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https://youtu.be/JzuamK0AyC4

According to this clip’s experiment, if the temperature of the third bottle increases, which raises the pressure, the water from the second bottle will then rise and could flow to the third bottle if the temperature is high enough.

Can anyone explain to me why that is?

enter image description here

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  • $\begingroup$ was that helpful ? $\endgroup$
    – Ankit
    Commented Sep 10, 2020 at 12:37
  • $\begingroup$ +1 for the video and your effort. $\endgroup$
    – Ankit
    Commented Sep 10, 2020 at 13:48

1 Answer 1

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Its more about creating vacuum than temperature difference. The same can be done using vacuum pipe.

Let us number the bottles first in the order they are arranged from left to right as 1, 2 and 3.

In the beginning , the pressure at both the ends of the tube connecting $1^{st}$ and $2^{nd}$ became equal and that's why the height of the liquid was stable.

$P_{o} + (density \; of \; liquid)gh = P_{air}+ (density \; of \; liquid)gh$

In the clip it is shown that when the burning candle is kept inside the $3^{rd}$ bottle , the oxygen of the third bottle is being used up and after sometime in the third bottle there is no air or it is vaccum (although there may be some sort of nitrogen gas I think , correct me if I am wrong here). Now due to this vacuum or lesser air, the air from the $2^{nd}$ bottle goes to the third bottle and due to this, the amount of air in the second bottle is reduced.

Now due to this reduction there is a reduction in the pressure acting on the fluid from the air i.e. $P_{air} $ in the above equation is reduced . And hence the pressure at the end of the tube in the second bottle is reduced. But the pressure in the first bottle is not altered and it is the same as earlier.

Since in the beginning the pressure difference at the ends of the joining tube was same , this means that now the end of the tube in the first bottle has more pressure than at the end in the second bottle. Due to this liquid flows from the first bottle to the second bottle and the liquid in the second bottle rises and due to this rise, the pressure at the end of the tube in the second bottle will increase and the pressure at the end of the tube in the first bottle will decrease untill the pressure at both the ends again equalises.

$P_{o} + (density \; of \;liquid)gh_1 = P_{air}+ (density \;of \;liquid)gh_2$

So the rise of liquid in the second bottle is restricted to a particular value and all the liquid can only come into the second bottle (i.e the end in the first bottle is open to only atmospheric pressure) if the $(pressure\; of\; raised\; liquid \;+ pressure \;of \;some\; remaining\; air\; equal \;the \;atmospheric\; pressure.)$

$ P_o = P_{gas} + (density \;of\; liquid)gh_3$

So liquid can go to the third bottle if :

1 : density of liquid is less which will increase the height raised by the liquid for getting the same pressure.

2: almost all of the air is taken out (it may or may not be helpful depending on the density essentially since highly dense liquid can get the same pressure with little rise only)

3: size of bottle is small.

Note : $P_o $ is used for atmospheric pressure.

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  • $\begingroup$ @Mac_79 if something was unclear then you can ask . $\endgroup$
    – Ankit
    Commented Sep 10, 2020 at 12:37
  • $\begingroup$ Thanks so much. That’s quite clear. I get the main concept now. For the equation, I’ll need to plug in some numbers in order to have a clearer picture. $\endgroup$
    – Mac_79
    Commented Sep 10, 2020 at 12:44

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