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A 0.500-mol sample of an ideal gas has an initial tem- perature of 298 K. It is compressed from an initial volume of 13.00 L to 5.00 L by an external pressure of 2.45 atm, at which point the internal and external pressures are equal. Calculate AS sys, AS sure and ASuniy for the process.

A 0.500-mol sample of an ideal gas has an initial tem- perature of 298 K. It is compressed from an initial volume of 13.00 L to 5.00 L by an external pressure of 2.45 atm, at which point the internal and external pressures are equal. Calculate AS sys, AS sure and ASuniy for the process.

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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A 0.500-mol sample of an ideal gas has an initial tem- perature of 298 K. It is compressed from an initial volume of 13.00 L to 5.00 L by an external pressure of 2.45 atm, at which point the internal and external pressures are equal. Calculate AS sys, AS sure and ASuniy for the process.
3.23. A 0.500-mol sample of an ideal gas has an initial tem- perature of 298 K. It is compressed from an initial volume of 13.00 L to 5.00 L by an external pressure of 2.45 atm, at which point the internal and external pressures are equal. Calculate ASSys AS surn and AS univ for the process.
Transcribed Image Text:3.23. A 0.500-mol sample of an ideal gas has an initial tem- perature of 298 K. It is compressed from an initial volume of 13.00 L to 5.00 L by an external pressure of 2.45 atm, at which point the internal and external pressures are equal. Calculate ASSys AS surn and AS univ for the process.
Expert Solution
Step 1

Given that-

External and internal pressure are equal so as Pext=Pint 

The gas is compressed at constant tempearture so as the process is isothermal.

Initial temperature, T1=298 K

Initial volume of gas, V1=13.00 L

Final volume of gas, V2=5.00 L

Change in volume, 

∆V=V2-V1       =13.00 L-5.00 L∆V=8.00 L

External pressure, Pext= 2.45 atm

Mol of gas, n=0.500 mol

Now, for isothermal process, the change in internal energy is zero. So, as per first law of thermodynamics,

∆U=Q+W    0=Q+W   W=-Q

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