1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Charles' Law Experiment

  1. Jul 24, 2013 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Jul 24, 2013 #2
  4. Jul 24, 2013 #3
    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?!
  5. Jul 24, 2013 #4
    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.
  6. Jul 25, 2013 #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: Jul 25, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted