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enter image description here

Now at A, there is pressure by mercury column which is nothing but it's weight. Then, at C there is pressure which is atmospheric pressure. Also, the density at A is more than at C. What they say according to the law is that the pressure at C and A are same. I am not able to understand why the pressures are same actually because of the difference I listed in my text. Also, the pressure at C is inside the barometer and at A is outside. So, how can they ever be same and if they are, what is the condition for it.

Pressure at A = Pressure at B +pgh. Now how is this possible? Pressure at A should be more than pressure at B since the height for pressure at is more. So, why is that? Acceding to me Z it should be Pressure at A + pgh= pressure at B.

Then, they also say that pgh is atmospheric pressure. Where they used denstiy of mercury. Now, This should be mercury pressure and not atmospheric right. Density for atmosphere is very low right

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Said by other words, you do not believe that $3 + 2 = 3 + 2$

  • There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom.
  • Both have the same height, so the pressure difference is the same.
  • Yet you have doubts the pressures at their tops are the same.

Pressure at A = Pressure at B +pgh

It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\rho$ is the density of mercury.


The vertical gradient of the hydrostatic pressure in liquid of the density $\rho$ at gravitational accelerationg $g$ is:

$$\mathrm{d}p = - g \cdot \rho \cdot \mathrm{d}h$$

As we can consider $\rho$ and $g$ as constants, the difference of the pressures is then :

$$p_C - p_A = - g \cdot \rho \cdot \int_{h_A}^{h_C} {\mathrm{d}h}$$

$$h_C=h_A \implies p_C=p_A$$


If pressure at the same altitude within the connected liquid of constant density was different, there would be the liquid flow until it is equal.

The pressures at the same altitude would be slightly different, if e.g. mercury at both columns had different temperature and density.


Feedback to comments:

In fact, it does not matter if it is mercury or water, the important thing is the density is constant across the liquid. What other than atmospheric pressure would you expect at C ( regardless of the liquid below ) ? Your scenario description is not exactly clear. Now at A, there is pressure by mercury column which is nothing but it's weight.

If the density along the thought path from A down and then up to C is constant, then pressure at A and C are the same, if the liquid is in the rest. Any pressure difference would cause motion of the liquid.


Let assume the liquid is mercury with density $\rho$.
Let assume the atmospheric pressure at C is $p_0$.
Let assign to C the relative height $h=0$.

Then pressure anywhere below C ( with negative $h$ ) is

$$ p = p_0 - \rho \cdot g \cdot h \gt p_0$$

The same is anywhere below A. At A is the same pressure as at C, i.e. $p_0$.

Between A and B, at height $h$, is the pressure:

$$ p = p_0 - \rho \cdot g \cdot h \lt p_0,$$

but this time $h$ is positive.

B is then at the height

$$ h_\mathrm{B}= \frac {p_0}{\rho \cdot g}$$

with $p=0$

( if we neglect very low vapour pressure of mercury.)

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  • $\begingroup$ Give me some recommendation of book , site pls so that I can improve my theory in this chapter till high school level. $\endgroup$ – srijan Sri Jan 23 at 8:13
  • $\begingroup$ That may be hard, as I did not study that from English books. A great site, written in the form of "cheat sheets", is hyperphysics.phy-astr.gsu.edu/hbase/index.html Try searching for educational stuff, there is plenty on internet. $\endgroup$ – Poutnik Jan 23 at 8:48
  • $\begingroup$ Ok.@Poutnik sir . . $\endgroup$ – srijan Sri Jan 23 at 12:07
  • $\begingroup$ the moderators are very strict in physics stack exchange. They suspend if the question is of a homework but has a conceptual doubt in it. $\endgroup$ – srijan Sri Jan 23 at 12:12
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    $\begingroup$ You can upload it to SE, you can upload it anywhere. Note that comments under questions or answers are not intended for chatting, but for improvement suggestions or asking for clarifications of Q or A. $\endgroup$ – Poutnik Jan 23 at 13:44
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Your terminology makes seeing the correct picture harder.

Part of the problem is that the pressure at A is could be at any point in the column of mercury and that is not constant. The pressure in the column of mercury varies with height (depending on the weight of mercury above that point). When you fill the tube with mercury and invert it there is no pressure above the tube of mercury (neglecting the tiny value of mercury vapour). So all that can support the mercury in the tube is external pressure from the surrounding atmosphere acting on the external surface of the mercury.

The mercury in the tube will fill up to the point where the total weight of mercury produces just enough pressure at the bottom of the tube to balance the external atmospheric pressure. The density is irrelevant as liquids are essentially incompressible.

So the pressure at the top of the mercury column or B will be zero as there is nothing in the space above the mercury to exert any pressure. The pressure will rise at points down the mercury column until the point where it balances the external pressure on the external surface of the mercury. (if your intention was to define A as the point in the tube equivalent to the external level of mercury in the bowl, then pressure at this point is atmospheric but this is not true at other points in the column).

The point is that the pressures are equal and the densities are irrelevant. If you used a liquid other than mercury the same effect would be seen but the height of column would be different (water is about 14 time less dense than mercury so the total column would be ~14 times higher for the weight of water to balance the pressure of the atmosphere). This is the only point where density matters: it doesn't vary in the liquid column.

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