# Isothermal pressure change in a U-shaped tube

## Homework Statement:

The U-tube in figure 2-18, of uniform cross section 1 cm^2, contains mercury to the depth shown. The barometric pressure is 750 Torr. The left side of the tube is now closed at the top and the right side is connected to a good vacuum pump. A) How far does the mercury level fall in the left side and b) what is the final pressure of the trapped air? The temperature remains constant.

## Relevant Equations:

Pv = RT

Hi, just reviewing some thermodynamics from the textbook by Sears and Salinger, having a hard time conceptualizing this one. It's an isothermal change in pressure, so the volumes of the mercury and the air both change to reach equilibrium, but if it's a "good vacuum pump", then won't the right side be completely evacuated, and the mercury sucked completely into the right side? Half of the air molecules will be evacuated, so all else being equal, the pressure of the air should be reduced by half, but because of this arrangement, we have one end of the mercury subject to zero pressure and the other subject to normal pressure. I just don't see how any end state can be specified besides all of the mercury sucked into the right side of the tube. Thanks in advance for any hints!

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Chestermiller
Mentor
Are you familiar with the equation ##\Delta p=\rho g h##? Do you think that it might apply to this problem?

Two hints: Mercury is effectively incompressible, and Mercury is very heavy.

256bits
Gold Member
The barometric pressure is 750 Torr
Which is how many mm that the atmospheric pressure can push a column of mercury to rise into a vacuum?
... mercury sucked completely into the right side...
Perhaps that statement needs to be thought about some more.

Chestermiller
I just don't see how any end state can be specified besides all of the mercury sucked into the right side of the tube.
Having heard nothing, another hint: Vacuum does not “suck”. The existence of a vacuum above one surface of the Mercury imparts no force on the Mercury.