- #1
StephenDoty
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- 0
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
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