Liquid mercury has a molar volume of 0.01465 dm3 mol-1 and solid mercury a molar volume of 0.01413 dm3 mol-1, both being measured at the melting point, -38.87°C under 101.3 kPa (atmospheric) pressure. The enthalpy of fusion for mercury is 1954 J mol-1. Using the setting out given in lectures (i.e. show all working), calculate the melting points of mercury under a pressure of (a) 10 times atmospheric (1013 kPa) and (b) 750 times atmospheric (75,975 kPa). N.B. Be careful of units. The following information will prove very useful to this problem. 1dm3 =1x10-3 m3 1Jm-3 =1Pa 0°C= 273.15 K we have to use the clapeyron equation
Liquid mercury has a molar volume of 0.01465 dm3 mol-1 and solid mercury a molar volume of 0.01413 dm3 mol-1, both being measured at the melting point, -38.87°C under 101.3 kPa (atmospheric) pressure. The enthalpy of fusion for mercury is 1954 J mol-1. Using the setting out given in lectures (i.e. show all working), calculate the melting points of mercury under a pressure of (a) 10 times atmospheric (1013 kPa) and (b) 750 times atmospheric (75,975 kPa). N.B. Be careful of units. The following information will prove very useful to this problem. 1dm3 =1x10-3 m3 1Jm-3 =1Pa 0°C= 273.15 K we have to use the clapeyron equation
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Liquid mercury has a molar volume of 0.01465 dm3 mol-1 and solid mercury a molar volume of 0.01413 dm3 mol-1, both being measured at the melting point, -38.87°C under 101.3 kPa (atmospheric) pressure. The enthalpy of fusion for mercury is 1954 J mol-1. Using the setting out given in lectures (i.e. show all working), calculate the melting points of mercury under a pressure of (a) 10 times atmospheric (1013 kPa) and (b) 750 times atmospheric (75,975 kPa).
N.B. Be careful of units. The following information will prove very useful to this problem.
1dm3 =1x10-3 m3
1Jm-3 =1Pa
0°C= 273.15 K
we have to use the clapeyron equation
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