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The state of an ideal gas with Cp= (5/2)R is changed from P=1 bar and V₁² = 12 m³ to P2 = 3 bar and V₂ = 4 m³. Calculate Q, W, AU, and AH¹ by the following a mechanically reversible process. The values of R are given in the following table: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. R = 8.314 J-mol-¹.K-¹8.314 m³-Pa-mol-¹-K-¹ = 83.14 cm³-bar-mol-¹-K-¹ = 8314 cm³-kPa-mol-¹-K-¹ 82.06 cm³-(atm)-mol-¹ K-¹ = 62,356 cm³-(torr).mol-¹.K-¹ 1.987 (cal)-mol-¹-K-¹ = 1.986 (Btu) (lb mole)-¹(R)-¹ -1 = 0.7302 (ft)³ (atm) (lb mol)-¹(R)-¹ = 10.73 (ft)³ (psia) (lb mol)-¹(R)-¹ -1 -1 1545 (ft) (lb) (lb mol)-¹(R)-¹ = Values of the Universal Gas Constant = = Consider the process of cooling at constant pressure followed by heating at constant volume. (Include a minus sign required.) The heat required is The work done is EkJ. The change in internal energy is The change in enthalpy is kJ. 0 kJ. 0 kJ.

The state of an ideal gas with Cp= (5/2)R is changed from P=1 bar and V₁² = 12 m³ to P2 = 3 bar and V₂ = 4 m³. Calculate Q, W, AU, and AH¹ by the following a mechanically reversible process. The values of R are given in the following table: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. R = 8.314 J-mol-¹.K-¹8.314 m³-Pa-mol-¹-K-¹ = 83.14 cm³-bar-mol-¹-K-¹ = 8314 cm³-kPa-mol-¹-K-¹ 82.06 cm³-(atm)-mol-¹ K-¹ = 62,356 cm³-(torr).mol-¹.K-¹ 1.987 (cal)-mol-¹-K-¹ = 1.986 (Btu) (lb mole)-¹(R)-¹ -1 = 0.7302 (ft)³ (atm) (lb mol)-¹(R)-¹ = 10.73 (ft)³ (psia) (lb mol)-¹(R)-¹ -1 -1 1545 (ft) (lb) (lb mol)-¹(R)-¹ = Values of the Universal Gas Constant = = Consider the process of cooling at constant pressure followed by heating at constant volume. (Include a minus sign required.) The heat required is The work done is EkJ. The change in internal energy is The change in enthalpy is kJ. 0 kJ. 0 kJ.

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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- Required information The state of an ideal gas with Cp= (5/2)R is changed from P=1 bar and V₁ = 12 m²³ to P2 = 3 bar and V₂t = 4 m³. Calculate Q, W, Aut, and AH by the following a mechanically reversible process. The values of R are given in the following table: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. R = 8.314 J-mol-¹.K-18.314 m³ Pa-mol-¹.K-1 83.14 cm³-bar-mol-¹.K-1 8314 cm³ kPa-mol-¹.K-1 82.06 cm³ (atm)-mol-¹ K-¹ = 62, 356 cm³-(torr).mol-¹.K-¹ 1.987 (cal)-mol¹K-¹ = 1.986 (Btu) (lb mole)-¹(R)-¹ 0.7302 (ft)³ (atm) (lb mol)-¹(R)-¹ = 10.73 (ft)³ (psia) (lb mol)-¹(R)-¹ -1 1545 (ft) (lbf) (lb mol)-¹(R)-¹ = = = Values of the Universal Gas Constant = = KJ. Consider the process of cooling at constant pressure followed by heating at constant volume. (Include a minus sign if required.) The heat required is The work done is[ TkJ. The change in internal energy is The change in enthalpy is = 0 kJ. 0 kJ.
Transcribed Image Text:- Required information The state of an ideal gas with Cp= (5/2)R is changed from P=1 bar and V₁ = 12 m²³ to P2 = 3 bar and V₂t = 4 m³. Calculate Q, W, Aut, and AH by the following a mechanically reversible process. The values of R are given in the following table: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. R = 8.314 J-mol-¹.K-18.314 m³ Pa-mol-¹.K-1 83.14 cm³-bar-mol-¹.K-1 8314 cm³ kPa-mol-¹.K-1 82.06 cm³ (atm)-mol-¹ K-¹ = 62, 356 cm³-(torr).mol-¹.K-¹ 1.987 (cal)-mol¹K-¹ = 1.986 (Btu) (lb mole)-¹(R)-¹ 0.7302 (ft)³ (atm) (lb mol)-¹(R)-¹ = 10.73 (ft)³ (psia) (lb mol)-¹(R)-¹ -1 1545 (ft) (lbf) (lb mol)-¹(R)-¹ = = = Values of the Universal Gas Constant = = KJ. Consider the process of cooling at constant pressure followed by heating at constant volume. (Include a minus sign if required.) The heat required is The work done is[ TkJ. The change in internal energy is The change in enthalpy is = 0 kJ. 0 kJ.
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