# Volume of water consumed by vacuum

• I
• northmop
In summary, Boyle's Law explains how the volume of a gas or liquid changes when pressure is applied. In this experiment, the water in the container would have to reduce the volume of gas in the container by a factor of 96/98.5 in order to achieve a pressure of 98.5 kPa.
northmop
I'm trying to do the math for a backyard thermodynamic pump experiment, and getting stuck. Suppose an empty 3.79 liter container was airlocked by a vertical 25 cm long empty tube whose end sits at the top of an unlimited supply of water in a basin. The tube protrudes through the container such that there is a length of tube above the base of the container so that any water entering the container cannot escape back down the tube and into the basin. If the temperature of the external environment, water, and gas are all constant (295°K), what volume of water (density of 1000 kg/m3) would enter the container if the pressure differential between of the external environment (100 kPa) and the inside the container (96 kPa) is 4 kPa?

I know that some of the vacuum pressure will be consumed to maintain the 25 cm height of the water column in the tube. This means that 4 kPa - (9.80665 m/s2 * 1000 kg/m3 * 0.25 m) = 1.54834 kPa is "available" to vacuum up the water from the basin into the container. What I'm getting stuck with is how to calculate the hypothetical volume of water that will get sucked up and stored in the container: my head is mush after thinking about this too long. Would you please be so kind as to share the formulas used to figure this out?

Last edited:
I wanted to post this separately so that I don't confuse the question. Here's the short of my experiment. During the day the container warms up. As night passes, the container cools down and sucks up liquid. Using the Ideal Gas Law, one can determine a change in volume from the hottest point in the day (H) and the coolest point in the day (C). If the container was malleable, the volume of the container would change, either by expanding or collapsing, until the internal pressure of the container was equalized with the pressure of its external environment. That change in volume, ΔV, can be calculated as:

ΔV = VH – VC
VC = VH – ΔV
PH VH = n R TH
n = ( PH VH ) / ( R TH )
PC VC = n R TC
PC ( VH – ΔV ) = [ ( PH VH ) / ( R TH ) ] ( R TC )
PC VH – PC ΔV = ( PH VH R TC ) / ( R TH )
PC VH – PC ΔV = ( PH VH TC ) / ( TH )
– PC ΔV = ( PH VH TC ) / ( TH ) – ( PC VH )
PC ΔV = – ( PH VH TC ) / ( TH ) + ( PC VH )
ΔV = [ – ( PH VH TC ) / ( TH ) + ( PC VH ) ] / PC
ΔV = – ( PH VH TC ) / ( PC TH ) + ( PC VH / PC )
ΔV = – ( PH VH TC ) / ( PC TH ) + VH
ΔV = VH – ( PH VH TC ) / ( PC TH )
ΔV = VH [ 1 – ( PH TC ) / ( PC TH ) ]
ΔV = VH [ 1 – ( PH / PC ) ( TC / TH ) ]​

However, the jug is not malleable meaning that VH=VC and V is therefore a constant. A volume of matter will enter the jug when ΔV > 0 and exist the jug when ΔV < 0 in order to equalize its pressure with that of the external environment, PE. As my concern is when ΔV > 0, PC = PE and I get a final equation of:

ΔV = V [ 1 – ( PH / PE ) ( TC / TH ) ]​

What gives me the willies is that, though this theoretically will yield the volume of water sucked into the straw and container, it does not take into account having to overcome gravity to lift the water from the basin and maintain the water column as the container cools and fills; and hence why my head is mush and my question above.

Can you please provide a diagram (sketch)?

Delta2
Chestermiller said:
Can you please provide a diagram (sketch)?
How's this?

Delta2
The internal pressure would have to rise from 96 kPa to 98.5 kPa in order to stop the flow up the tube. So the liquid in the container would have to reduce the volume of gas in the container by a factor of 96/98.5.

Delta2 and northmop
Chestermiller said:
The internal pressure would have to rise from 96 kPa to 98.5 kPa in order to stop the flow up the tube. So the liquid in the container would have to reduce the volume of gas in the container by a factor of 96/98.5.
Now that you point it out, the answer is pretty obvious: Boyle's Law to the rescue!

P1V1= P2V2
96 kPa * 3.79 L = 98.5 kPa * V
V = 96 kPa * 3.79 L / 98.5 kPa
V = 3.69 L​

About 100 mL of water would enter the container. I was really overthinking the simplest thing. Thank you, Chestermiller, for putting my feet back on the ground.

Chestermiller and Delta2

## 1. What is the volume of water consumed by a vacuum?

The volume of water consumed by a vacuum depends on the size and strength of the vacuum. Generally, a vacuum will consume a small amount of water, usually less than a milliliter, as it is used to create suction and remove air from a space.

## 2. How does a vacuum consume water?

A vacuum consumes water through the process of suction. As the vacuum removes air from a space, it also pulls in any liquid that may be present, including water. This water is then collected in a tank or filter within the vacuum.

## 3. Is it safe to consume the water collected by a vacuum?

No, it is not safe to consume the water collected by a vacuum. This water may contain dirt, debris, and bacteria that have been picked up from surfaces and floors. It is important to dispose of this water properly and not use it for drinking or cooking.

## 4. How much water does a vacuum use compared to other cleaning methods?

Vacuums use significantly less water compared to other cleaning methods, such as mopping or steam cleaning. This is because vacuums primarily use suction to remove dirt and debris, rather than water. This makes them a more efficient and environmentally friendly option for cleaning.

## 5. Can a vacuum consume too much water?

Yes, a vacuum can consume too much water if it is not properly maintained. If the tank or filter is not emptied regularly, it can become full and cause the vacuum to malfunction. It is important to follow the manufacturer's instructions for emptying and cleaning the vacuum to prevent this from happening.

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