COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 15, Problem 100QAP
To determine

To calculate:

The work done, absorbed heat, change in internal energy, and entropy change of an ideal gas of an engine.

Expert Solution & Answer
Check Mark

Answer to Problem 100QAP

Thus, the work done, absorbed heat, change in internal energy, and entropy change of an ideal gas of an engine have calculated.

Explanation of Solution

Given data:

n =1.23 mol

Specific heat ratio, γ =1.41

Temperature, T1 =300K

Temperature, T2 =600K

Pressure, P1 =15 atm

Pressure, P2 =3 atm

Formula used:

Work done is calculated by,

  W=

  nR(ΔT)

Change in internal energy is calculated by,

  ΔU=nCvΔT

Heat absorbed is calculated by,

  Q=nCpΔT

Change in entropy is calculated by,

  ΔS=QΔT

Calculation:

Work done for each part of the cycle is,

  Wa=P1(V2V1)=P1V2P1V1we know, Pv=nRTWa=nRT2nRT1=nR(T2T)Wa=(1.23)(8.314)(600300)Wa=3.06kJWb=nRT2ln( P 2 P 1 )=(1.23)(8.314)(600)ln(3 15)Wb=9.87kJ

  Wc=(1.23)(8.314)(300600)Wa=3.06kJWd=nRT1ln( P 2 P 1 )=(1.23)(8.314)(300)ln(3 15)Wd=4.93kJ

Work done for complete cycle is,

  Wtotal=3.069.873.06+4.93Wtotal=4.94kJ

Absorbed heat for each part of the cycle is,

  Q=nCpΔTCp=γRγ1=(1.41)(8.314)1.411=28.59=28.6Qa=(1.23)(28.6)(600300)Qa=10.55kJQb=U+Wb=0+Wb(isothermal process)Qb=9.87kJQc=(1.23)(28.6)(600300)Qc=10.55kJQd=U+Wd=0+Wd(isothermal process)Qd=4.93kJ

Absorbed heat for complete cycle is,

  Qtotal=10.559.87+10.55+4.93Qtotal=16.16kJ

Change in internal energy for each part of the cycle is,

  ΔU=nCVΔTCV=Cpγ=28.61.41=20.28

  ΔUa=(1.23)(20.28)(600300)ΔUa=7.48kJΔUb=0(constant temperature, ΔT=0)ΔUc=(1.23)(20.28)(300600)ΔUc=7.48kJΔUd=0(constant temperature, ΔT=0)

Change in internal energy for complete cycle is,

  ΔUtotal=7.48k+07.48k+0ΔUtotal=0J

Change in entropy for each part of th ecycel si,

  ΔS=QaΔTΔSa=10.55× 103600300=35.16JK1ΔSb=0       (constant temperature, ΔT=0)ΔSc=10.55× 103300600=35.16JK1ΔSd=0       (constant temperature, ΔT=0)

Change in entropy of for complete cycle is,

  ΔStotal=35.16+035.16+0ΔStotal=0JK1

Conclusion:

Work done for complete cycle is, Wtotal=4.94kJ

Absorbed heat for complete cycle is, Qtotal=16.16kJ

Change in internal energy for complete cycle is, ΔUtotal=0J

Change in entropy of for complete cycle is, ΔStotal=0JK-1

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Chapter 15 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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