# Blah Pressure Blah Blah

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Thread starter 4 years ago
#1
Hello, this is quite a basic question but nevertheless one that I need to grasp the concept off. You see in this question, one tube is inverted and one is upright, how comes the total pressure for the inverted tube = atmospheric pressure - exerted pressure and the total pressure for the upright tube = atmospheric pressure + exerted pressure. Is it linked with it P being proportional to 1/V ?

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4 years ago
#2
The trapped dry air pushes outward on the tube, and the atmospheric pressure pushes inward on the tube, hence it will be atmospheric minus the exerted pressure.

For the second, the weight of the thread is no longer compressing the trapped air, and instead when the mercury slips downward, a vacuum will form, and this will keep the mercury upward. Atmospheric pressure acts on the tube and pushes on the mercury, while the dry air pulls inward on the tube and mercury.

Hence that's my reasoning. Hope I helped!
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Thread starter 4 years ago
#3
(Original post by Callicious)
The trapped dry air pushes outward on the tube, and the atmospheric pressure pushes inward on the tube, hence it will be atmospheric minus the exerted pressure.

For the second, the weight of the thread is no longer compressing the trapped air, and instead when the mercury slips downward, a vacuum will form, and this will keep the mercury upward. Atmospheric pressure acts on the tube and pushes on the mercury, while the dry air pulls inward on the tube and mercury.

Hence that's my reasoning. Hope I helped!

That's what I thought you see, as pressure is acting in different ways you'd have to take away exerted pressure from the atmospheric pressure but on the mark scheme it shows this.

BTW: the exerted pressure from the mercury is 0.2 x 10^5 so for the first one, they added the exerted pressure and the atmospheric pressure
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