Air is compressed in a cylinder

  • Thread starter Karol
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  • #1
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Homework Statement


A cylinder of height 1[m] has a weightless piston on it's head. mercury is spilled slowly on the piston and it goes in. at what distance will the mercury start to spill over the cylinder. the process is at constant temperature.

Homework Equations


$$PV=nRT$$
At constant temperature: P1V1=P2V2.
The density of mercury: ρ=13.6[gr/cm3]
Hydrostatic pressure: ##P=h\rho g##
Definition of 1[atm]=101,325[pa=N/m[SUP]3[/SUP]]

3. The Attempt at a Solution

The pressure of the mercury: ##P=h\cdot 13,600\left[\frac{kg}{m^3}\right]\cdot 9.8\left[\frac{m}{sec^2}\right]=133,280\cdot h##
A is the base area of the cylinder.
$$101,325[pa]\cdot A\cdot 1[m]=133,280\cdot h\cdot (1-h)\cdot A$$
$$\rightarrow h^2-h+0.76=0$$
And it gives a negative discriminant
 

Answers and Replies

  • #2
TSny
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$$101,325[pa]\cdot A\cdot 1[m]=133,280\cdot h\cdot (1-h)\cdot A$$
Hello. What was the reasoning that led you to this equation?
 
  • #3
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You're dealing with mercury and air here. One is a liquid, and the other a gas. Can you really use the equation ##p_1V_1 = p_2V_2##?
 
  • #4
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The mercury doesn't mix with the air, it doesn't enter inside the cylinder. the piston goes down and a cup, a vessel is formed on the top of the cylinder with the piston as the bottom of the cup.
$$101,325[pa]\cdot A\cdot 1[m]=(133,280\cdot h)((1-h)\cdot A)$$
##(133,280\cdot h)## is the new pressure, the pressure of the mercury, P2, and ##((1[m]-h)\cdot A)## is the new volume, V2.
I forgot to state at the beginning that the absolute pressure in the cylinder at the beginning was 1[atm]
 
  • #5
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I think i solved. the absolute pressure at the end is the pressure of the mercury+1[atm]
 
  • #6
TSny
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I think i solved. the absolute pressure at the end is the pressure of the mercury+1[atm]
OK.
 

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