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Q.

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure  Pi=105Pa and volume  Vi=10−3m3  changes to final state at Pf=132×105Pa  &  Vf=8×10−3m3 in an adiabatic quasi-static process, such that  P3V5=constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately

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a

112 J

b

294 J

c

588 J

d

813 J

answer is C.

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Detailed Solution

Case 1 : PV53=constant​⇒γ=53 , Monoatomic gas​Process is adibatic.​W=PfVf−PiVi1−γ=132×105×8×10−3−105×10−31−53=112.5J​⇒△U=−W=−112.5JCase 2 : ​Wisobaric=PiVf−Vi=1058×10−3−10−3=700J​Wisochoric=0​⇒Wnet=700J​First law of thermodynamics,​Q=△U+Wnet=−112.5+700=587.5J=588 J nearly
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