Solving Air Pressure Concepts in Torricelli Experiment

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SUMMARY

The discussion focuses on the calculation of pressure change (ΔP) in a Torricelli experiment involving air and water. The formula presented, ΔP = μ(water).g.h, where μ represents the density of water, is confirmed to be correct for determining the pressure difference due to the height of water in the tube. The participant highlights the importance of considering temperature changes using Clapayron's Equation (p.v=R.n.T) to understand the behavior of the air in the vessel. The conclusion emphasizes that the negative pressure difference arises from the contraction of hot air as it cools and the rapid heat transfer with water.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly the Torricelli theorem.
  • Knowledge of thermodynamic equations, specifically Clapayron's Equation.
  • Familiarity with pressure concepts in gases and liquids.
  • Basic grasp of density and its role in fluid dynamics.
NEXT STEPS
  • Study the derivation and applications of the Torricelli theorem in fluid dynamics.
  • Learn about Clapayron's Equation and its implications in thermodynamics.
  • Explore the relationship between temperature, pressure, and volume in gases using the Ideal Gas Law.
  • Investigate the effects of heat transfer on fluid behavior in thermodynamic systems.
USEFUL FOR

Students and professionals in physics, engineering, and thermodynamics, particularly those interested in fluid mechanics and pressure dynamics in experimental setups.

Pedro Lemos
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Well, I'm trying to grasp the concepts behind the following problem :

"A guy puts some air in a vassel, the air's temperature was t0 the vessel's basis area is 50 cm³ and the height is 20. Afterwards the vassel was set onto a container full of water at 300k (Similar to Torricelli experiment), part of the water entered the tube reaching 4 cm of height and then the system acquires equilibrium. The Change in pressure is asked"

The awser is : ΔP = μ(water).g.h ->I don`t understand what formula is that nor that creepy μ.
=-1.10³.10.4.10^-2
=-4,0.10² N/m²
There is a change in temperature as well, so shouldn't this change to be considered as a factor of pressure through Clapayron's Equation p.v=R.n.T?
 

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Apperently I find there's something IS wrong with the solution given.

How is the change in pressure related to μ(water).g.h...the upthrust?
 
That first number should be the density of water. Normally I see it written as the greek letter rho (looks like a rounded lowercase p). P1 - P2 = ΔP = dgΔh.

It looks like in this situation the pressure difference is negative, as the hot expanded air is being used to suck water into the tube as it contracts. Heat transfer in water is fairly rapid, so the air in the tube should change to match the temperature of the water, thus causing it to cool and contract by Pv=nRT.
 

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