Isothermal pressure change in a U-shaped tube

In summary, the conversation revolves around an isothermal change in pressure involving a vacuum pump and mercury. The question is raised about the end state and the possibility of the mercury being completely sucked into the right side of the tube due to the vacuum pump. The concept of atmospheric pressure and the equation ##\Delta p=\rho g h## are suggested as hints to help understand the problem. It is also mentioned that vacuum does not suck and the weight of the mercury and its incompressibility should be taken into consideration. Finally, it is noted that the barometric pressure is 750 Torr and the question is posed about the distance the atmospheric pressure can push a column of mercury into a vacuum.
  • #1
obstinatus
12
0
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
IMG_20200212_153819730.jpg


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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  • #2
Are you familiar with the equation ##\Delta p=\rho g h##? Do you think that it might apply to this problem?
 
  • #3
Two hints: Mercury is effectively incompressible, and Mercury is very heavy.
 
  • #4
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?
obstinatus said:
... mercury sucked completely into the right side...
Perhaps that statement needs to be thought about some more.
 
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  • #5
obstinatus said:
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.
 

1. What is an isothermal pressure change in a U-shaped tube?

An isothermal pressure change in a U-shaped tube refers to a change in the pressure of a gas or liquid that occurs while the temperature of the system remains constant. This is typically observed when a gas or liquid is confined to a U-shaped tube and one side of the tube is subjected to a change in pressure.

2. How does an isothermal pressure change occur in a U-shaped tube?

An isothermal pressure change in a U-shaped tube occurs when there is a difference in pressure between the two arms of the tube. This could be due to a change in the volume of gas or liquid on one side, or a change in the external pressure on one side of the tube.

3. What factors can affect an isothermal pressure change in a U-shaped tube?

The main factors that can affect an isothermal pressure change in a U-shaped tube include the temperature of the system, the volume of the gas or liquid in the tube, and the external pressure acting on one side of the tube. Additionally, the properties of the gas or liquid, such as its compressibility and density, can also play a role in the pressure change.

4. How is an isothermal pressure change in a U-shaped tube measured?

An isothermal pressure change in a U-shaped tube can be measured using a manometer, which is a device that measures the difference in pressure between two points. The manometer is connected to one arm of the U-shaped tube and the other arm is connected to the system being studied. The difference in pressure between the two arms of the tube can then be read on the manometer's scale.

5. What are the practical applications of studying isothermal pressure changes in U-shaped tubes?

Studying isothermal pressure changes in U-shaped tubes has practical applications in various industries, such as chemistry, physics, and engineering. It can be used to measure the pressure of gases or liquids in closed systems, as well as to understand the behavior of gases and liquids under different conditions. This knowledge can be applied in the design and optimization of systems and processes involving gases and liquids, such as in chemical reactions and fluid flow systems.

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