Mercury barometer problem

  • #1
utkarshakash
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Homework Statement


To construct a barometer, a tube of length 1m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76cm. Suppose a 1m tube is filled with mercury up to 76cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of mercury column in the tube over the surface in the
cup will be

a)zero b)76cm c)>76cm d)<76cm

Homework Equations



The Attempt at a Solution


Since the tube has been partially filled thus air is present in it. When it is inverted the air exerts a pressure on the mercury filled in it. Also the atm .pressure is exerted at the surface of mercury present in cup. But which one is greater?
 

Answers and Replies

  • #2
The volume of the 24 cm column of air will probably increase once the tube is inverted.
 
  • #4
utkarshakash
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855
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The volume of the 24 cm column of air will probably increase once the tube is inverted.
Why?
 
  • #5
The downwards pressure exected by the air and mercury column is balanced by the upwards atmospheric pressure.
 
Last edited:
  • #6
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To answer the question in my previous post, you need to use the ideal gas law. You end up with the equation

100 - x + (24)(76)/x = 76

where x is the distance that the 24 cm increases to. The answer is x = 56.4 cm. That leaves a Hg column of 43.6 cm.
 
  • #7
utkarshakash
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100 - x + (24)(76)/x = 76

.
Can you please make it more clear by stating what those numbers mean?
 
  • #8
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Can you please make it more clear by stating what those numbers mean?
The (100 - x) is the hydrostatic pressure of the Hg column in cm of Hg.

The (24)(76)/x is the pressure of the air bubble in cm of Hg. The 76 in this expression is the pressure of the air in the bubble before the bubble expands downward.

The pressure of the Hg column plus the pressure of the air bubble must equal the pressure of the atmosphere outside the tube at the Hg surface, which is equal to 76 cm of Hg.
 
  • #9
utkarshakash
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.

The (24)(76)/x is the pressure of the air bubble in cm of Hg. The 76 in this expression is the pressure of the air in the bubble before the bubble expands downward.
Are you using PV=const equation?
 
  • #10
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Are you using PV=const equation?
Yes. Before the Hg level drops, the pressure in the air bubble is 76 cm Hg, and the length of the air bubble is 24 cm. If the area of the cylinder is A, the initial volume of the air bubble is 24A. After the Hg level drops, the length of the air bubble is x, and the volume of the air bubble is xA. So the volume ratio is (24A)/(xA). The new pressure in the expanded air bubble is

(76)(24A)/(xA) = (76)(24)/x cm Hg.
 
  • #11
utkarshakash
Gold Member
855
13
Yes. Before the Hg level drops, the pressure in the air bubble is 76 cm Hg, and the length of the air bubble is 24 cm. If the area of the cylinder is A, the initial volume of the air bubble is 24A. After the Hg level drops, the length of the air bubble is x, and the volume of the air bubble is xA. So the volume ratio is (24A)/(xA). The new pressure in the expanded air bubble is

(76)(24A)/(xA) = (76)(24)/x cm Hg.
OK I have understood.
 

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