A living plant leaf is an open system that exchanges energy (sunligh.pdf

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A living plant leaf is an open system that exchanges energy (sunlight and thermal flows) + matter (O2, CO2, water vapor, etc.) with the surroundings (the exterior). Even with irreversible processes (energy dissipation) we find that the net entropy change for the system + surroundings can decrease; i.e. dS < 0. Using equations and your own critical thinking: a) Explain how this is possible in terms of deS and diS. b) Explain how the Second Law remains valid for this case. Solution Kindly go through below details for entropy to clear concepts about entropy. According to the principle of entropy , the entropy change of the universe i.e system + surroundings can never have a negative value. i.e [(dS ) system + (dS) surroundings] >or = 0 always (dS) universe > 0 => Irreversible process (dS) universe = 0 => reversible process (dS) universe < 0 => Impossible So, it is impossible to have dS<0 for system and universe unlike mentioned by you in the question. Now let us analyse entropy change of system Entropy change of the system is summation of entropy change due to internal irreversibilty and entropy change due to external interaction. i.e dS = (dS) iir +(dS) ei = S gen (entopy generation) + dQ/T CASE I : If sytem is reversible than value of entropy generation is zero and Entropy change of system is given by dS= dQ/T ( A ) if heat is supplied to system then dQ/T >0 and hence entropy change of system is having positive value i.e dS >0 (B)if heat is rejected from system then dQ/T <0 and hence entropy change of system is having negative value i.e dS <0 (C) if system is adiabatic or well insulated system then dQ/T =0 and hence entropy change of system is having zero value i.e dS =0 NOTE : ENTROPY CHANGE OF THE SYSTEM MAY HAVE +VE, -VE OR ZERO VALUE BUT ENTROPY CHANGE OF THE UNIVERSE CAN NEVER HAVE NEGATIVE VALUE. CASE II :IF system is irreversible than the entropy change is given by dS = (dS)iir + (dS) ei or dS = S (gen ) + dQ/T (A) if the system is adibatic then dQ=0 therefore dS = (dS)iir =S (gen) and we know that the value of entropy generation can never be negative. (B) If the system is not adiabatic and the increase in internal ireversibilty is compensated by decrease in external interaction than the value of entropy changes can be zero i.e iscentropic dS = (dS)iir { increases) + (dQ/T)ei { decreases} e.g. 0 = 5 - 5 Now after discussing you must be clear that what is being wrongly/ mistakenly mentioned in the question. Consider the leaf as an irreversible sytem due to various process occuring for the conversion for one form of energy into another and also the transfer of heat occurs between systen ( leaf here) and surroundings and we can say heat is absorbed by the system here therefore entropy change of the system, dS > 0 since {(dS)iir >0 and also (dQ/T) ei >0 } and (dS) system + surrunding > 0 always for an irreversible process.

Education

(A) if the system is adibatic then dQ=0 therefore dS = (dS)iir =S (gen)
and we know that the value of entropy generation can never be negative.
(B) If the system is not adiabatic and the increase in internal ireversibilty is
compensated by decrease in external interaction than the value of entropy changes can be zero i.e
iscentropic
dS = (dS)iir { increases) + (dQ/T)ei { decreases}
e.g. 0 = 5 - 5
Now after discussing you must be clear that what is being wrongly/ mistakenly mentioned in the
question.
Consider the leaf as an irreversible sytem due to various process occuring for the conversion for
one form of energy into another and also the transfer of heat occurs between systen ( leaf here)
and surroundings and we can say heat is absorbed by the system here
therefore entropy change of the system, dS > 0 since {(dS)iir >0 and also (dQ/T) ei >0 }
and (dS) system + surrunding > 0 always for an irreversible process

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