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Consider picture below where a force applied to a water by a piston and suppose the gauge reads an arbitrary pressure. If I lock the piston to its position and remove the force applied to piston, will the pressure on the gauge read the same? Consider that the manometer is below water level so it is filled already.

enter image description here

EnumaElish
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  • Why you think additional pressure should go away without force applied ? Pressure increases, because of squeezed water volume, so to say due to water bulk modulus feature : $ K=-V{\frac {dP}{dV}}$. So if position of piston stay fixed, pressure must stay fixed too, unless water solidifies into ice, then situation can be different, depending on ice bulk modulus, phase conversion nuances, etc. – Agnius Vasiliauskas Jan 09 '24 at 13:10
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    Can the piston be deformed? – my2cts Jan 09 '24 at 13:51
  • I assumed it as incompressible. – EnumaElish Jan 09 '24 at 13:51
  • By 'locking' the piston in place you are merely moving the force applied from the weight to whatever locks it in place when the weight is removed. – Jon Custer Jan 09 '24 at 14:14

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Assuming that heat flow into/out of the water is negligible, then the water remains at the same pressure. The pressure of the water is determined at a given temperature and volume by an equation of state, and in the situation you're describing the temperature and volume are remaining the same.

Note that if you "lock the piston into its position" then the locking mechanism will end up applying the same amount of force on the piston as was originally being exerted on the piston by the external agency. It has to, otherwise the piston would move.

Michael Seifert
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  • Okay. Thank you for your great explanation. I have another question if we suppose it is an incompressible liquid? Does pressure still remain or will it vanish since it does not cause any enthalpy change via volume change? – EnumaElish Jan 10 '24 at 08:08
  • @EnumaElish: I'm not quite sure what you're asking, and generally new questions should be posted as new questions rather than addressed in comments. So I'd encourage you to post a new question expanding on what you're asking; link to this question for context if it helps. – Michael Seifert Jan 10 '24 at 12:57
  • It isnt much of a new question to begin with. Same conditions and everything remains the same except this time water is not compressible at all. After locking the piston will gauge read the same? – EnumaElish Jan 11 '24 at 16:29
  • @EnumaElish: I think I see what you're asking now, but I still think it's worth asking as a new question. Basically, my answer would boil down to the fact that there's no such thing as an "incompressible" fluid; there are merely situations in which the compression of fluids can be neglected. This is not such a situation. There's not enough space in this comment box to elaborate on this statement, but if you post a new question I'll try to remember to write a more detailed answer. – Michael Seifert Jan 11 '24 at 16:37
  • Related: If water is incompressible, how can sound propagate underwater? – Michael Seifert Jan 11 '24 at 16:39
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When you lock the piston and remove the force(ie pressure) in the process of locking the piston, you are still applying the force required to hold the piston in its current volume, and if the piston is unable to lock itself of course volume and pressure will change. Assuming the lock is strong enough, i think the pressure remains the same. (all this assuming walls are adiabatic)

Chethas Pai
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