Answer :

From graph we can write


VA =VD = 200cc=200×10-6m3


VB=VC = 400cc= 400×10-6m3


PB=PD=155kPa= 155×103Pa


PA=PC=50kPa=50×103Pa


Given


Heat absorbed in process ABC ΔQ1=50cal= 50×4.2 J=210J


Let heat absorbed into the system during process ADC =ΔQ2


We know that work done by the gas is given as


ΔW=PΔV


Work done in path ACB WACB =ΔW1= WAC+WBC


ΔW1=PA(VC-VA) + 0 (WBC=0 because VB=VC)


=50×103×(400-200)×10-6


=10J


Work done in path ADB WADB=ΔW2=WAD + WDB


ΔW2 = PB (VD-VB) +0 ( WAD=0 because VA=VD)


=155×103×(400-200)×10-6


=31 J


Now initial point A and final point C is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.


Therefore,


ΔU1=ΔU2=ΔU ……(i)


From first law of thermodynamics, we know that,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


Using first law of thermodynamics for process ABC


ΔQ1=ΔU1+ΔW1


ΔU1=ΔQ1-ΔW1=210-10=200J


Using first law of thermodynamics for process adc


ΔQ2=ΔU2+ΔW2


=ΔU1+ΔW2 (from (i))


=200+31=231J.


heat supplied to the system during process ADC=231J.


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