Answer:
ΔU = 45.814 KJ
ΔH = 46.4375 KJ
ΔS = 18.76 J/K
Explanation:
H2O(l) → H2O(l) → H2O(steam)
298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm
∴ ΔHp = Qp = nCpΔT
∴ n H2O = 1 mol
∴ Cp,n = 75.3 J/mol.K
∴ ΔT = 373.25 - 298.15 = 75 K
⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J
⇒ ΔHp = 5647.5 J = 5.6475 KJ
⇒ ΔH = ΔHp + ΔHv
∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ
⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ
ideal gas:
∴ ΔH = ΔU + PΔV
∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L
∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L
⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³
∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa
⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))
⇒ ΔU = 46.4375 KJ - 0.623 KJ
⇒ ΔU = 45.814 KJ
∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)
⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)
⇒ ΔS = 16.896 J/K + 1.863 J/K
⇒ ΔS = 18.76 J/K
