# An interesting spin on the mercury u tube

Suppose I have a u-shaped tube of diameter d. I put mercury in the tube so everything is nice and even up to a height, h, with a pressure, P, above the mercury. Now suppose I seal off one side of the tube, and attach a vacuum pump to the other, reducing the pressure on this side to zero. Assuming a constant temperature how can I determine the final height of the mercury?

Initially I thought to approach the problem in a manner similar to the typical mercury u-tube, with some other fluid introduced. But the sealing off of one end means that the pressure on the left is decreasing as the gas cavity expands. So using the ideal gas law:

\begin{align} P_fV_f = P_iP_f \\ P_f = \frac{P_iV_i}{V_f} \end{align}

Then considering that the pressure from the gas must balance the pressure from the column, which is,

$$P_f = \rho g\Delta h$$

I then get,

$$\frac{P_iV_i}{V_f} = \rho g\Delta h$$

and now let $$\Delta y$$ be the height drop of the left hand side, so $$2\Delta y = h$$, then since $$V_f = \Delta y A + V_i$$, I end up with a quadratic:

$$2\rho gA\Delta y^2 + 2\rho gV_i\Delta y - P_iV_i = 0$$

Which can be solved for $$\Delta y$$. I'm not sure if the reasoning I've used is correct, or if there is a more simple way to do this.

• I have a problem with your $V_f=\Delta h A$, as it assumes that $V_i=0$. Maybe this is so, but it is not obvious from your question. – akhmeteli Feb 23 at 11:10
• I have a problem with your $V_f = \Delta h A + V_i$ as well even after you corrected it. It should be $V_f = 0.5 \Delta h A + V_i$ – user115350 Feb 24 at 23:14
• Thank you, corrected. Does that look right? – monkeyofscience Feb 24 at 23:16