Mercury barometer with a small amount of air

[plain stupid]

Homework Statement

We have a faulty mercury barometer with a small amount of air trapped at the very top, above the mercury. For an atmospheric pressure of 768 mmHg, our barometer shows 761 mmHg, and for an outside pressure of 740 mmHg, it shows 736 mmHg. What's the length of the barometer tube?

Relevant Equations

p = ρgh

$$\rho_{Hg} gh_{actual} = \rho_{Hg} gh_{measured} + \rho_{air} gh_{air}$$

Note: by "actual", I mean "theoretical", i.e. what the barometer would measure were there no air inside it. By "measured" I mean "as measured by the faulty barometer, i.e. with some air introduced".

I believe this assumption should hold, because those are the only things in that tube. The pressures should be in an equilibrium, meaning the pressure from the outside (actual pressure) should correspond to the pressure exerted by the column of mercury + the tiny bit of air. I'm assuming here that because air is a gas (or multiple gasses) that it should spread out evenly, so its "height" should correspond to the height of the remaining part of the tube. I don't think I should assume its density is known, because it should change (since its mass remains the same, but its volume varies).

I can cross out the gravitational constant from both sides, and replace the density of air with its mass over its volume, assuming the tube's cross-sectional area is constant throughout:

$$\rho_{Hg} h_{actual} = \rho_{Hg} h_{measured} + \frac{m_{air}}{A_{tube} h_{air}} h_{air}$$

simplifying yields

$$\rho_{Hg} \left(h_{actual} - h_{measured} \right) = \frac{m_{air}}{A_{tube}}$$

Now, I'm pretty sure this ratio of air mass and barometer cross-sectional area should be constant, and so should be the density of mercury, so I'm not at all sure what I've arrived at here. What am I even looking for here? I've been looking at this for a couple of days now, but am not sure what I'm missing.

Also, it's not clear what's exactly given (known) aside from the obvious, but I think we could assume the density of mercury could be read from a table, and even the average density of air, if that could be of help.

I'm not really sure how to proceed from here, even though the problem statement suggests there should somehow be two equations with two unknowns...

Any hints? Thank you!

 

[haruspex]

Your first equation is wrong. Air pressure in a closed space isn't just a question of its weight, e.g. what would happen if the air were heated?
Also, your use of "actual" is a bit misleading. The measured height is the actual height. You mean the height it would be if there were no air. Maybe "theoretical"?

 

[plain stupid]

Your first equation is wrong. Air pressure in a closed space isn't just a question of its weight, e.g. what would happen if the air were heated?
Also, your use of "actual" is a bit misleading. The measured height is the actual height. You mean the height it would be if there were no air. Maybe "theoretical"?

I don't think there are any indications that the air is heated or that the temperature changes (I assumed an equilibrium state). Do you think this is actually pertinent considering how relatively basic this problem is?

Yeah, by "actual" I did mean "theoretical", but if the barometer is faulty, then we can reasonably talk about an actual pressure (in terms of the millimeters of mercury) vs. a measured one (on the faulty barometer). I'll try to edit my post to clarify what I meant regardless.

 

[plain stupid]

Should I try to play with my setup and get the result? I.e. am I on the right way (perhaps extracting that height of air and cancelling those terms in particular wasn't the right idea, but I should continue playing with something similar)?

 

[haruspex]

I don't think there are any indications that the air is heated

You are missing the point. The way you have calculated the pressure the air exerts is as though the tube continues up into space and is open at the top. It would be minute.
Pressure in a closed volume is given by PV=nRT.

 

[Lnewqban]

Once the vacuum is broken, the column may reach a very low level.
If we imagine the air bubble as being big enough, it could completely fill up the space within the tube while being at atmospheric pressure, for which condition the height of the column of mercury would be zero.

If we then apply a vacuum pump at the top end of the vertical tube, the level of the column would increasely go up as mass of air is being removed from the cavity and its absolute pressure decreases.

When a high degree of vacuum is reached inside the top cavity of the tube, for an atmospheric pressure of 768 mm Hg, our barometer would show 768 mm above the open free surface of mercury.

 

[plain stupid]

You are missing the point. The way you have calculated the pressure the air exerts is as though the tube continues up into space and is open at the top. It would be minute.
Pressure in a closed volume is given by PV=nRT.

Okay, I see. Thank you very much! I'm sorry for not realizing the hint in your first reply, and having you spell it out for me.

Taking that into account, I get the right solution. I'll show a few steps below just in case someone else is interested.

$$\rho_{Hg}gh_{theoretical} = \rho_{Hg} gh_{measured} + \frac{nRT}{V_{air}}$$

$$\rho_{Hg}g(h_{theoretical}-h_{measured}) = \frac{1}{h_{air}} \frac{nRT}{A_{tube}}$$

Simplifying ($C$ is a constant and doesn't change, assuming the temperature of air is the same for both cases, $R$ is a constant, and the number of particles of air is conserved)...

$$\rho_{Hg} \cdot g \cdot (h_{theoretical}-h_{measured}) \cdot h_{air} = C$$

The above has to be evaluated for the two cases (measurements), and from those we get the ratio between two heights of air columns.

$$\frac{h_{air,1}}{h_{air,2}} = ratio$$

There's also a second equation we can easily come up with, since the height of the whole tube doesn't change:

$$h_{measured,1} + h_{air,1} = h_{measured,2} + h_{air,2}$$

These two equations are enough to get the solution.

 

[haruspex]

Okay, I see. Thank you very much! I'm sorry for not realizing the hint in your first reply, and having you spell it out for me.

Taking that into account, I get the right solution. I'll show a few steps below just in case someone else is interested.

$$\rho_{Hg}gh_{theoretical} = \rho_{Hg} gh_{measured} + \frac{nRT}{V_{air}}$$

$$\rho_{Hg}g(h_{theoretical}-h_{measured}) = \frac{1}{h_{air}} \frac{nRT}{A_{tube}}$$

Simplifying ($C$ is a constant and doesn't change, assuming the temperature of air is the same for both cases, $R$ is a constant, and the number of particles of air is conserved)...

$$\rho_{Hg} \cdot g \cdot (h_{theoretical}-h_{measured}) \cdot h_{air} = C$$

The above has to be evaluated for the two cases (measurements), and from those we get the ratio between two heights of air columns.

$$h_{air,1} \cdot h_{air,2} = ratio$$

There's also a second equation we can easily come up with, since the height of the whole tube doesn't change:

$$h_{measured,1} + h_{air,1} = h_{measured,2} + h_{air,2}$$

These two equations are enough to get the solution.

Very good. Thanks for posting your solution.