Chapter Thirteen - Entropy Production and Lost Work

what do Clausius inequalites suggest

change in entropy = Qrev/T = Qirr/T + hidden variable greater than 0

what is the equation to express that entropy changes in two ways

dS = d(dash)eS + d(dash)is where the first term is entropy change of a system due to exchange with surroundings and the second is entropy production due to irreversible processes which is > or equal to 0

why is diS the uncompensated transformation

dStot = dSsys + dSres = d(dash)Q/T + d(dash)iS - d(dash)Q/T = d(dasH)iS > or equal to 0 so etropy production accounts for the portion of the systems entropy change that isnt cancelled by the reservoir

how does clausius theorem relate to entropy production

for a cyclic process integral of dS = 0 = integral of d(dash)Q/T + integral of diS so integral of d(dash)Q/T < or equal to 0 in a cyclic process

what is the net irreversible entropy production

change in entropy for irreversible = change in entropy of system -Q/T

how is net entropy change in irreversible reaction calculated

change in irreversible entropy = integral of d(dash)iS = change in entropy of system + change in entropy of reservoir

what is the connection between work lost and reversible and irreversible expansion

T change in S irr = Wrev-Wirr

how is the lost work eqn found

if a piston moves quasistatcally against series of pressures at fixed T, dU = CvdT= 0, dS= (P/T)dV, change in Ssys = integral of P/T dV = nR ln(V1/V2), change in U =0 so Q = W so for irreversible Q = Wirr = P2(V2-V1), so change in Sirr = nRln(v1/V2) - P2(V2-V1)/T

what is the sign of lost work

Wrev-Wirr = T change in Sirr > 0

how do you derive the sign of lost work

for reversible dU= TdS - d(dash)Wrev, for irreversible dU = d(dash)Qirr - d(dash)Wirr, since U is a state function Tds - d(dash)Wrev = d(dash)Qirr - d(dash) W irr, d(dash)Qirr/T = deS and d(dash)Wrev - d(dash)Wirr/t = dis so second term is postive

what causes lost work

entropy production in the irreversible process

why does entropy production cause loss of work

entropy increase of the system is due to internal entropy produce traceable to irreversibility so less heat is absorbed, for isothermal process this heat is turned to work so Wirr<Wrev

what is an isentropic system

constant entropy

what are the two possibilities for dS=0

deS=diS=0 or deS = -diS

what happens when deS = -diS

because diS >0 then dQ/T + diS =0 and so dQ<0- the system has to export heat to its surroundings

how is isentropic system achieved for U= U(T)

dU = -PdV and since dV = 0, dU =0

what is the free energy of an isochoric and isentropic system and why

constant as dU=0

why is d(dash)Q = 0 not the same as isentropic

there is only no heat exchange if diS =0 ( all processes are reversible)