1 mole of ideal gas follows cyclic rule justify
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Minimum temperature is at point D. (=330K).
ab & cd are isentropic process.
∴TA=600K;TB=1200K;Tc=600K
Using PV=nRT at any point PV=RT (as n=1)
Work done in isentropic process is given by:
W=P1V1ln(V1V2)=nRTln(V1V2)
Work along DA & BC is zero (const vol.)
Work Done along AB
WAB=RTln(VAVB)=600Rln(2)
Word Done along CD:
WCD=RTln(V2VD)=300Rln(21)=−300Rln(2)
Therefore net work done is 300Rln(2).
As this is a cyclic process so there is no change in internal engery (initial and final points are same)
Therefore using first law: Q=δU+W. Net heat exchange is 300Rln(2)
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