Accounting for Water Vapor Condensation in Charles' Law Experiment Measurements

In summary, the experiment tested Charles' Law by observing the change in water level in a plastic dropper submerged in salty water and then placed in a freezer. It was noted that air contains water vapor which can condense in cold temperatures, potentially affecting the water level measurements. The significance of this condensation needs to be estimated and corrected for in order to accurately test Charles' Law. This can be done by considering the change in number of moles of gas and the partial pressure of air within the gas.
  • #1
sunflowerzz
25
0

Homework Statement


Background information: I did an experiment trying to support the validity of Charles' Law - we put a plastic dropper in a container and submerged it in salty water and then put it in the freezer. We watched the water level rise in the plastic dropper.

The question is: since we know that air contains water vapour which condenses in cold temperature, can you estimate the significance of this condensation on your measurements?


Homework Equations




The Attempt at a Solution



I'm just not sure what this means? I know that the change in the water column is due to the combined change of the volume of air inside the dropper and the volume of the plastic dropper itself (which I found to be 6 mL)

But I'm not sure how that helps me? Is there supposed to be a numerical answer or just an explanation?

So, air contains water vapour that condenses in cold temperature - which means it changes from a gaseous state to a liquid state - does that mean my water level might not be from the salty water that it is submerged in? That is, part of it could come from the condensation of the water vapour in the air?
 
Physics news on Phys.org
  • #2
sunflowerzz said:
So, air contains water vapour that condenses in cold temperature - which means it changes from a gaseous state to a liquid state - does that mean my water level might not be from the salty water that it is submerged in? That is, part of it could come from the condensation of the water vapour in the air?

Yes.
 
  • #3
voko said:
Yes.

Is that all the question was asking? Just saying that additional water could have come from the water vapour in the air? That's the only significance?

It feels like there should be more to it?!
 
  • #4
sunflowerzz said:
Is that all the question was asking? Just saying that additional water could have come from the water vapour in the air? That's the only significance?

It feels like there should be more to it?!

The question is asking "estimate the significance". That is, how much water that could be and how that compares with the primary effect under the experiment.
 
  • #5
In the experiments you are doing, the water vapor pressure within the gas is always in equilibrium with the liquid water at the temperature of the system. When you raise the temperature, some of the liquid water evaporates and raises the number of moles of gas; when you lower the temperature, some of the water vapor in the gas condenses, which lowers the number of moles of gas. Charles' law is based on assuming that the number of moles of gas is constant. So you need to figure out how to correct for the change in the number of moles of gas when you change the temperature.

More importantly, the partial pressure of the air is changing, even though the total pressure of the gas is constant. The partial pressure of the air within the gas plus the partial pressure of the water water vapor within the gas adds up to the total pressure of the gas. When water condenses out, the partial pressure of the water within the gas decreases, and the partial pressure of air within the gas increases. If you apply the full ideal gas law to the air within the gas, you can readily calculate the correction factor you need to test Charles' law for this experiment.
 
Last edited:

Related to Accounting for Water Vapor Condensation in Charles' Law Experiment Measurements

What is Charles' Law Experiment?

Charles' Law Experiment is an experiment that illustrates the relationship between the volume and temperature of a gas at constant pressure. It states that the volume of a gas is directly proportional to its temperature, as long as the pressure and amount of gas remain constant.

What materials are needed for the Charles' Law Experiment?

The materials needed for the Charles' Law Experiment include a gas supply, a gas syringe or a container with a movable piston, a thermometer, and a heat source such as a Bunsen burner or hot plate.

What is the procedure for conducting the Charles' Law Experiment?

The procedure for conducting the Charles' Law Experiment involves setting up the gas syringe or container with a movable piston, measuring the initial volume and temperature of the gas, heating the gas to increase its temperature, and then measuring the new volume and temperature. The data can then be plotted on a graph to observe the direct relationship between volume and temperature.

What are the possible sources of error in the Charles' Law Experiment?

Some possible sources of error in the Charles' Law Experiment include heat loss to the surroundings, errors in measuring the volume and temperature, and presence of impurities in the gas. It is important to control these variables in order to obtain accurate results.

What are the real-world applications of Charles' Law?

Charles' Law has various real-world applications, such as in hot air balloons, where the air inside the balloon is heated to increase its volume and lift the balloon. It is also used in refrigeration systems, where the volume of the gas is decreased by cooling it in order to lower the temperature. Additionally, Charles' Law is also applied in the production of aerosol cans and in diving, where changes in pressure and temperature affect the volume of gases in the body.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
396
  • Introductory Physics Homework Help
Replies
5
Views
629
  • Introductory Physics Homework Help
Replies
4
Views
969
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Mechanical Engineering
Replies
28
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
911
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top