# A U-shaped Tube problem….

1. The problem statement, all variables and given/known data

A u-shaped tube with a 1 square centimeter cross-section is filled with mercury to a depth of 50 centimeters on each side.

One side is capped and the air pressure is 750 Torr. The other side is connected to a vacuum pump.

a) how far does the mercury in the dupe drop on the left side? b) what is the final pressure of the air there?

(We are assuming a constant temperature)

2. Relevant equations

I’m using the old perfect gas law, PV = constant.

3. The attempt at a solution

OK, so I know that the mercury will move h centimeters on one side (down) and h on the other (up). So the total distance the mercury moves is 2h.

If I assume that the pressure in the part of the tube where the air is starts at P0V0$and goes to PfinalVfinal I can do this: [itex]P_0V_0 = P_{final}V_{final}$ because both PVs = nRT.

I also know that the pressure exerted by the mercury is $50{\rho}_{Hg}$ because there are 50 centimeters worth of it at the start.

Since the pressure of the air increases the volume on the left side I know that

$P_{final}V_{final} = P_{final}(V_0+h)$ and since the cross section is 1 cm2 we don’t have to worry about area here, it’s just 1.

Meanwhile, the mercury will (once everything settles to equilibrium) exerts the same pressure, but it will be ${\rho}_{Hg}(50+2h)$.

That should give me $P_{final}V_{final} = P_{final}(V_0+h) = {\rho}_{Hg}(50+2h)(50+2h)$ and since we know V0 (It’s 50) then:

$P_{final}V_{final} = P_{final}(50+h) = (50{\rho}_{Hg}+2h{\rho}_{Hg})(50+2h)$

And this is where I suspect I am going wrong. Or maybe I just did something algebraically that messed it up. I don’t thnik I should be getting h2 factors.

I know the answer it supposed to be 25 centimeters and 500 Torr for the pressure of the air. And sort of intuitively I know it. But I am trying to work this out correctly. I’ve seen similar problems on the forum here but none quite like this.