Tutorial 7 Notes - Nov 6th 10th Tutorial Notes Exercise 4.4...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Nov. 6 th , 10 th Tutorial Notes Exercise 4.4 a) We are given an equation that describes how the vapour pressure of a pure substance changes with inverse temperature: log p Torr ! " # $ % & = 7.960 1780 T / K ( ) a) We are first asked to calculate the enthalpy of vaporization. This can be done by noting that this equation is in the same form as the Clausius Clapeyron equation in the form that was derived without limits: ln vap m vap m H S p RT R ! ! = " + where -1780 is the slope of the line vap m H R ! " and 7.960 is the y-intercept vap m S R ! . Since ln log 2.303 p p = the slope and intercept must also be divided by 2.303. Therefore the enthalpy of vaporization can be calculated from: ! " vap H m 2.303 # R = ! 1780 K " vap H m = ! 2.303 ! 1780 K ( ) 8.314 J K -1 mol -1 ( ) 10 ! 3 kJ J -1 ( ) = 34.08 kJ mol -1 Therefore the enthalpy of vaporization for this liquid is 34.08 kJ mol -1 . b) We are then asked to calculate the ‘normal’ boiling point for this substance - in other words, what is the temperature required to boil this substance at 1 atm? In order for a substance to boil its vapour pressure must be equal that of the external pressure. Since we were given an equation that describes how the vapour pressure of this substance changes with inverse temperature, we can substitute in the 1 atm pressure and solve for T. Note that the pressure in this equation is in units of Torr, therefore, the 1 atm pressure must be entered in terms of Torr. First we must rearrange the given equation to solve for T:
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
T ! log p Torr " # $ % & = T ! 7.960 ( 1780 K -1 ( ) T log p Torr " # $ % & ( 7.960 " # $ % & = ( 1780 K -1 ( ) T = ( 1780 K -1 ( ) log p Torr " # $ % & ( 7.960 Now substitute in 1 atm = 760 Torr: T = !
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern