What is the entropy change of the system in the following processes?
Isothermal reversible expansion of an ideal gas from volume V 1 to V 2 at temperature T.
Isothermal irreversible expansion of an ideal gas from V 1 to V 2.
An adiabatic reversible expansion from V 1 to V 2.
An adiabatic irreversible expansion from V 1 to V 2.
A reversible cyclic process.
An irreversible cyclic process.
21.5)
Freezing of water at 0oC is a reversible process because water and ice are in equilibrium at this temperature at 1 atm pressure. H(0 o C) = -1436 cal/mol. What is the entropy change for this process? Here H = q rev.
Liquid water can be supercooled below 0o C. The supercooled liquid is unstable and can be easily frozen into ice, and the process not spontaneous. Calculate the entropy change in the spontaneous freezing of water at -20o C.
Calculate the entropy change of the surroundings and Suniverse.Suniverse = Ssystem + Ssurroundings.
Hint:Calculation ofS can not be done by dividing q irr =H (-20 o C) by T = 253K (= -20 0 C). CalculateSin three reversible steps.
S1 = entropy change of water from 253K to 273K.S =Cp ln T 2 /T 1 , Cp = 18 cal/deg mol .
S2 = entropy of freezing at 273 K
S3 = entropy change of ice from 273 K to 253K. Use Cp (ice) = 9 cal/deg mol. To calculateHsurroundings, useHsurr = -Hsys ,Hsys (253K) =Hsys (273K) +Cp (T 2 -T 1 ). AssumeCp to be independent of temperature.Cp = Cp (ice) – Cp (water).
