25. Calculate the change in the molar enthalpy, entropy, and internal energy for
ID: 1059443 • Letter: 2
Question
25. Calculate the change in the molar enthalpy, entropy, and internal energy for water at atmospheric pressure (1.013 x 105 Pa) in the following processes: At the liquid--gas transition, assume that the volume of the liquid is negligible compared with that of the gas and that the ideal gas law applies to the gas. The molar specific heat cp is for water 75.4 J/(K mole) and for steam 35.7 J/(K mole), and the latent heat of vaporization for water is 4.07 x 104 J/mole.) (a) going from a liquid at 0 C to a liquid at 100 o C; (b) going from a liquid at 100 oC to a gas at 100 oC; (c) going from a gas at 100 oC to a gas at 200 oC.Explanation / Answer
Answer a.
dH = Cp*dT = 75.4 J/(K mole) * (373K - 273K) = 7540 J/mole.
Cp - Cv = R ;
Cv (for water) = Cp - R = 75.4 J/(K mole) - 8.314 J/(K mole) = 67.086 J/(K mole)
Internal energy change dU = Cv * dT = 67.086 J/(K mole) * (373K -273K) = 6708.6 J/mole .
Answer (b)
It is given that volume of liquid is negligible as compared to gas , thus dV = Volume of gas (V)
dU = dH - PV = dH - nRT (here n will be one mole since we need to find out change in molar internal energy )
dU = 40700 J / mole - 8.314 J / (K mole) * 373 K = 40700 J / mole - 3101.12 J / mole = 37598.8 J / mole.
Answer (c):
dH = Cp*dT = 35.7 J / (K mole) * (473K -373K) = 3570 J / mole.
dS = dH/dT = 3570 J / mole / 100 K = 35.7 J / (K mole)
dU = Cv*dT = (Cp - R) * dT = (35.7 J / (K mole) - 8.314 J / (K mole)) * 100 K = 2738.6 J / mole.
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