How does atmospheric pressure affect the readings in a mercury tube manometer?

[Pengwuino]

Ok I got this problem here...

http://img.photobucket.com/albums/v81/Veto1024/mercury.jpg

How is this possible?!

If its pushing the mercury down the tube and back up into the atmosphere, the pressure has to be greater then 1atm yet the answer ends up being less then 1 atm inside the flask.

Also, we were tough to just add whatever the displacement was to 760 torr to find the correct pressure. Now I added (14.3in * 25.4 in/mm) + 760 torr to get 1122 torr which, when divided by 760, yields 1.48atm. Now even if I was wrong and was suppose to subtract, I would get (760-363.22) = 396.78 torr = .522atm yet even that's not what the program accepted.

What the crap? Is it wrong to use 760torr as standard atmospheric pressure?

 

[Andrew Mason]

Ok I got this problem here...

http://img.photobucket.com/albums/v81/Veto1024/mercury.jpg

How is this possible?!

If its pushing the mercury down the tube and back up into the atmosphere, the pressure has to be greater then 1atm yet the answer ends up being less then 1 atm inside the flask.

There is a vacuum above the mercury column. The difference in height of the mercury gives the pressure in the flask:
$$\rho gh = P$$

Since 14.3 in = 36.3 cm, P = 36.3/76 = .478 atm.

AM

 

[Pengwuino]

But that would mean the vacuum is inside the flask...

 

[Andrew Mason]

But that would mean the vacuum is inside the flask...

No. The vacuum is above the mercury only. That is why the height of mercury measures the pressure on the bottom of the mercury column.

AM

 

[Pengwuino]

Ok if the vacuum is at the top of the column... wouldn't that mean the pressure inside hte flask is higher then the outside?

 

[HallsofIvy]

The "pressure outside the flask" has nothing to do with this problem because the flask is sealed. There are two forces acting on the column of mercury- the weight of the mercury pressing it down and the force upward due to the pressure of the gas in the flask. Since the mercury is in equilibrium, those two forces are equal. The weight of the mercury is, of course, its density times volume: letting A be the cross section area of the mercury in square inches and $\rho$ the (weight) density of mercury, the weight of the mercury is $14.3\rhoA$. Letting P be the pressure of the gas in the flask, the force on the mercury is PA. Set those equal and solve for P, being careful to convert units.

 

[Pengwuino]

The "pressure outside the flask" has nothing to do with this problem because the flask is sealed. There are two forces acting on the column of mercury- the weight of the mercury pressing it down and the force upward due to the pressure of the gas in the flask. Since the mercury is in equilibrium, those two forces are equal. The weight of the mercury is, of course, its density times volume: letting A be the cross section area of the mercury in square inches and $\rho$ the (weight) density of mercury, the weight of the mercury is $14.3\rhoA$. Letting P be the pressure of the gas in the flask, the force on the mercury is PA. Set those equal and solve for P, being careful to convert units.

I thought the left side of the tubing, past the mercury, was open to the atmosphere? Thats what the chapter was about, measuring the air pressure in the flask vs. the atmospheric temperature.

 

[lightgrav]

The tube is round on the top, so it is like a test-tube bottom.
Besides, they *call* it a sealed-tube manometer.

Physics concepts are always about *one item* ...
your chapter was about air pressure in the flask
as a function of Temperature IN THE FLASK,
and the amount of material IN THE FLASK
compared to the Volume OF the FLASK.

 

[Pengwuino]

Oh well this is weird...

In the previous problem, it was open to the atmosphere and they called it a sealed-tube manometer and the equation used for open-atmosphere manometers worked correctly.