Self-test 5.1 At 25°C, the density of a 50 per cent by mass ethanol/water solution is 0.914 g cm. Given that the partial molar volume of water in the solution is 17.4 cm' mol-, what is the partial molar volume of the ethanol? [56.4 cm' mol-]
Self-test 5.1 At 25°C, the density of a 50 per cent by mass ethanol/water solution is 0.914 g cm. Given that the partial molar volume of water in the solution is 17.4 cm' mol-, what is the partial molar volume of the ethanol? [56.4 cm' mol-]
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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![Illustration 5.1 The determination of partial molar volume
A polynomial fit to measurements of the total volume of a water/ethanol mixture
at 25°C that contains 1.000 kg of water is
v = 1002.93+54.6664x-0.363 94x2 +0.028 256x
where v = Vlcm', x = ng/mol, and ng is the amount of CH,CH,OH present. The
partial molar volume of ethanol, Vp, is therefore
a(Vicm)
av
VE =
ang
cm' mol-
dx
p.Ttw
cm' mol-=
%3D
a(m/mol) p.Tw
p.Tw
Then, because
dv
== 54.6664 – 2(0.36394)x+ 3(0.028256)x?
dx
we can conclude that
V/(cm'mol-)= 54.6664 - 0.72788x+0.084768x
Figure 5.3 is a graph of this function.
Self-test 5.1 At 25°C, the density of a 50 per cent by mass ethanol/water solution
is 0.914 g cm, Given that the partial molar volume of water in the solution is
17.4 cm mol-, what is the partial molar volume of the ethanol?
[56.4 cm' mol-]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F462788d4-ee38-4fe6-8cb2-385677096736%2F190c332c-d347-49bb-a897-140ceb870696%2Fjbcyvba_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Illustration 5.1 The determination of partial molar volume
A polynomial fit to measurements of the total volume of a water/ethanol mixture
at 25°C that contains 1.000 kg of water is
v = 1002.93+54.6664x-0.363 94x2 +0.028 256x
where v = Vlcm', x = ng/mol, and ng is the amount of CH,CH,OH present. The
partial molar volume of ethanol, Vp, is therefore
a(Vicm)
av
VE =
ang
cm' mol-
dx
p.Ttw
cm' mol-=
%3D
a(m/mol) p.Tw
p.Tw
Then, because
dv
== 54.6664 – 2(0.36394)x+ 3(0.028256)x?
dx
we can conclude that
V/(cm'mol-)= 54.6664 - 0.72788x+0.084768x
Figure 5.3 is a graph of this function.
Self-test 5.1 At 25°C, the density of a 50 per cent by mass ethanol/water solution
is 0.914 g cm, Given that the partial molar volume of water in the solution is
17.4 cm mol-, what is the partial molar volume of the ethanol?
[56.4 cm' mol-]
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