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What will be pressure of gas measured by manometer if mercury starts to rise on the other side?

Manometer Image

The $P_a$ side is open to the gas container and the $P_0$ side is open to the atmosphere with atmospheric pressure $P_0$.

The pressure of gas $P$ here is $P_0 + \rho g H$ where $H$ is as given in image and other symbols have usual meanings.

Now here is my question,

What would be the pressure of a gas if the pressure of gas was high enough to cause the liquid (mercury here) to move down the U shaped tube and rise up on the other side?

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closed as off-topic by John Rennie, ZeroTheHero, Yashas, AccidentalFourierTransform, Jon Custer Mar 19 '17 at 20:58

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  • $\begingroup$ One possible solution I can think of is that if the liquid rises up, the more it rises the value of $H$ keeps increasing. The formula remains same. $\endgroup$ – Agile_Eagle Mar 17 '17 at 11:32
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    $\begingroup$ If it was any higher than $\rho g$ times the total length of the liquid column, it would continue to accelerate until it blew out the top of the tube. $\endgroup$ – Chet Miller Mar 17 '17 at 12:04
  • $\begingroup$ @ChesterMiller Can you clarify a bit? $\endgroup$ – Agile_Eagle Mar 17 '17 at 12:05
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    $\begingroup$ Here is a force balance on the column of liquid if the pressure pa were high enough to push all the fluid into the left column of the manometer: $$p_a-\rho g L-p_0=ma$$ where L is the total length of liquid column. What does it tell you? $\endgroup$ – Chet Miller Mar 17 '17 at 12:12
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    $\begingroup$ No, it shouldn't. $\endgroup$ – Chet Miller Mar 17 '17 at 17:10
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If the gas pressure $P_a$ increases from the diagram, the mercury level on the right will fall and that on the left will rise. This will increase the value of $H$ so the pressure of the mercury pushing back on the gas also increases, untill $P_0+\rho gH=P_a$ again.

If the gas pressure $P_a$ is strong enough to push the mercury level to the bottom of the manometer tube, then $H$ has reached its maximum possible value $H_{max}$, which is the length of mercury in the manometer (straight plus curved sections). Even if the mercury rises further up on the left the difference $H$ between the 2 levels of mercury cannot increase any more. So the pressure pushing back on the gas cannot increase any more.

If $P_0+\rho gH_{max} \lt P_a$ then the mercury will be pushed further and further up on the left until all of it is forced out of the top of the manometer tube, however high that is.

In short, if the gas can push the mercury down to the very bottom of the bend in the manometer, then it will continue pushing the mercury all the way out of the manometer.

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  • $\begingroup$ No, this was not exactly my question. What would happen if the pressure is high enough to cause the liquid to move down the U shape tube and rise up on the left? You gave the case when it rises up on the right. $\endgroup$ – Agile_Eagle Mar 18 '17 at 4:20
  • $\begingroup$ Anyways, Thanks for the insight! Could you answer the original question now? $\endgroup$ – Agile_Eagle Mar 18 '17 at 4:32
  • $\begingroup$ @sammygerbil The OP meant that the mercury rises in the left tube due to high pressure of injected gas in the right tube, such that no part of the mercury column remains in the right tube. In such a situation there will be gas present in the whole of the right tube as well as the bottom part of the left tube (we assume that the mercury won't spill out from the left tube). $\endgroup$ – user102705 Mar 18 '17 at 5:02
  • $\begingroup$ I agree that the diagram in the question is misleading. $\endgroup$ – user102705 Mar 18 '17 at 5:03

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