10 g of neon initinally at a pressure of 506.625 kPa and temperature of 473 K expandadiabaticallyto a pressure of 202.65 kPa .Calculateentropyofthe system and oftotalentropy changefor the following waysof carryingout is this expamsion .(i) Expansion is carriedout expansion . (ii) Expansionoccursaganista constant external pressureof 202.65 kPa . (iii) Expansion is a free expansion.

10 g of neon initinally at a pressure of 506.625 kPa and temperature o

1.	10 g of neon initinally at a pressure of 506.625 kPa and temperature of 473 K expandadiabaticallyto a pressure of 202.65 kPa .Calculateentropyofthe system and oftotalentropy changefor the following waysof carryingout is this expamsion .(i) Expansion is carriedout expansion . (ii) Expansionoccursaganista constant external pressureof 202.65 kPa . (iii) Expansion is a free expansion.
Answer» Solution :(i)For reversibleadiabaticprocess `DeltaS_(surr)= 0""DeltaS_(sure)= - DeltaS_(SYSTEM) = 0` `DeltaS_("total") = DeltaS _(SYS)+DeltaS = 0 ` Hence `DeltaS_(sys)= DeltaS_(surr)= DeltaS_("total") =0` (ii)`DeltaS_(surr)= 0` `DeltaS_(sys)= nC_(V)In ((T_(2))/(T_(1)))+ nR In ((V_(2))/(V_(1)))` `= nC_(V) In ((T_(2))/(T_(1))) + nR In ((P_(1))/(P_(2))(T_(1))/(T_(2)))` `= N[C_(V) In ((T_(2))/(T_(1))) + nR In ((T_(2))/(T_(1)))+ In((P_(1))/(P_(2)))]` `= n[C_(V) In ((T_(2))/(T_(1))) + nR In((P_(1))/(P_(2)))]""[T_(2)" of irreverrsubile adiabatic will be calculated "]` `= 0.957 J//K""byw = DeltaU rArr - P_(ext)(V_(2)-V_(1))=nC_(V,m)(T_(1)-T_(2))`