DGD 6

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University of Ottawa *

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Course

2132

Subject

Chemistry

Date

Feb 20, 2024

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pdf

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3

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CHM2132 DGD Assignment #6 One copy of your group’s solutions must be uploaded to Brightspace on your Group page by 8 pm on Wednesday Nov 3 rd . You must show your work to receive full credit. Appropriate sig figs and units are required in the final answer. Assignments that are not legible cannot be marked. After the DGD, every student must record participation marks for each of their group members using the link: https://forms.gle/qDdtZn1C2ZD2reuX7 Students who do not complete the form will not receive credit for participation at that DGD. 1) The enthalpy of vaporization for a liquid is 32.6 kJ mol -1 at its normal boiling point of 368.4 K. The molar volumes of the liquid and the vapour at the boiling point are are 18.6 mL mol -1 , and 30.56 L mol -1 , respectively. a) Use the Clapeyron equation to estimage the temperature dependence of the vapour pressure (dp/dT) at the normal boiling point. b) Use the Clausius-Clapeyron equation to estimate dp/dT. c) Does the Clausius-Clapeyron equation overestimate or underestimate the temperature dependence of the vapour pressure? What is the percent error of this estimation? 2) Suppose the partial pressure of water is 245 Pa on a fall morning at -8.38 o C. Use a calculation to predict whether frost will form under these conditions. You may need to use some of the data below, and assume that the enthalpy values are temperature independent. Data for H 2 O Δ vap H m o = 44.0 kJ mol -1 Δ fus H m o = 6.01 kJ mol -1 C p , m o l ( ) = 75.3 J K -1 mol -1 C p , m o g ( ) = 33.6 J K -1 mol -1 T triple point = 273.16 K p triple point = 611.2 Pa M = 18.02 g mol -1
I Buap Ho 32.6 kt molt Trap 368.4K Vm l 18.6 10 310 Um g 30.56 It ftp.nlgj.my 32600J mot 368 412 30.5640 18.6 100 1110 73 2.897 0 Pak b CLAUSIUS CLAPEYRON DUM Ving day Duepit Trap Um g 32600J mot 368 411 30.5641 110 1 7 2.896 103 Pa K l C CLAUSIUS CLAPEYRON UNDERESTIMATES It ERROR DIFFERENCE ACTUAL dog 4100 2.897 103 2.896 103 2 897 103 100 0.06
2 Pino 245 Pa T 8.388 264.77k NEED TO FIND PHO EQUILIBRIUM THIS TEMPERATURE E degli t y Asb Amphorae Ho 44.0 6.01 KI 50.0 II Pz p É FROM TABLE TRIPLE point 27.3.16 K PTRIPLE Point 64.2Pa P2 611.2Pa expf III É ata 304Pa SINCE ACTUAL Pao 7245 Pa LESS THAN VAPOUR PRESSURE OF H2O EQUILIBRIUM FROST WILL NOT FORM g I S EQUILIBRIUM WILL PROCEED IN DIRECTION OF GAS SO IF ANY FROST IS PRESENT IT WILL SUBLIME
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