Vapor Pressure Help: Solving \DeltaHvap w/ Assumptions

AI Thread Summary
The discussion revolves around calculating the enthalpy of vaporization (\DeltaHvap) using vapor pressure data at different temperatures. The user, Stephen, applies the formula Ln(P2) - Ln(P1) = -\DeltaH/R (1/T2 - 1/T1) to derive a value of \DeltaHvap as 3.18 * 10^4 J/mol. Stephen also outlines three assumptions made during the lab: ideal gas behavior, constant atmospheric pressure, and homogeneity of air in the bubble. The validity of these assumptions is confirmed by other participants, and they express confidence in the calculated value of \DeltaHvap. Overall, the thread emphasizes the importance of assumptions in thermodynamic calculations.
StephenDoty
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Hello all

I completed a lab last week where we filled a graduated cylinder up 90% of the way with water and inverted it into a water bath and then we heated the bath to 80 degrees Celsius then measured the volume of the air bubble as it cooled to 50 degrees Celsius and at 1 degree Celsius.

My questions are given vapor pressure at 39.5 degrees Celsius is 400 torr. at 7.7 degrees Celsius the vapor pressure is 100 torr. What is the value for \DeltaHvap?

Ln(P2) - Ln(P1) = -\DeltaH/R ( !/T2 - 1/T1)

\DeltaHvap= 3.18 * 10^4 J/mol
right??

And three assumptions made in obtain results in this lab would be:
1. the bubble of air follows the ideal gas laws and not other equations like the van deer walls equation.
2. there isn't any other pressures besides the atmosphere pressure, the water vapor pressure, and the pressure of the air. or that the atmosphere pressure stays the same.
3. that the air is homogeneous and will fill the volume evenly with an exact number of particles, mols, in the bubble.

Are these three assumptions ok?

Thanks for the help everyone.
Stephen
 
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StephenDoty said:
Are these three assumptions ok?

Can't think of better ones.
 
great! thanks.
but is Delta Hvap= 3.18 * 10^4 J/mol
right??
 
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