Given data is
mass of the mixture (m) = 1 kg
Pressure of the mixture P1=1bar;400C
P2=5bar
Compression index n = 1.2
Specific heat at constant pressure of nitrogen , Cp = 1.04 kJ/kgK
Oxygen, Cp = 0.924 kJ/kgK
Carbon Dioxide, Cp = 0.85 kJ/Kgk
molecular weight M = mass/no of moles
from equation of state of gases, PV = mRT
TPN2V=mN2RN2........(1)
RN2=MN2Rˉ , Here M= molecular weight and Rˉ is the universal gas constant
Similarly for the oxygen and carbon dioxide
PO2V=mO2RO2T........(2)
PCO2V=mCO2RCO2T........(3)
From equations 1, 2 and 3
TV(PN2+PO2+PCO2)=MN2mN2Rˉ+MO2mO2Rˉ+MCO2mCO2Rˉ
TVP=[MN2mN2+MO2mO2+MCO2mCO2]Rˉ........(4)
Where P=PN2+PO2+PCO2 pressure of mixture
R=[MN20.6+MO20.3+MCO20.1]Rˉ
R=[250.6+320.3+440.1]×8314
R=274.86J/kgK
Gas constant, R=274.86J/kgK
Molecular weight of mixture
M =MN2+MO2+MCO2
M=nN2mN2+nO2mO2+nCO2mCO2
M=20.6+20.3+30.1
M=0.483
Work done of the mixture at 40o C
∫10pdv⟹(P2P1)nn−1=T2T1
(51)1.21.2−1=T240+273
T2=409.298K
W=∫12Pdv=P(v2−v1)=mR(T2−T1)
W=0.483×274.80(409.298−313)=12.784kJ
From first law of thermodynamics
Q=△U+W
△U=CvdT
However, Cv=n−1R=1.2−1278.86=1.374kJ/kgK
△U=1.374(409.298−313)=132.313kJ
Q=△U+W=132.313+12.784=145.1kJ
Change in entropy , △s=CplnT1T2−RlnP1P2
However, Cp=n−1nR=1.2−11.2×278.86=1.649kJ/kgK
△s=1.649ln313409.298−0.27486ln15=0.44kJ/k
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