NEET Physics Thermodynamics Questions Solved


A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are : PA=3×104Pa,PB=8×104Pa and VA=2×103m3,VD=5×103m3. In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be 

(1) 560 J

(2) 800 J

(3) 600 J

(4) 640 J

(1) By adjoining graph WAB=0 and

WBC=8×104[52]×103=240J

WAC=WAB+WBC=0+240=240J

Now, ΔQAC=ΔQAB+ΔQBC=600+200=800J

From FLOT ΔQAC=ΔUAC+ΔWAC

800=ΔUAC+240ΔUAC=560J.

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