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a mercury barometer reads 75cm. Now 3cm^3 of atmospheric air is introduced into the tube. The mercury level falls to a height of 65cm and the length of the air column above the mercury is found to be 15cm. Calculate the cross-sectional area of the barometer tube.

Hello Jordan,

Converting the pressure of 75 cm Hg to Pascals:

75 cm Hg X (1333 Pa / 1 cm Hg) = 1x10^5 Pa = 100 kPa

Then 3 cm^3 of air at that pressure is added to the vacuum that was above the Hg. So the added air pushed the Hg level down 10 cm. The pressure of the added air above the Hg is exerting a downward force on the column of Hg equal to the weight of the Hg that was between the 65 cm and 75 cm points before the air was added. Let the cross-sectional area be called A and the units be meters^2.

So the volume of the missing Hg, the Hg that had been between the 65 and 75 cm points, is A*10 cm. The mass, M, of that Hg is the density of Hg times A*10 cm or A*0.1 m.

M = (13.6x10^3 Kg/m^3)*A*0.1 m

And the weight of that Hg is

W = M*g = (13.6x10^3 Kg/m^3)*A*0.1 m*g = 13.33x10^3*A kg/(m.s^2)

We know that A has units of m^2. Let's let A' be the magnitude of A so that we can include the area's m^2 units in the simplification of units. If we look at all the units involved here, we find that it reduces to kg.m/s^2 which is equivalent to Newtons. So the weight, W, is 13.33x10^3*A' Newtons.

As I said above, that weight is equal to the downward force of the pressure of the added air above the Hg. That force, F, is the pressure of that air times the cross-sectional area of the column of Hg.

F = P*A = 100 kPa*A

So the weight, W, equals the force, F.

13.33x10^3*A' N = 100 kPa*A

Hmmm. The magnitude of A cancels out. Ahh, I haven't used the 15 cm detail yet.

The pressure of the air was 100 kPa. Now the pressure above the Hg must be W/A.

13.33x10^3*A' N / A = 13.33x10^3 N/m^2 = 13.33x10^3 Pa

From Boyle's Law:

P1*V1 = P2*V2

We need the original volume in m^3

3 cm^3 (1 m^3 / (100 cm/1 m)^3 = 3x10^-6 m^3

100 kPa*3x10^-6 m^3= 13.33 kPa*V2

Now, I'll leave the rest to you. Solve for V2 and use that to find the cross-sectional area. Also, go through my arithmetic carefully!

I hope this helps,

Steve

Physics

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