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Question:

A reversible cyclic process for an ideal gas is shown. Here, P, V and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively. The correct option(s) is/are?

qAC=ΔUBCandwAB=P2(V2−V1)

ΔHCA<ΔUCAandqAC=ΔUBC

qBC=ΔHACandΔHCA>ΔUCA

wBC=P2(V2−V1)andqBC=ΔHAC

Solution:

AC→IsochoricAB→IsothermalBC→Isobaric(A) qAC=ΔUBC=nCv(T2−T1)WAB=nRT1 ln(V2V1)∴A is wrong (B) qBC=ΔHAC=nCP(T2−T1)WBC=−P2(V1−V2)∴B is correct (C) nCP(T1−T2)<nCv(T1−T2)∴C is correct (D) ΔHCA<ΔUCA∴D is wrong.