You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and L21 5 1.1. Estimate the value of the parameters at 508C

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

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 =          ------------ equation (1)

where:

= Wilson parameter = ???

= liquid molar volume of component 2 = 75.14 cm³/mol

= 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:

To calculate the temperature dependent parameters of the Wilson equation  ∧₂₁

∧₂₁ =          ---------- equation (2)

From equation (1) ; let replace  178.98 K for

75.14 cm³/mol

= 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ =

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for

75.14 cm³/mol

= 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ =

∧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

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

it is a solid

explanation: