Q. 9

50 cal of h

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

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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