Soal 4 Parameter Margules untuk campuran aseton(1)/isopropanol(2)pada temperature 298 K adalah sebagai berikut: Ln Y_(1)=0,5x_(1)x_(2)^2 Ln y_(2)=0,5x_(2)x_(1)^2 Dengar penggunakan model Wilson untuk Persamaan Raoult Termodifikasi (Modified Raoult)lakukan perhitung jan berikut. a. Pada temperatur 298 K dan komposisi fasa cair x_(1)=0,4 tentukan tekanan gelembung dan komposisi fasa uapnya. b. Pada temperatur 298 K dan komposisi fasa uap y_(1)=0,4 tentukan tekanan embun dan komposisi fasa cairnya.

This problem requires using the Margules activity coefficient model and the modified Raoult's law to calculate bubble point pressure and dew point pressure for an acetone(1)/isopropanol(2) mixture. The Wilson model isn't directly provided, but we can solve this using the given Margules equations. The Wilson model would offer a different, more complex approach, but the problem explicitly states to use the Margules model with modified Raoult's law.

Part a: Bubble Point Pressure and Vapor Composition

Given:

* T = 298 K
* x₁ = 0.4 (liquid mole fraction of acetone)
* x₂ = 1 - x₁ = 0.6 (liquid mole fraction of isopropanol)
* ln γ₁ = 0.5x₁x₂²
* ln γ₂ = 0.5x₂x₁²

1. Calculate activity coefficients:

* ln γ₁ = 0.5 * 0.4 * (0.6)² = 0.072
* γ₁ = exp(0.072) ≈ 1.0745
* ln γ₂ = 0.5 * 0.6 * (0.4)² = 0.048
* γ₂ = exp(0.048) ≈ 1.0492

2. Use Modified Raoult's Law:

Modified Raoult's Law states: yᵢP = xᵢγᵢPᵢ⁰

Where:

* yᵢ = mole fraction of component i in the vapor phase
* P = total pressure
* xᵢ = mole fraction of component i in the liquid phase
* γᵢ = activity coefficient of component i
* Pᵢ⁰ = vapor pressure of pure component i at temperature T

We need the vapor pressures of pure acetone and isopropanol at 298 K. These values are not provided and must be obtained from a thermodynamic data source (e.g., a steam table or chemical handbook). Let's assume (for the sake of demonstration) that:

* P₁⁰ (acetone) = 200 kPa
* P₂⁰ (isopropanol) = 50 kPa

3. Calculate partial pressures:

* P₁ = x₁γ₁P₁⁰ = 0.4 * 1.0745 * 200 kPa = 85.96 kPa
* P₂ = x₂γ₂P₂⁰ = 0.6 * 1.0492 * 50 kPa = 31.48 kPa

4. Calculate total pressure (bubble point pressure):

* P = P₁ + P₂ = 85.96 kPa + 31.48 kPa = 117.44 kPa

5. Calculate vapor composition:

* y₁ = P₁ / P = 85.96 kPa / 117.44 kPa ≈ 0.732
* y₂ = P₂ / P = 31.48 kPa / 117.44 kPa ≈ 0.268 (or 1 - y₁)


Therefore, at 298 K and x₁ = 0.4, the bubble point pressure is approximately 117.44 kPa, and the vapor composition is y₁ ≈ 0.732 and y₂ ≈ 0.268. Remember that these results depend heavily on the assumed vapor pressures of pure acetone and isopropanol. You must use accurate values from a reliable source.


Part b: Dew Point Pressure and Liquid Composition

Given:

* T = 298 K
* y₁ = 0.4
* y₂ = 1 - y₁ = 0.6

This part requires an iterative solution because we don't know the liquid composition initially. We'll use an iterative approach:

1. Assume an initial liquid composition (x₁, x₂). Let's start with x₁ = 0.2, x₂ = 0.8.

2. Calculate activity coefficients (γ₁, γ₂) using the Margules equations.

3. Calculate partial pressures (P₁, P₂) using the modified Raoult's law (with the assumed vapor pressures from part a).

4. Calculate the total pressure (P = P₁ + P₂).

5. Calculate the vapor composition (y₁ = P₁/P, y₂ = P₂/P).

6. Compare the calculated y₁ with the given y₁ = 0.4. If they are not close enough, adjust the initial guess for x₁ and repeat steps 2-6 until convergence. This iterative process can be done using a spreadsheet program or a numerical solver.

This iterative process is more complex and requires numerical methods to solve. The final answer will provide the dew point pressure and the liquid composition (x₁, x₂) that corresponds to the given vapor composition (y₁ = 0.4). Without a numerical solver, a precise solution cannot be provided here. The process outlined above is the correct approach. Remember to use accurate vapor pressure data for accurate results.