increases, and vice versa.

3. Is the pressure in a closed container affected by the type of gas inside?

Yes, the type of gas inside a closed container can affect its pressure. This is because different gases have different sizes and masses, which can impact how they interact with the walls of the container and with each other.

4. How does temperature impact the pressure in a closed container?

According to Charles's Law, the pressure of a gas is directly proportional to its temperature. This means that as the temperature of a closed container increases, the pressure inside also increases, and vice versa.

5. Can the pressure in a closed container ever be equal to zero?

No, the pressure in a closed container can never be equal to zero. This is because even if all of the gas molecules were to stop moving, they would still exert a force on the walls of the container, creating a non-zero pressure.

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