1) a. What is the entropy change if two mole of liquid water is heated from 20 C
ID: 771405 • Letter: 1
Question
1) a. What is the entropy change if two mole of liquid water is heated from 20 C to 100 C under constant pressure? Cp=75.5 J/mol*K over this temperature interval. b. Calculate (delta)r S^o for following process. H2O (G,100 C, 1 atm) --> H2O (s,0 C, 1 atm) (delta)r H fus= 6025 J/mol (delta)r Hvap= 40.7 kj/mol 2) Calculate the following a. Two moles of an ideal gas at 0 C expand (isothermally and reversibly from 10 to 40 liters. Calculate q,w, (delta) U, (DELTA) h AND (DELTA) S. B. Two moles of an ideal gas expand isothermically against a constant opposing pressure of 1 bar from 5 to 45 liters. Calculate q,w, (delta) U, (DELTA) h AND (DELTA) S.Explanation / Answer
1)
a. q=Cp(T2-T1) =75.5*80 =6040 J
?S = Cp*ln(T_final/T_initial) = 75.5*ln(373.15/293.15) =18.217J/K
b. Entropy at transition from gas to liquid =?S(g to l) =( delta)rHvap/Tb = 40.7*10^3/283.15=143.74
Entropy from 100C to 0C =?S' = Cp*ln(273.15/373.15) =75.5*ln(273.15/373.15) =-23.55
Entropy at transition from liquid to solid =? S(l to s) =( delta)r Hfus/Tm = 6025/273.15=22.057
(delta)r S^o =? S(g to l) +?S' +? S(l to s) =143.74-23.55+22.057 =142.247 J/K
2)
A. q=0,
w=-PdV=-nRTln(V2/V1) =- 2*8.314*273.15*ln(40/20) = -3148.23 J
deltaU= q-PdV =-3148.23 J
deltaS=0
B.
q=0 ..(isothermal process)
w=-Pext(V2-V1) = -10^5*(45-5)*0.001=-4000 J
deltaU= q-PdV =-4000 J
deltaS=0
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