# Thermodynamics

• ANS A

• @You-Knowwhere
Process CBC B is done at minimum temperature, while process DAD A is done at maximum temperature Process A→b(A \rightarrow b( isobaric
VAVB=TATB=800400=2\frac{V_{A}}{V_{B}}=\frac{T_{A}}{T_{B}}=\frac{800}{400}=2
hence VB=2V0V_{B}=2 V_{0}
B→C( isothermal )B \rightarrow C(\text { isothermal })
C→D(C \rightarrow D( isobaric
VDVC=TDTC⇒VD=2V0\frac{V_{D}}{V_{C}}=\frac{T_{D}}{T_{C}} \Rightarrow V_{D}=2 V_{0}
D→(D \rightarrow( isothermal WAB=P(VB−VA)=nRDTW_{A B}=P\left(V_{B}-V_{A}\right)=n R D T
=nR(−400)=−400nR=n R(-400)=-400 n R
WBC=nRT1nVCVB=nR4001nV02V0W_{B C}=n R T 1 n \frac{V_{C}}{V_{B}}=n R 4001 n \frac{V_{0}}{2 V_{0}}
=−400nR1n2=-400 n R 1 n 2
WCD=P(VD−VC)=nR(400)=400nRW_{C D}=P\left(V_{D}-V_{C}\right)=n R(400)=400 n R
WDA=nRTlnVAVDW_{D A}=n R T \ln \frac{V_{A}}{V_{D}}
=nR×800×1n2V0V0=800nR1n2=n R \times 800 \times 1 n \frac{2 V_{0}}{V_{0}}=800 n R 1 n 2
ΔW=400nR1n2\Delta W=400 n R 1 n 2
ΔQ=\Delta Q= heat is extracted QCD=nCpΔT=n54R400=1000nRQ_{C D}=n C_{p} \Delta T=n \frac{5}{4} R 400=1000 n R
=800nR1n2=1000nR+800nR1n2=800 n R 1 n 2=1000 n R+800 n R 1 n 2
Efficiency =ΔWΔQ=\frac{\Delta W}{\Delta Q}
η=21n25+41n2×100\eta=\frac{21 n 2}{5+41 n 2} \times 100%

• @raj1213 But how Vb = Va and Vc = Vd