Question

Question 2 (Engineering EoS and Departure function): (a) Detailing your steps, and starting with 3). + V, show that the enthalpy departure function in terms of pressure integral is given as: " -1976), (b) Find the expression for C .; where 2 = 1 + such that B is a function of temperature, T (i.e. B = B(T)) (c) For an isothermal compression of methane at temperature of 293.15 K from an initial state with a density of 312.5 mol/m² and a Z value of 0.987 to a final state with a density of 933.7 mol/m² and a Z value of 0.961, applying the Virial equation in the enthalpy departure function (given as equation 8.33 in your book), calculate the change in enthalpy due to the change of state (i.e. from the initial state (1) to the final state (2)). Take R = 8.314 J/molk To = 190.6 K; P = 46.04 bar; and w = 0.011. (Hint: the ideal gas part in the equation AH = (H2 - H',') + (9-H)(H1-H) is zero since the change of state is isothermal).

Answer 1

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page 2

dH) = Tds + V dp/ aphy PV=ZRT – Real gas condition. Pro RT – ideal gas. According to Maxwell's equations. Pa p azient for realgas, P avis) = R. de ideal gas. - ( 2) trava) + (0-1) (8 -3,4-fv-19 - RT ZRT - RT@Z) + V.-19 ppp dT /p = yor - W BI 2Z + X - vg ( era -- Free d JT

page 3

we get b) 2 = It BP RT (22) = 0 + P dB - BP dT p RT dT RT2 a IP PV = nRT V ZRT A ñ - P n molar density. = 1 = molas density, I ZRT 2 = 0.987 T, = 293.15% S = 312.5 moym P = S, ZRT Z = 0,961 T = 293.15 (isothermal) & 2 = 933.7 molsmo P z ₂ Zz RT₂ - 2186, 908kla = 751. 790kPa HI=0 = (p. 41 (sothermal proces) Po, = 1 = 0.16 327 Pop = 0.475 Toi = Tile = 1.5380 Tra 1.5380

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