**Problem 5:**When the Poynting correction factor is almost unity, Equation (10.1-25) reduces to:\[ H_i = \gamma_i^\infty P_i^{\text{vap}} \phi_i^\text{V} (T, P_i^{\text{vap}}) \]Under atmospheric pressure, Miyano (2004) reported the infinite dilution activity coefficients of n-butane (2) in isobutanol (1) at different temperatures as follows:| \( T \) (K) | \( \gamma_2^\infty \) ||-------------|-----------------------|| 250 | 4.03 || 270 | 4.00 || 290 | 3.97 |Estimate Henry’s law constant for n-butane in isobutanol using Equation (1). Use the virial equation of state to determine fugacity coefficients of pure n-butane.---**Problem 6:**A binary liquid mixture of 28 mol% benzene (1) and 72% n-heptane (2) is at a temperature of 318.15 K. Estimate the composition of the vapor phase in equilibrium with this liquid mixture if the system is represented by the three-suffix Margules equation.**Data: At 318.15 K**- Henry’s law constant for benzene in n-heptane = 0.522 bar.- Henry’s law constant for n-heptane in benzene = 0.285 bar.

Hi and thanks for the question. Since you have posted multiple questions and have not specified which question is to be answered. We will answer the first question for you. If you need help with the other question, please post it separately mentioning that question to be answered. Given Equation, Hi = γi∞ Pivap ϕiV(T, Pivap) It is required to calculate Henry's law constant for n-butane in isobutanol using the above equation1) Temperature T = 250 K Critical Temperature of n-butane Tc = 425 K Critical Pressure of n-butane Pc = 37.96 bar Reduced Temperature of n-butane Tr = TTc = 250425 = 0.588 Accentric factor of n-butane ω = 0.199 Calculation of Pivap using Antoine Equation ln P2vap (KPa) = 13.6608 - 2154.70TCo + 238.789ln P2vap (KPa) = 13.6608 - 2154.70250 - 273 + 238.789ln P2vap (KPa) = 3.676 P2vap = e3.676P2vap = 39.49 KPaP2vap = 0.3949 bar Calculation of the fugacity coefficient of n-butane using Virial equation of state: Virial equation of state: Z = 1 + BPRTB = Bo + ωB1Bo = 0.083 - 0.422Tr1.6 = 0.083 - 0.4220.5881.6 = -0.904B1 = 0.139 - 0.172Tr4.2 = 0.139 - 0.1720.5884.2 = -1.461B = -0.904 + 0.199 × -1.461 = -1.195 lnϕ2vap = ∫0P2vapZ - 1P2vapdP2vaplnϕ2vap = ∫0P2vap1 + BP2vapRT - 1P2vapdP2vaplnϕ2vap = ∫0P2vapBRTdP2vap lnϕ2vap = BP2vapRTlnϕ2vap = -1.195 × 0.3949 bar8.314 × 10-2 bar.Lmol.K × 250 Klnϕ2vap = -0.0227ϕ2vap = e-0.0227ϕ2vap = 0.978 Calculation of Henry's law constant for n-butane in isobutanol: H2 = γ2∞ P2vap ϕ2V(T, P2vap)H2 = 4.03 × 0.3949 bar × 0.978H2 = 1.556 bar 2) Temperature T = 270 K Critical Temperature of n-butane Tc = 425 K Critical Pressure of n-butane Pc = 37.96 bar Reduced Temperature of n-butane Tr = TTc = 270425 = 0.635 Accentric factor of n-butane ω = 0.199 Calculation of Pivap using Antoine Equation ln P2vap (KPa) = 13.6608 - 2154.70TCo + 238.789ln P2vap (KPa) = 13.6608 - 2154.70270 - 273 + 238.789ln P2vap (KPa) = 4.523 P2vap = e4.523P2vap = 92.11 KPaP2vap = 0.9211 bar Calculation of the fugacity coefficient of n-butane using Virial equation of state: Virial equation of state: Z = 1 + BPRTB = Bo + ωB1Bo = 0.083 - 0.422Tr1.6 = 0.083 - 0.4220.6351.6 = -0.789B1 = 0.139 - 0.172Tr4.2 = 0.139 - 0.1720.6354.2 = -1.019B = -0.789 + 0.199 × -1.019 = -0.992 lnϕ2vap = ∫0P2vapZ - 1P2vapdP2vaplnϕ2vap = ∫0P2vap1 + BP2vapRT - 1P2vapdP2vaplnϕ2vap = ∫0P2vapBRTdP2vap lnϕ2vap = BP2vapRTlnϕ2vap = -0.992 × 0.9211 bar8.314 × 10-2 bar.Lmol.K × 270 Klnϕ2vap = -0.0407ϕ2vap = e-0.0407ϕ2vap = 0.960 Calculation of Henry's law constant for n-butane in isobutanol: H2 = γ2∞ P2vap ϕ2V(T, P2vap)H2 = 4.00 × 0.9211 bar × 0.960H2 = 3.537 bar3) Temperature T = 290 K Critical Temperature of n-butane Tc = 425 K Critical Pressure of n-butane Pc = 37.96 bar Reduced Temperature of n-butane Tr = TTc = 290425 = 0.682 Accentric factor of n-butane ω = 0.199 Calculation of Pivap using Antoine Equation ln P2vap (KPa) = 13.6608 - 2154.70TCo + 238.789ln P2vap (KPa) = 13.6608 - 2154.70290 - 273 + 238.789ln P2vap (KPa) = 5.237 P2vap = e5.237P2vap = 188.1 KPaP2vap = 1.881 bar Calculation of the fugacity coefficient of n-butane using Virial equation of state: Virial equation of state: Z = 1 + BPRTB = Bo + ωB1Bo = 0.083 - 0.422Tr1.6 = 0.083 - 0.4220.6821.6 = -0.695B1 = 0.139 - 0.172Tr4.2 = 0.139 - 0.1720.6824.2 = -0.719B = -0.695 + 0.199 × -0.719 = -0.838 lnϕ2vap = ∫0P2vapZ - 1P2vapdP2vaplnϕ2vap = ∫0P2vap1 + BP2vapRT - 1P2vapdP2vaplnϕ2vap = ∫0P2vapBRTdP2vap lnϕ2vap = BP2vapRTlnϕ2vap = -0.838 × 1.881 bar8.314 × 10-2 bar.Lmol.K × 290 Klnϕ2vap = -0.0654ϕ2vap = e-0.0654ϕ2vap = 0.937 Calculation of Henry's law constant for n-butane in isobutanol: H2 = γ2∞ P2vap ϕ2V(T, P2vap)H2 = 3.97 × 1.881 bar × 0.937H2 = 6.997 bar Answer: Henry's law constant for n-butane in isobutanol: At Temperature T = 250 K, H2 = 1.556 bar At Temperature T = 270 K, H2 = 3.537 bar At Temperature T = 290 K, H2 = 6.997 bar