Respuesta :
Here is the correct question.
You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C
Answer:
the values of the parameters at 50° C are 0.766 and 1.047
Explanation:
From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:
Compound Liquid molar volume (cm³/mol)
Ethanol (1) 58.68
1 - propanol (2) 75.14
To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.
∧₁₂ = [tex]\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )[/tex] ------------ equation (1)
where:
[tex]a_{12}/R[/tex] = Wilson parameter = ???
[tex]V_2[/tex] = liquid molar volume of component 2 = 75.14 cm³/mol
[tex]V_1[/tex] = liquid molar volume of component 1 = 58.68 cm³/mol
T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K
Replacing our values in the above equation ; we have:
[tex]0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]-a_{12}/R= 0.60 * 298.15 \ K[/tex]
[tex]-a_{12}/R= - 178.89 \ K[/tex]
[tex]a_{12}/R= 178.89 \ K[/tex]
To calculate the temperature dependent parameters of the Wilson equation ∧₂₁
∧₂₁ = [tex]\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )[/tex] ---------- equation (2)
[tex]1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )[/tex]
[tex]-a_{12}/R = 0.3426 * 298.15 \ K[/tex]
[tex]-a_{12}/R =102.15 \ K[/tex]
[tex]a_{12}/R = -102.15 \ K[/tex]
From equation (1) ; let replace 178.98 K for [tex]a_{12}/R[/tex]
[tex]V_2 =[/tex] 75.14 cm³/mol
[tex]V_1[/tex] = 58.68 cm³/mol
T = 50° C = ( 50 + 273.15 ) K = 348.15 K
So;
∧₁₂ = [tex]\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )[/tex]
∧₁₂ = 1.281 exp(-0.5138)
∧₁₂ = 1.281 × 0.5982
∧₁₂ =0.766
From equation 2; let replace 102.15 K for [tex]a_{12}/R[/tex]
[tex]V_2 =[/tex] 75.14 cm³/mol
[tex]V_1[/tex] = 58.68 cm³/mol
T = 50° C = ( 50 + 273.15 ) K = 348.15 K
So;
∧₂₁ = [tex]\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )[/tex]
∧₂₁ = 0.7809 exp (0.2934)
∧₂₁ = 0.7809 × 1.3410
∧₂₁ = 1.047
Thus, the values of the parameters at 50° C are 0.766 and 1.047
