V = Vo (1 + a₂t + a₂t² + azt³) Vo 1.228 x 10-³ m³/kg, a₁ = 1.3240 x10-³ a₂ = 3.8090 x 10-6, az - 0.87983 x 10-8 == Calculate (#) and (25) at T = 300 K T Solution dH= Cal+V (1-7 dP -Bv dp ds= C Он ӘР Tutorial T = V(1-BT), as ӘР = - BV TI At 300 K or 26.85 °C, V = 1.275 x 10-³ m³/kg The coefficient of thermal expansion for liquid ß at 26.85 °C B = - (+) (dr) (#) - 3a,1² + 2a,t + a₂ azt³ + a₂t² + a₁t+1 = 1.454 x 10-³K-1

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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Related questions
Question
What was the math used to get the coefficient of thermal expansion for B?
V = Vo(1 + a₁t + a₂t² + a3t³)
Vo 1.228 x 10-³ m³/kg, a₁ = 1.3240 x10-³
a₂ = 3.8090 x 10-6, a3 = - 0.87983 x 10-8
Calculate
Solution
(3H), and (3),²
T
Tutorial
dH= C +V (1-37 dp
dS = C
Он
ӘР
T
-BV dp
= V(1-BT),
at T = 300 K
Tutorial
as
ӘР
= -Bv
I
At 300 K or 26.85 °C, V = 1.275 x 10-³ m³/kg
The coefficient of thermal expansion for liquid ß at 26.85 °C
B =
-)))
- (+) (dr) (dt)_ _3a₂t² + 2a₂t + a₂
=
a3t² + a₂t² + a₁t+1
= 1.454 x 10-³K-¹
Transcribed Image Text:V = Vo(1 + a₁t + a₂t² + a3t³) Vo 1.228 x 10-³ m³/kg, a₁ = 1.3240 x10-³ a₂ = 3.8090 x 10-6, a3 = - 0.87983 x 10-8 Calculate Solution (3H), and (3),² T Tutorial dH= C +V (1-37 dp dS = C Он ӘР T -BV dp = V(1-BT), at T = 300 K Tutorial as ӘР = -Bv I At 300 K or 26.85 °C, V = 1.275 x 10-³ m³/kg The coefficient of thermal expansion for liquid ß at 26.85 °C B = -))) - (+) (dr) (dt)_ _3a₂t² + 2a₂t + a₂ = a3t² + a₂t² + a₁t+1 = 1.454 x 10-³K-¹
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