Define Relative Volatility and derive the expression XA = YA/α - (α - 1)YA from the ratio of partial pressure to mole fraction. (b) Compute the vapour-liquid equilibria at constant pressure of 750mmHg for a mixture of n-heptane and n-octane which may be expected to form ideal solutions. Complete the table below and plot (i) vapour composition y as a function of liquid composition x at constant pressure. (ii) temperature composition diagram

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Asked on March 19, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-19T00:21:09.613Z

Part (a) Step 1: Definition of relative volatility. The relative volatility _xy (n-heptane = x, n-octane = y) is _xy = (y_x / x_x)/(y_y / x_y) where x_x, y_x are mole fractions of n-heptane in liquid and vapor. Step 2: Derivation for ideal solution (Raoult's law). Partial pressure: p_x = y_x P = x_x P_x^, so (y_x)/(x_x) = (P_x^)/(P). Similarly, (y_y)/(x_y) = (P_y^)/(P). Thus, _xy = (P_x^ / P)/(P_y^ / P) = (P_x^)/(P_y^). Part (b) For ideal VLE at constant P = 760 mmHg: Bubble point: x_x P_x^ + (1 - x_x) P_y^ = 760. Formula first: x_x = (760 - P_y^)/(P_x^ - P_y^). Then y_x = x_x (P_x^)/(760), _xy = (P_x^)/(P_y^). Step 3: T = 95^C, P_x^ = 750 mmHg, P_y^ = 320 mmHg. _xy = (750)/(320) = 2.34375 ≈ 2.34. x_x = (760 - 320)/(750 - 320) = (440)/(430) ≈ 1.02 ≈ 1.0. y_x = 1.0 × (750)/(760) ≈ 0.99 ≈ 1.0. Step 4: T = 115^C, P_x^ = 1150 mmHg, P_y^ = 650 mmHg. _xy = (1150)/(650) = 1.769 ≈ 1.77. x_x = (760 - 650)/(1150 - 650) = (110)/(500) = 0.22. y_x = 0.22 × (1150)/(760) = 0.22 × 1.513 = 0.333. Step 5: T = 123^C, P_x^ = 1230 mmHg, P_y^ = 750 mmHg. _xy = (1230)/(750) = 1.64. x_x = (760 - 750)/(1230 - 750) = (10)/(480) = 0.0208 ≈ 0.021. y_x = 0.0208 × (1230)/(760) = 0.0208 × 1.618 = 0.0337 ≈ 0.034. Completed table: |c|c|c|c|c|c| T (^C) & P_x^ (mmHg) & P_y^ (mmHg) & x_x & y_x & _xy \\ 95 & 750 & 320 & 1.0 & 1.0 & 2.34 \\ 115 & 1150 & 650 & 0.22 & 0.33 & 1.77 \\ 123 & 1230 & 750 & 0.021 & 0.034 & 1.64 \\ |c|c|c|c|c|c| 95 & 750 & 320 & 1.0 & 1.0 & 2.34 \\ 115 & 1150 & 650 & 0.22 & 0.33 & 1.77 \\ 123 & 1230 & 750 & 0.021 & 0.034 & 1.64 \\

Accepted Answer · 2026-03-19T00:21:09.613Z

Part (a) Step 1: Definition of relative volatility. The relative volatility _xy (n-heptane = x, n-octane = y) is _xy = (y_x / x_x)/(y_y / x_y) where x_x, y_x are mole fractions of n-heptane in liquid and vapor. Step 2: Derivation for ideal solution (Raoult's law). Partial pressure: p_x = y_x P = x_x P_x^, so (y_x)/(x_x) = (P_x^)/(P). Similarly, (y_y)/(x_y) = (P_y^)/(P). Thus, _xy = (P_x^ / P)/(P_y^ / P) = (P_x^)/(P_y^). Part (b) For ideal VLE at constant P = 760 mmHg: Bubble point: x_x P_x^ + (1 - x_x) P_y^ = 760. Formula first: x_x = (760 - P_y^)/(P_x^ - P_y^). Then y_x = x_x (P_x^)/(760), _xy = (P_x^)/(P_y^). Step 3: T = 95^C, P_x^ = 750 mmHg, P_y^ = 320 mmHg. _xy = (750)/(320) = 2.34375 ≈ 2.34. x_x = (760 - 320)/(750 - 320) = (440)/(430) ≈ 1.02 ≈ 1.0. y_x = 1.0 × (750)/(760) ≈ 0.99 ≈ 1.0. Step 4: T = 115^C, P_x^ = 1150 mmHg, P_y^ = 650 mmHg. _xy = (1150)/(650) = 1.769 ≈ 1.77. x_x = (760 - 650)/(1150 - 650) = (110)/(500) = 0.22. y_x = 0.22 × (1150)/(760) = 0.22 × 1.513 = 0.333. Step 5: T = 123^C, P_x^ = 1230 mmHg, P_y^ = 750 mmHg. _xy = (1230)/(750) = 1.64. x_x = (760 - 750)/(1230 - 750) = (10)/(480) = 0.0208 ≈ 0.021. y_x = 0.0208 × (1230)/(760) = 0.0208 × 1.618 = 0.0337 ≈ 0.034. Completed table: |c|c|c|c|c|c| T (^C) & P_x^ (mmHg) & P_y^ (mmHg) & x_x & y_x & _xy \\ 95 & 750 & 320 & 1.0 & 1.0 & 2.34 \\ 115 & 1150 & 650 & 0.22 & 0.33 & 1.77 \\ 123 & 1230 & 750 & 0.021 & 0.034 & 1.64 \\ |c|c|c|c|c|c| 95 & 750 & 320 & 1.0 & 1.0 & 2.34 \\ 115 & 1150 & 650 & 0.22 & 0.33 & 1.77 \\ 123 & 1230 & 750 & 0.021 & 0.034 & 1.64 \\

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