[Chemistry] Vapor pressure of water at temperature

  • Thread starter RossH
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Homework Statement


Hi. I'm working on a lab in which I must find the vapor pressure of water. Data that I know is the temperature of the system, the volume of trapped air (2.35 mL), and the atmospheric pressure (1.024 atm). This means that I have 1.1x10^-4 mol of trapped air.

Beyond this, I took temperature and volume readings of the system at 5 deg C intervals as it cooled from 80 deg C. So at 80 deg C I know that the volume of the trapped air was 5.85 mL. The problem is that I am not sure how to differentiate between pressure from the temperature of the air changing and pressure from the temperature of the water changing.


Homework Equations


I know that pv=nRT. I am using a value of R=0/082Latm/Kmol.
I know that Pair=Ptotal-PH20.


The Attempt at a Solution



So my big question is if the temperature of the water outside of the system is changing, does this mean that the pressure of the air and the pressure of the water will change? Or is the pressure a function of pv=nrt as with the information that I found in the beginning? I'm obviously not allowed to use any information from the internet other than these equations, since this is a lab. Thank you for any help that you ca give.
 

Answers and Replies

  • #2
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I think I might have figured it out. Is this it: atmospheric pressure is always equal to total pressure inside the bubble? Then I would use pv-nrt to figure out P_air and P_h20=Atmospheric pressure - P_air. Is this correct?
 

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