EnthalpyConc-inclass_example

1161 no heat of solution since xa 00 then h cpb

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Unformatted text preview: e this is at 80.1 and for toluene it is at 110.6 oC). Benzene (A) has the lower boiling point than Toluene (B) so let To (ref temp) = 80.1 oC Find h and H for (xa, ya = 1.0) So h(xa = 1.0) = 0 kJ/mol H(ya = 1.0) = ? H = yA[λA + cpyA(T ­T0)] + (1 ­yA)[λ B + cpyB(T ­T0)] (Eq. 11.6 ­2) Have ya = 1.0 and are given CpyA and we know T=To. So Eq. 11.6 ­2 reduces to H(ya = 1.0) = λA (latent heat of component A at To) ??? H(ya = 1.0) = ______________ For xa, ya = 0.0 (pure toluene) h(xa = 0.0) = ? h = xAcpA(T ­T0) + (1 ­ xA) cpB (T ­T0) + ΔHsol (Eq. 11.6 ­1) (no heat of solution) Since xa = 0.0 then h = cpB (T ­T0) Q: What is T? h(xa = 0.0) = _______________ H(ya = 0.0) = ? H = yA[λA + cpyA(T ­T0)] + (1 ­yA)[λ B + cpyB(T ­T0)] (Eq. 11.6 ­2) H = [λ B + cpyB(T ­T0)] Q: What is λB ?...
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