Solution.
Ti=1073K;
Pi=1.50⋅106Pa;
Pf=3⋅105Pa;
v=80.0kg/min;
a)(TfPfVf)γ=(TiPiVi)γ;
Tf=Ti(PiPf)(γ−1)/γ;
γ=35 for Argon,
Tf=1073K⋅(1.50⋅106Pa3⋅105Pa)0.4=564K;
b)ΔEint=nCVΔT=Q−Weng=0−Weng⟹Weng=−nCvΔT;
P=tWeng=t−nCVΔT;
P=2.12⋅105W;
c)ec=1−TiTf=1−1073K564K=0.475 or 47.5%;
Answer: a)Tf=564K;
b)P=2.12⋅105W;
c)ec=47.5%.
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