Question

1 mole of ideal gas follows cyclic rule justify​

Answer 1

Answer:

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)

thnx...

God bless us all