The latent heat of vaporization for n-pentane at 25°C is 366.3 J-g¹. Take the value of the ideal gas constant to be 8.314 J-mol-1-K-1. (The characteristic properties of n-pentane are given in the table below.) n- TABLE B.1 Characteristics Properties of Pure Species Vc w Tak TK Pobar Zc cm3 mol Tn/K 1 Molar mass 72.150 0.252 469.7 33.70 0.270 313. 309.2 Pentane Evaluate the latent heat of vaporization AHn for the following equations.
The latent heat of vaporization for n-pentane at 25°C is 366.3 J-g¹. Take the value of the ideal gas constant to be 8.314 J-mol-1-K-1. (The characteristic properties of n-pentane are given in the table below.) n- TABLE B.1 Characteristics Properties of Pure Species Vc w Tak TK Pobar Zc cm3 mol Tn/K 1 Molar mass 72.150 0.252 469.7 33.70 0.270 313. 309.2 Pentane Evaluate the latent heat of vaporization AHn for the following equations.
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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The latent heat of vaporization for n-pentane at 25°C is 366.3 J-g-1. Take the value of the ideal gas
constant to be 8.314 J-mol-¹1-K-1. (The characteristic properties of n-pentane are given in the table
below.)
n-
TABLE B.1 Characteristics Properties of Pure Species
Vc
W Tdk Pobar Zc cm³ mol Tn/k
1
Molar
mass
72.150 0.252 469.7 33.70 0.270 313.
309.2
Pentane
Evaluate the latent heat of vaporization AHn for the following equations.
Consider the equation AH = RT
The latent heat of vaporization AHn is
1.092(InPc - 1.013)
0.930 - Tr
J-g-1
Expert Solution
Step 1
Given that-
The latent heat of vapourization for n-heptane is calculated using the equation-
From the given data table-
R=8.314 J/mol.K
Tc=469.7 K
Pc =33.70 bar
Tn=309.2 K
Hence, reduced tempearture (Tm) at the normal boiling point is calulated as-
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