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dane: przemiana izobaryczna, p=10(5)Pa, dV=0,5m3, Cp/Cv=1,4, cząsteczki
.......gazu 2 atomowego, Cv=2,5R, Cp=Cv+R=3,5R, R=8,31J/molK
szukane: dU
dU = Qp - pdV......Qp=n*Cp*dT.....Cp=Cv+R=2,5R+R=3,5R.....pV=nRT......
.......................p*dV=nRdT.......nRdT=p*dV
dU = n*Cp*dT - p*dV = 3,5*n*R*dT - p*dV = 3,5p*dV - p*dV = 2,5p*dV
dU = 2,5*10(5)N/m2*0,5m3 = 1,25*10(5)Nm = 125 kJ
..........................pozdrawiam