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For work done in an isobaric process, there is 2 equations equal to each other to find the answer (seen from the photo) but as I used those 2 equations I got 2 different values. Can someone help? 2) There is a phase change from solid to liquid which is isothermal. However, is it also an isobaric process? If not, what is the pressure at the solid state?

For work done in an isobaric process, there is 2 equations equal to each other to find the answer (seen from the photo) but as I used those 2 equations I got 2 different values. Can someone help? 2) There is a phase change from solid to liquid which is isothermal. However, is it also an isobaric process? If not, what is the pressure at the solid state?

Introduction to Chemical Engineering Thermodynamics
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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1) For work done in an isobaric process, there is 2 equations equal to each other to find the answer (seen from the photo) but as I used those 2 equations I got 2 different values. Can someone help? 2) There is a phase change from solid to liquid which is isothermal. However, is it also an isobaric process? If not, what is the pressure at the solid state?
0.5kg M=0.05 Solia T₁ = 0 °C T₁ = 0°C P₁₂₁² = 1 Isothermal Isobaric? (W) Work done for isuharit mests. Liquid T₂ = 0°C P₂ = 1 bay V₂ = 0.00049 m³ Isobaric 2 V2 Wokpa = V₂ = T₂= 25°C V₂ 7. Liquid T3 = 35% → Pz = Vs = 0.0005 m³ T₂. 0.00049 pcves = P(V3 - V₂) = -0.001 = I bar density at = -20.355 1000 kg/m³ -MR (T3-T₂) (0.5) (8.142) (5) 5% του 306 O.S Ka
Transcribed Image Text:0.5kg M=0.05 Solia T₁ = 0 °C T₁ = 0°C P₁₂₁² = 1 Isothermal Isobaric? (W) Work done for isuharit mests. Liquid T₂ = 0°C P₂ = 1 bay V₂ = 0.00049 m³ Isobaric 2 V2 Wokpa = V₂ = T₂= 25°C V₂ 7. Liquid T3 = 35% → Pz = Vs = 0.0005 m³ T₂. 0.00049 pcves = P(V3 - V₂) = -0.001 = I bar density at = -20.355 1000 kg/m³ -MR (T3-T₂) (0.5) (8.142) (5) 5% του 306 O.S Ka
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