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I am quite familiar with the pascal’s law but i still think that being on the same points still they should have different pressures as above on one point is atmosphere and on the other is only the column of mercury can anyone give physical explanation for this and also if the pressure inside the torricelli vacuum is only due to mercury vapours which is quite low than why the vacuum does not get crushed under the atmospheric pressure?

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    $\begingroup$ Atmospheric pressure is $1~kg/cm^2$. If the tube's surface area is $1cm^2$, it needs to be strong enough to not break when you put $1kg$ over it's entire surface. $\endgroup$ – AgentS Dec 2 '19 at 5:40
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The pressure inside the tube at height is not equal to the pressure outside. It is a vacuum, or partial vacuum created by the weight of the column of mercury in the tube. Higher atmospheric pressure at the open pool of mercury at the base of the tube will push the column of mercury higher in the tube. Lower atmospheric pressure at the base pool will let the pool level rise, making the column of mercury drop. So the pressures equal at the surface of the mercury pool at the bottom, not at height inside the tube. see https://en.wikipedia.org/wiki/Barometer#/media/File:MercuryBarometer.svg

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  • $\begingroup$ No what i meant to say was pressure of mercury inside the tube at any point should not be equal to the pressure of mercury in the trough at the same levels $\endgroup$ – Vasu Kaura Dec 2 '19 at 6:22
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    $\begingroup$ Where the tube is under the surface of the pool the pressure is equal inside and out, it has to be, the tube is open to the pool. Just as pressure increases with depth in liquids, because the weight of the liquid above pressing down, raising liquid above the surface lowers pressure from the weight of the liquid pulling down. Once the weight of the liquid pulling down equals atmospheric pressure, it starts forming a vacuum above the liquid. $\endgroup$ – Adrian Howard Dec 2 '19 at 6:40
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enter image description here

Let understand this by the figure, if we notice that pressure at surface of mercury would be the atmospheric pressure. And same pressure at the bottom of the mercury tube and if we go a height h above.

Now the pressure in the space above the mercury is less than atmospheric pressure and equilibrium is attained when $ Mg$ (where M is the new mass of mercury which is less than initial mass in the mercury column) + P*A (where P is less than atmospheric pressure) = (Atmospheric pressure) $ A$ where A is area of cross section of tube. Or as $ρ g h$ (where h is the new height of column) = P + Atmospheric pressure

So we get. $$ \rho g h = P_\text{atm} - P $$

where $\rho$ is the density of the mercury.

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