Inverted the pipet into cold water

  • Thread starter MercuryRising
  • Start date
  • Tags
    Cold Water
In summary, when you heat a gas (such as air inside the pipet), what happens to it? If the container it is in (the pipet) is open to the outside air, what direction does the gas flow, in or out, of the container when it is being heated? When you cool this previously heated pipet, what happens to the air inside it?
  • #1
MercuryRising
28
0
ok in this lab, we heated a plastic pipet in water so that it expandes. then we covered the top so that the gases are trapped inside the pipet. then we inverted the pipet into cold water, and releases our thunb that covered the hole of the pipet. thus water got sucked in...but why? :uhh:
thank you in advance :cool:
 
Chemistry news on Phys.org
  • #2
Here are some hints:
When you heat a gas (such as air inside the pipet), what happens to it?
If the container it is in (the pipet) is open to the outside air, what direction does the gas flow, in or out, of the container when it is being heated?
When you cool this previously heated pipet, what happens to the air inside it?
 
  • #3
Moonbear, have you done this experiment before?
 
  • #4
well, the volume of the pipet expands, and when it is inverted into cold water, it shrinks again...
i still don't know how that caused the water to be sucked in
 
  • #5
MercuryRising said:
well, the volume of the pipet expands, and when it is inverted into cold water, it shrinks again...
i still don't know how that caused the water to be sucked in
What about the air in the pipet ? What happens to it ?
 
  • #6
MercuryRising said:
well, the volume of the pipet expands, and when it is inverted into cold water, it shrinks again...
i still don't know how that caused the water to be sucked in

I'm a little confused on this. Did you see your pipet visibly change shape/size? Or are you just assuming the pipet expanded? If you did not see the pipet actually change shape, don't assume it did.
 
  • #7
GeneralChemTutor said:
Moonbear, have you done this experiment before?

Not this one, but the concept is common to several experiments (when I was in school, we did something similar but sucked a hardboiled egg into a flask instead of water into a pipet...all fine and good until the teacher forgot to clean up the egg before leaving for the weekend :yuck:)
 
  • #8
The boiled egg is nicer with a clear plastic bottle and some liquid nitrogen.
 
  • #9
I would honestly doubt that the volume of the pipet (made outa plastic,if I'm not mistaking) would change such as to produce visible effects.
One can then assume that the gas from the pipet undergoes an isochore transformation...

Daniel.
 
  • #10
I had seen a similar case where low proof ethanol was poured on to a plate, a lighted match was thrown on top, and was concealed with a drinking glass...the liquid rose up instantly almost to the top. Certainly this is not a case where the volume expands and such, however it seems to have the same concept in mind.

A rather pathetic suggestion of mine is in terms of adhesivity and cohensivity of water; the increase in temperature decreases the surface tension of water and allows the water to climb up the glass from the sides somewhat engulfing the air inside the container.

I tried searching for this experiment on the net to no avail.
 
  • #11
dextercioby said:
I would honestly doubt that the volume of the pipet (made outa plastic,if I'm not mistaking) would change such as to produce visible effects.

That's my initial assumption. Your typical cheap plastic transfer pipet is what I'm guessing was used. But it would be good to clarify this so we don't lead MercuryRising astray with an incorrect assumption.

If this assumption is correct, then my understanding of the experiment is that the pipet was left open at one end, heated, then sealed while still hot (stick your finger on the end), and not cooled until placed in the water and seal (finger) removed. If that's true, then there's a simple explanation to consider. The answer lies in the effect of temperature changes on the air inside the pipet (or any vessel). My guess at this point is the experiment is intended to demonstrate Charles' Law, or generally ideal gas law.
 
  • #12
GeneralChemTutor said:
A rather pathetic suggestion of mine is in terms of adhesivity and cohensivity of water; the increase in temperature decreases the surface tension of water and allows the water to climb up the glass from the sides somewhat engulfing the air inside the container.

If the intention of the experiment were to demonstrate adhesivity/cohesivity properties, it would have been better done with an open, glass capillary tube (narrow diameter). A plastic pipet with a closed bulb at the top wouldn't demonstrate this property well.

I can't envision the experiment working at all if the pipet bulb expanded during heating.
 
  • #13
This is a very simple experiment with a very elementary explanation.

Let's not confuse things for Mercury, and bring in second (or higher) order effects. Mercury, have you heard of Gay Lussac's Law (or even the Ideal Gas Equation) ?
 
  • #14
MercuryRising said:
ok in this lab, we heated a plastic pipet in water so that it expandes. then we covered the top so that the gases are trapped inside the pipet. then we inverted the pipet into cold water, and releases our thunb that covered the hole of the pipet. thus water got sucked in...but why? :uhh:
thank you in advance :cool:


I'm assuming you allowed the inside of the pipet to 'start' release the gas right?

This way the gas pushed out the outside pressure and once you put your thumb on the hole you essentially closed the contaire - which was now at a lower pressure than the outside world. Once you opened the pipet the water should rush in under air's outside pressure until the two reach equilibrium (there is an equation for this, but essentially it should reach about 80 percent of the pipet with water)

PV = nRT - by keeping n and T constant you get
P1V1 = P2V2

So in outside you have p (atm) * V1 (a lot of air) = P2 (about 0.8 of atm) * V2 of bottle
Once you opened the bottle the two will try to reach equilibrium, and so you should end up with water inside the bottle
 
  • #15
Ouch ! This needs to be cleared up.

Here's what happens. First you heat up the pipet and the air in it, to some temperature T1. Now you put your thumb over it and contain this hot air. The volume of the pipet V is nearly constant. Next you lower the sealed pipet into cold water and reduce the temperature of the air to T2 (T2 < T1). Since the volume enclosed is roughly constant, the pressure drops (P1/T1 = P2/T2). But since the pressure of the water is atmospheric, and the air pressure is lower, water will rush into the pipet.

Going further, there is no way to tell how much water will enter the pipet unless you know the volume of the pipet and the temperatures of the hot and cold water.

If the volume of the pipet is V, and calling atmospheric pressure Po, the number of moles of air trapped in the pipet is given by :

[tex]n=\frac{P_oV}{RT_1} [/tex]

The pressure upon cooling to a lower temperature [itex]T_2~ (< T_1) [/itex] is given by :

[tex]P_2 = \frac {nRT_2}{V} = \frac {P_oT_2}{T_1} < P_o[/tex]

After water fills the pipet, the volume of air in the pipet (V') must be such that its pressure is atmospheric. So :

[tex]P_o = \frac {nRT_2}{V'} = P_o \frac {T_2V}{T_1V'} [/tex]

[tex]=> V' = V \frac {T_2}{T_1} [/tex]

Thus the fraction of water in the pipet is simply :

[tex]V_w = \frac {V - V'}{V} = 1- \frac {T_2}{T_1} [/tex]

Even if you used boiling water (T1 = 373K) and icy water (T2 = 273K), you can only get about 27% of the pipet filled with water (neglecting the thermal expansion of the pipet material compared to that of the air). If you do get more than this amount of water into the pipet, it can only result from not being very careful (ie : allowing air to bubble out) or performing the experiment deep underwater. To get the pipet filled 80% with water, you will need

[tex]V'/V = \frac {P_oT_2}{P_2T_1} < 0.2 [/tex]

For this, you'd have to be under at least 120 feet of (icy) water.
 
  • #16
I don't believe that Pinitial is equal to the atmospheric pressure, as long as the temperature of the gas is greater within the container. When the container is exposed -an open system-to the atmosphere, depending on the shape/area of the opening, a flow of air will be established with the constant maintenance of the temperature; air will appear to be flowing out of the pipet to a degree.

Anyways...

When you place any container in an inverted manner to a certain depth, depending upon the shape/volume of the container the liquid will rise up the container until the water encounters an pressure equal to the pressure at that specific depth of the water-the point of entry into the pipet (which takes into account both the atmospheric pressure as well as the gravitation force due to the upper mass of water). At greater depths you'll get more water into the pipet, even if w/0 heating the pipet. And even with a room temperature pipet, you'll have water move up the column to a certain degree in the case where the pipet were to be placed completely in the water

So what caused the water to dramtically fill the pipet? Perhaps it's due to the shape of the pipet; the volume of the bulb at the end takes up a significant proportion.

Despite all of this I am not able to see the utility of this experiment, it seems to be a very cheap way of explaining the ideal gas law.
 
  • #17
GeneralChemTutor said:
I don't believe that Pinitial is equal to the atmospheric pressure, as long as the temperature of the gas is greater within the container. When the container is exposed -an open system-to the atmosphere, depending on the shape/area of the opening, a flow of air will be established with the constant maintenance of the temperature; air will appear to be flowing out of the pipet to a degree.
Yes there will be some air that leaves the pipet as it gets warm. But it will equilibrate with Po in a fraction of a second (at least for any pipet that I'm familiar with). So, the initial pressure will be atmospheric. And even if it is not, there will be either a (i)pressure drop, if the pipet is stiff (isochoric reduction of pressure), or (ii) reduction in volume if the pipet has a soft rubber bulb (isobaric compression). Either way, equilibrium is restored by liquid entering the pipet.


When you place any container in an inverted manner to a certain depth, depending upon the shape/volume of the container the liquid will rise up the container until the water encounters an pressure equal to the pressure at that specific depth of the water-the point of entry into the pipet (which takes into account both the atmospheric pressure as well as the gravitation force due to the upper mass of water). At greater depths you'll get more water into the pipet, even if w/0 heating the pipet.
The hydrostatic pressure from a few inches of depth is negligible compared to an atmosphere.


And even with a room temperature pipet, you'll have water move up the column to a certain degree in the case where the pipet were to be placed completely in the water
For a standard cylidrical pipet (one with a plunger, not a bulb) that's about 4 inches long, even if you completely submerge the pipet in water, the hydrostatic pressure will force the water up by only 1 millimeter. (Even with a bulb, it would go up only 2 or 3 mm.) If the diameter of the pipet is small (which is typical), capillarity will have a much greater effect than hydrostatic pressure.
 
Last edited:
  • #18
hey Gokul...what does urs signature statement mean...what would appear as infinite/?
 
  • #19
Yes there will be some air that leaves the pipet as it gets warm. But it will equilibrate with Po in a fraction of a second (at least for any pipet that I'm familiar with). So, the initial pressure will be atmospheric. And even if it is not, there will be either a (i)pressure drop, if the pipet is stiff (isochoric reduction of pressure), or (ii) reduction in volume if the pipet has a soft rubber bulb (isobaric compression). Either way, equilibrium is restored by liquid entering the pipet.

At this point I believe that your conclusions is neither consistent with an open or closed system. It seems that you are assuming that the system (meaning the pipet and surroundings) equilibriates through a change in the number of moles (and thus your account of "some of the air" leaving as the pipet gets warm). But note that this is clearly an open system. And that there is a constant work input on the system, which makes it a dynamic system. I should review a standard text to clear things up conclusively , however no time right now.

I'm not trying to undermine your final conclusion that you arrived at, certainly the above is just technical and your conclusion (the "clearing up" post) remains very helpful.

The hydrostatic pressure from a few inches of depth is negligible compared to an atmosphere.

I'll get back to this later, or perhaps you can perform a order of magnitude calculation.

For a standard cylidrical pipet (one with a plunger, not a bulb) that's about 4 inches long, even if you completely submerge the pipet in water, the hydrostatic pressure will force the water up by only 1 millimeter. (Even with a bulb, it would go up only 2 or 3 mm.) If the diameter of the pipet is small (which is typical), capillarity will have a much greater effect than hydrostatic pressure.

And thus the I don't really see the significance of this experiment.
 
  • #20
Biology said:
hey Gokul...what does urs signature statement mean...what would appear as infinite/?

blurry :uhh:
 
  • #21
What are you guys getting on about with standard pipets with plungers instead of bulbs? I think you've all over-complicated this immensely! Of course it would really help if MercuryRising came back and clarified some of the experimental set-up.

I had in mind a simple, plastic, 1 ml transfer pipet. The cheap kind you'll buy in a box of a 1000, like any high school or college freshman chem lab might be likely to use, not a serological pipet or syringe or whatever it is you're thinking about with a plunger. With a 1 ml transfer pipet, the approximately 10 cm long "shaft" contains 1 ml of volume...rough guesstimate of diameter would be 2.5 or 3 mm (guesstimates here just to give you an idea of the pipet I'm envisioning being used). The bulb on it is larger. I've never measured the volume of the bulb, but again, as a guesstimate, it would be about 3 cm high and about 1 cm in diameter.

If you heat the bulb, let's say you immerse it in boiling water, which is probably the best available heat source that won't melt plastic in a typical gen chem type lab, leaving one end open, the air (gas) inside the pipet expands and escapes the open end (I don't think you'd be heating it fully immersed for this experiment, otherwise how do you hold the end after heating?) Now, you stick your finger over the end of the pipet, and prevent the air from flowing back into equilibrate. If you're really slow, you'd see the pipet bulb contract as if being squeezed during the cooling (which of course we know is due to the decrease in volume to equilibrate pressure with the surrounding atmosphere once cooling occurs). Now you plunge your pipet, including the open end now, into cold water (or even just stick the tip in water and leave the top chilled in some other way...under running water, just room temp, etc). Now the water gets sucked into restore volume and pressure. If you use too large or uniform of a pipet, this experiment probably wouldn't show much, but if you use a 1 ml plastic transfer pipet, due to the relative volume of the shaft to the bulb, you'll be able to detect even Gokul's estimate of a 27% volume change as a pretty full shaft of water...at least enough to impress the students.

So, I'm just putting into words pretty much what Gokul showed in equations in post #15.

It's easy to overthink this one with more advanced knowledge, but it's important to keep in mind what the demonstration is likely to be used for. You wouldn't use such a simple set-up for demonstration of a more complex principle.
 
  • #22
GeneralChemTutor said:
At this point I believe that your conclusions is neither consistent with an open or closed system. It seems that you are assuming that the system (meaning the pipet and surroundings) equilibriates through a change in the number of moles (and thus your account of "some of the air" leaving as the pipet gets warm). But note that this is clearly an open system. And that there is a constant work input on the system, which makes it a dynamic system. I should review a standard text to clear things up conclusively , however no time right now.
Yes, the system is open until the thumb is pressed to the end of the pipet. But my calculation begins only after this point. It is not essential that the initial temperature of the gas be equal to that of the liquid the pipet is immersed in. Once my thumb is placed over the pipet, the system is closed and n is conserved.

Until this point we have a system that is trying to equilibrate with the surroundings through exchange of mass and energy. However, the time scale for mass exchange is much smaller than that for temperature exchange, since the mass-flow impedance (essentially determined by the diameter of the opening) is tiny compared to the thermal impedance (determined by conductivities of plastic, convective heat transfer coefficient of the air, as well as the mass flow). So, it is possible to sustain a thermal gradient much longer than you can sustain a pressure gradient.


I'll get back to this later, or perhaps you can perform a order of magnitude calculation.
At a depth of 10 cm
[tex]P_{hydro} = h \rho g = 0.1 * 10^3 * 10 = 10^3 Pa [/tex]
[itex]1~ atm = 10^5~[/itex] Pa, which is 100 times bigger. 1% of 10 cm (roughly 4") is 1 mm. If there's a large, soft bulb (typically the bulb volume is a little more than the stem volume) with twice the stem volume, the rise from hydrostatic pressure is about 3 mm.

As for capillarity, this depends on the material of the pipet and the inside radius. Water has the greatest capillary action with glass, and with most plastics the effect is nearly an order of magnitude smaller. Look at the contact angles http://www.lib.umich.edu/dentlib/Dental_tables/Contangle.html .Let's Assume the pipet is a glass cylinder of 0.25" ID (pretty small) and the contact angle is 0. Then plugging in the surface tension of water, you get a capillary rise of only about 2 mm. (A little less than I'd imagined.)

So, the total rise from hydrostatic and capillary effects is no more than 3 mm (hard) to 5 mm (soft).

On the other hand, if you can get a temperature difference of even 30K, that's a 10% effect on the volume or pressure (depending on the design of the pipet - hard or soft). So, in the hard cylindrical pipet, this makes the water rise by at least 10mm, and in the bulbed pipette, it rises by 30 mm. This number is at least thrice (in the hard case) the height from the other effects and as much as 6 times the height (in the soft case) .

So, it wouldn't be unfair to say that the dominant cause for the water rise is the isochoric/isobaric behavior of the trapped air.

Also, the surface tension of water changes by no more than 10% over a comparable range of temperatures. So, that affects the capillary rise by only a small fraction of a millimeter.
 
Last edited by a moderator:
  • #23
Moonbear said:
If you use too large or uniform of a pipet, this experiment probably wouldn't show much, but if you use a 1 ml plastic transfer pipet, due to the relative volume of the shaft to the bulb, you'll be able to detect even Gokul's estimate of a 27% volume change as a pretty full shaft of water...at least enough to impress the students.

So, I'm just putting into words pretty much what Gokul showed in equations in post #15.

Umm...yeah ! Same difference ! :biggrin:

And you know how us guys like to overthink things. Boys with toys...just ignore us. I went ahead with the overthinking only because it looked like the OP was long gone.
 
  • #24
Perhaps it's because other posts are not so interesting as this one.
 
Last edited:
  • #25
I was going to respond to your post when I found more time, which is now...but the post is gone. My calculation does take into account the initial heating. Assuming the pressure at start is Po does not imply that I neglected the initial heating.

Anyway, if you want to drop it, I probably shouldn't go on and on.
 
  • #26
Yeah,that's partly why I dropped the post (your smaller n will account for the P drop in the end) and due to the fact that this topic is a bit fuzzy to begin with.
 
Last edited:

1. What is the purpose of inverting the pipet into cold water?

The purpose of inverting a pipet into cold water is to create a vacuum that allows for accurate and precise measurements of liquid. This is especially important when working with small volumes of liquid in laboratory experiments.

2. How do you properly invert a pipet into cold water?

To properly invert a pipet into cold water, you should first fill the pipet with the liquid you want to measure. Then, place the tip of the pipet into a container of cold water and slowly release the liquid until the water level is equal to the desired volume. Finally, remove the pipet from the water and wipe off any excess liquid from the tip.

3. Can any type of liquid be used when inverting a pipet into cold water?

Yes, any type of liquid can be used when inverting a pipet into cold water. However, it is important to make sure that the liquid is compatible with the materials of the pipet and will not react with the cold water, as this can affect the accuracy of the measurement.

4. What are the benefits of using this method for measuring liquids?

The benefits of using this method for measuring liquids include increased accuracy and precision, as well as the ability to measure small volumes of liquid. It also helps to reduce the risk of human error, as the vacuum created by inverting the pipet allows for a consistent and controlled release of liquid.

5. Are there any potential risks associated with inverting a pipet into cold water?

There are some potential risks associated with inverting a pipet into cold water, such as breakage of the pipet due to temperature shock or contamination of the liquid being measured. It is important to follow proper safety procedures and use appropriate materials to minimize these risks.

Similar threads

  • Chemistry
Replies
1
Views
3K
Replies
23
Views
1K
Replies
1
Views
921
  • Mechanical Engineering
Replies
30
Views
2K
Replies
11
Views
1K
  • Classical Physics
Replies
1
Views
989
  • Astronomy and Astrophysics
Replies
21
Views
5K
  • Biology and Medical
Replies
2
Views
886
Replies
1
Views
982
Replies
11
Views
1K
Back
Top