Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure:
If for the process AB, PT2 = constant Then the work done (in kcal) on the system for the process AB is
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Solution
From the graph, T1=1000K T2=500K Given for the process PT2=C1 and we know PV = nRT So P(PVnR)2=C1⇒P3V2=C ⇒P=(CV2)13 W=−∫V2V1PdV=−∫V2V1C13dVV23 =−C133(V132−V131) =−3(P2V2−P1V1) =3nR(T1−T2) =3×2×2×500 =6000Cal=6kCal