Ideal gas law: nRT=PV
Isobaric process W=PΔV
y
Q
1st law of thermodynamics: ΔU=Q‐W
Recall that ΔU=nCV(Tb‐Ta)=n(R/(γ‐1)) (Tb‐Ta)=1/(γ‐1)(PaVa‐PbVb) in the second step we have used CV=R/(γ‐1) and in the third step the ideal gas law
γ=1.4 for diatomic gas γ=1.667 for monoatomic gas. If we are given the adiabatic process then we can find γ=‐ln(Pb/Pa)/ln(Vb/Va) nCV(Tb‐Ta)=
1/(γ‐1)(PaVa‐PbVb) nCV(Tb‐TTa))=
1/(γ‐1)(PaVa‐PbVb) nCV(Tb‐Ta)=
1/(γ‐1)(PaVa‐PbVb) 0 since T does not change
=ΔU
0
0 since V does not change
=‐ΔU
ΔU
PVγ=const.
t
=P(Va‐Vb)
Q=ΔU+W
=nRTln(Vb/Va)
Q=W
Since we are given the adiabatic process we first compute the γ for this gas: γ=‐
ln(Pb/Pa)/ln(Vb/Va) =‐ln(5)/ln(1/3)=1.465 So we have a mostly diatomic gas with a little bit of monoatomic gas.
Wtot=Wab+Wbc+Wcd+Wda=Wab+0+Wcd+0=
=‐(γ‐1)‐1(PbVb‐PaVa)‐(γ‐1)‐1(PdVd‐PcVc)=
= ‐(γ‐1)‐1(PdVd‐PcVc+PbVb‐PaVa)=
=‐(γ‐1)‐1(PcVc(Vc/Vd)γ‐1‐PcVc+PbVb‐PbVb(Vb/Va)γ‐1)
= ‐(γ‐1)‐1(PcVc‐PbVb)((Vc/Vd)γ‐1‐1)=
= (γ‐1)‐1(Pc‐Pb)Vb(1‐(Vb/Va)γ‐1)
3(1‐(1/3)
0 465)
=(0.465)
(
)‐11(10‐5)×100kPa×0.333E‐4m
(
)
( ( / )0.465
=14.3 J
Not very realistic (Ta is close to 1 Kelvin) but let’s go with these numbers as an example)
Qin=(γ‐1)‐1(PcVc‐PbVb) =(γ‐1)‐1(Pc‐Pb)Vb=35.8 J
e=Wtot/Qin=1‐(Vb/Va)γ‐1=1‐(1/3)0.465=0.40=40%