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Question on normal boiling point and enthalpy of vaporization

  • #1
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Hello all

I performed a lab where we put a graduated cylinder filled to 90% with water upside down in a large beaker filled with water and measured the volume of the air bubble inside the graduated cylinder at different temperatures. When I graphed Ln(PH20) vs. 1/T I got a slope of -10412. Multiply this by -8.314 and you get the enthalpy of vaporization, which was -10412*-8.314=86565.368 for me. Is this right??
Using the enthalpy of vaporization how do I find the normal boiling point of water? I do not understand how to use the formula ln(P2/P1) =-Hvap/R * ((1/T2)-(1/T1)) to find the normal boiling point of water. Any help would be greatly appreciated.

I have attached the graph from the experiment. Please let me know if you guys want to see all of the data.

Thanks so much everyone.
Stephen
 

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Answers and Replies

  • #2
Mapes
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Boiling occurs when the partial pressure of a material equals the ambient pressure. Does this help?

EDIT: Hey, watch your significant figures; I doubt you measured volume to 7 decimal places!
 
  • #3
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if both pressures are the same then ln(P2/P1) in the formula ln(P2/P1)=(-Hvap/R)*((1/T2)-(1/T1)) would be zero and when finding the temperature T=(-hvap/R)*(1/ln(P)) ln(P) would equal ln(1) thus there would be a zero in the denominator. So how would that work?

Thanks for the help.
Stephen
 
  • #4
Mapes
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How are you deriving T=(-hvap/R)*(1/ln(P))?
 
  • #5
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My professor wrote ln(P)=(-Hvap/R)*(1/T)
on my paper then I just solved for T.

Is this is not the correct formula to find the normal boiling point? What is? Oh, and does my Hvap look ok?

Thanks for all of the help. I appreciate it.
Stephen
 
  • #6
Mapes
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I'd ask your professor where this equation comes from. I'm not sure how one can take the logarithm of 1 atm, for example (i.e., a number with units).
 

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