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Example 5.3 Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature. The constants from Table C.1 are A = 1.702 B= 9.81×10 ³ : ; C = -2.164 × 106

Example 5.3 Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature. The constants from Table C.1 are A = 1.702 B= 9.81×10 ³ : ; C = -2.164 × 106

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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Example 5.3 Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature. The constants from Table C.1 are A = 1.702 B= 9.81×10³ ; C=-2.164 × 106 Solution: AS's = 0, and Eq. (5.14) becomes: depends on T2, after rearrange this equation: (c#), T2 In R P2 In = 1 -1.6094 = P1 Because (C) depends on T2, we rearrange this equation for iterative solution: T2_ In P2/P. In " (CF)!R (In P2/P1 T; = T, expc#)!R %3D or (A)
Transcribed Image Text:Example 5.3 Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature. The constants from Table C.1 are A = 1.702 B= 9.81×10³ ; C=-2.164 × 106 Solution: AS's = 0, and Eq. (5.14) becomes: depends on T2, after rearrange this equation: (c#), T2 In R P2 In = 1 -1.6094 = P1 Because (C) depends on T2, we rearrange this equation for iterative solution: T2_ In P2/P. In " (CF)!R (In P2/P1 T; = T, expc#)!R %3D or (A)
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