Accounting for Water Vapor Condensation in Charles' Law Experiment Measurements

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Homework Help Overview

The discussion revolves around an experiment designed to validate Charles' Law, where a plastic dropper submerged in salty water is placed in a freezer, leading to a rise in the water level within the dropper. The participants are examining the impact of water vapor condensation on the measurements taken during the experiment.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to understand the significance of water vapor condensation on the water level observed in the dropper. Questions are raised about whether the observed changes are solely due to the salty water or if condensation from the air contributes to the measurements.

Discussion Status

Some participants have acknowledged the possibility that the water level change may include contributions from condensed water vapor. Others are probing deeper into the implications of this condensation and how it relates to the overall experiment, seeking clarification on the extent of its significance.

Contextual Notes

There is an emphasis on estimating the significance of the condensation effect relative to the primary changes expected from the experiment. Participants are also considering the equilibrium of water vapor pressure and its implications for the number of moles of gas involved in the experiment.

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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?
 
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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.
 
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?!
 
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.
 
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.
 
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