[Chemistry] Trying to find the vapor pressure of water with pv=nRT

  • Thread starter RossH
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  • #1
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Homework Statement


A student found that at 750 torr atmospheric pressure and 0.1 deg C, the corrected volume of the trapped air (in a graduated cylinder filled with water inverted in beaker) was 2.20 mL. Under these conditions, how many moles of trapped air are present?

This student then heated the beaker and found that at 70 deg C, the corrected volume of the bubble was 4.05 mL. What should she report as the vappor pressure of water at this temperature?

My problem is that I am finding that pressure air, from the first part, exceed pressure total, from the second part.


Homework Equations


PV=nRT (ideal gas law)
R=0.082L*atm/K*mol


The Attempt at a Solution



Part 1: PV=nRT. 2.20 mL=0.0022 L. R as above. 750 torr=0.987 atm. 0.1 deg C=273.3K
0.987 atm*0.0022L = n moles*(0.082L*atm/(K*mol))*273.3K
solve for n. n=1*10^(-4) moles.

Part 2: PV=nRT.
P_air=0.987 atm.
4.05mL=0.00405 L.
n=1*10^(-4) moles.
P*0.00405L=0.0001 moles*0.082(L*atm/K*mol)*343.15K
Solve for P: P_total=0.7 atm.

How can P_total be less than P_air? I should be able to get the partial pressure of the water vapor at 70 deg C from this, but that would come out negative. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
Borek
Mentor
28,600
3,081
What you have calculated is the pressure of air, not total pressure. Total is still 750 torr.

Don't round down intermediate results - that is, list them as rounded, but use them in calculations with full accuracy.
 

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