Tutorial 7 Notes

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

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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:

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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 = !
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