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a vertical cyclinder of height 100cm contains air at a constant temperature the top is closed by a frictionless light piston the atmospheric pressure equal to 75cm of mercury mercury is slowly poured over the piston find the maximum height of the mercury column that can be put on the piston

a vertical cyclinder of height 100cm contains air at a constant temperature the top is closed by a frictionless light piston the atmospheric pressure equal to 75cm of mercury mercury is slowly poured over the piston find the maximum height of the mercury column that can be put on the piston

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2 Answers

chua kok tong
29 Points
13 years ago

i may have idea on how to calculate it anyway is the answer is 25cm? If it is correct, i will show my working but if it is not than i try reconsider another formula.

chua kok tong
29 Points
13 years ago

since the atmospheric pressure is 75cm mercury than the pressure inside the cylinder must be same as atmospheric pressure. So the pressure inside the cylinder=pgh,p=density of mercury,g=9.81h=75cm. At 1st you assume a drop of mercury is apply on the piston and it does not leak and compress the piston downward by a small distance so more mercury can be apply to it and it wont leak. The question didn't mention it have an insulated wall and the process is slow so can be treated as isothermal process.

When 1cm of mercury is apply to it, the pressure inside the cyllinder increase become 76cm*gh
Applying p1v1=p2v2
volume of cylinder initially=h(pir^2)=100cm*pir^2
P1=pg(75cm)

pg(75cm)(100pir^2)=(76cm)pg(bpir^2) where b is the distance from the base of cylinder to the piston.
b=98.68 total height occupy by mercury and the distance from base to piston=98.68+1=99.68 cm which is less than 100 cm so mercury can be continue added to it.
From here you can see that when added 1cm mercury it compress the piston by 100-98.68=1.32cm.
However, when more and more mercury added, the pressure increse and you need more and more mercury to compress the piston by a unit cm.

A formula can be form from here
pg(75)(100pie^2)=(75+h)pg(100-h)(pir^2)
h=0 and h=25 discard 0.

or b+h=100
b=(100*75/(75+h))+h=100


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