Q. 9

50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in the figure. Find the quantity of heat to be supplied to take it from A to B via ADB.

Answer :

From graph we can write

VA =VD = 200cc=200×10-6m3

VB=VC = 400cc= 400×10-6m3

PB=PD=155kPa= 155×103Pa



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


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

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



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

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


=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.


ΔU1=ΔU2=ΔU ……(i)

From first law of thermodynamics, we know that,


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



Using first law of thermodynamics for process adc


=ΔU1+ΔW2 (from (i))


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

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