Finding the height of a column of mercury

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Homework Statement

The U-shaped tube in the figure has a total length of 1.0 m. It is open at one end, closed at the other, and is initially filled with air at 20C and 1.0 atm pressure. Mercury is poured slowly into the open end without letting any air escape, thus compressing the air. This is continued until the open side of the tube is completely filled with mercury.

What is the length of the column of mercury?

The density of mercury is 13,600 kg/m^3. Hint: it might help to define a variable for the cross-sectional area of the tube, then find your answer for L. You will find that the area drops out of your equations.

Homework Equations

(PV/T)_1 =(PV/T)_2 or
PV = nRT

The Attempt at a Solution

We know that the tube length is 1 m, that it is filled with 20 C air at 1 atm. We know that whatever amount of mercury must create a pressure equal to the pressure created by the air. So mg/A= . We know that the volume of the mercury is equal to length L plus a cross sectional area of the tube, so DALg/A= DLg=P_2. I assume constant temperature so (PV)_1= (PV)_2 = (DLgV)_2.

V_1 = (1)A and V_2 = (1-L)A
(PA)_1= (DLg(1-L)A)_2
(P)_1= (DLg(1-L))_2

P = 1 atm, D is the density listed in question, L is the length of mercury, g is gravitational acceleration. Then solve for L I guess. Is this correct?

 

[Daruf Sambid]

Looks like a good setup

 

[haruspex]

DALg/A= DLg=P_2

You appear to be making an assumption about the final arrangement, and I suspect it is wrong. If so, you should get a crazy answer.

Also, an excellent habit to get into is to ignore all the numerical values you are given (like, 1m here) and work entirely symbolically. Amongst the many advantages is that it makes your working much easier to follow. You did that for all except the length of the tube.