A cylindrical vessel with rigid adiabatic walls is separated intotwo parts by a

Question

A cylindrical vessel with rigid adiabatic walls is separated intotwo parts by a frictionless adiabatic piston. Each part contains 50L of an ideal monatomic gas withCv,m=3R/2,Initially,Ti=325K andPi=2.5*105Pa in each part. Heat is slowelyintroduced into the left part using an electrical heater until thepiston has moved sufficiently to the right to result in a finalpressure Pf=7.5 bar in the right part. Consider thecompression of the gas in the right part to be a reversibleprocess.....a)calculate the work done on the right part in thisprocess and the final temperature in the rightpart.........b)calculate the final temperature in the left part andthe amount of heat that flowed into this part.

Explanation / Answer

Cv = 3R/2, Cp = Cv + R = 5R/2, = Cp/Cv = 5/3 initial volume V = 50 L, T = 325 K, P = 2.5 bar final pressure P' = 7.5 bar a) The right part: adiabatic process, PV = constant PV = P'Vr, so the final volume of theright part Vr = V(P/P')1/ = 25.9 L the final temperature = Tr TV-1 = constant TV-1 =TrVr-1 , so Tr =T(V/Vr)-1 = 504 K heat = 0, change in internal energy Er =nCvT = n(3R/2)(Tr - T) =(3/2)(nRTr - nRT) = 1.5(P'Vr - PV) =1.04*104 J work done on the right part = Er =1.04*104 J b) the left part: final volume VL = 2V - Vr = 74.1 L the final temperature = TL PV/T = P'VL/TL TL= T(P'/P)(VL/V) = 1445 K heat flowed into the left part = change in internal energy in bothparts = nCvTL+ nCvTr =nCv(TL - T) + nCv(Tr -T) = n(3R/2)(TL - T + Tr - T) =1.5(P'VL + P'Vr - 2PV) = 1.5*[P'(VL + Vr) - 2PV] = 1.5*[P'*(2V) -2PV] = 1.5*2*(P' - P)V = 7.5*104 J

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