Question

An ideal solution is formed by mixing two volatile liquids A and B, X A and X B are the mole fractions of A and B respectively in the solution and Y A and Y B are the mole fraction of A and B respectively in the vapour phase ,a plot of 1 Y A along y-axis against 1 X A along x-axis gives a straight lines.what is the slope of the straight line?

Answer 1

P_B^0/P_A^0

Answer 2

An ideal solution formed by mixing two volatile liquids A and B follows Raoult's Law and Dalton's Law of Partial Pressures.

1. Raoult's Law: The partial pressure of component A in the solution ($P_A$) is given by: $P_A = X_A P_A^0$ where $X_A$ is the mole fraction of A in the liquid phase and $P_A^0$ is the vapor pressure of pure A. Similarly, for component B: $P_B = X_B P_B^0$

2. Dalton's Law of Partial Pressures: The total vapor pressure ($P_{total}$) above the solution is the sum of the partial pressures: $P_{total} = P_A + P_B$ Also, the mole fraction of component A in the vapor phase ($Y_A$) is given by: $Y_A = \frac{P_A}{P_{total}}$

3. Combining the laws: Substitute $P_A = X_A P_A^0$ into the equation for $Y_A$: $Y_A = \frac{X_A P_A^0}{P_{total}}$ Now, substitute $P_{total} = X_A P_A^0 + X_B P_B^0$: $Y_A = \frac{X_A P_A^0}{X_A P_A^0 + X_B P_B^0}$

4. Expressing $X_B$ in terms of $X_A$: Since $X_A + X_B = 1$, we have $X_B = 1 - X_A$. Substitute this into the equation for $Y_A$: $Y_A = \frac{X_A P_A^0}{X_A P_A^0 + (1 - X_A) P_B^0}$ $Y_A = \frac{X_A P_A^0}{X_A P_A^0 + P_B^0 - X_A P_B^0}$ $Y_A = \frac{X_A P_A^0}{P_B^0 + X_A (P_A^0 - P_B^0)}$

5. Finding the expression for $1/Y_A$: To get the desired form for the plot ($1/Y_A$ vs $1/X_A$), take the reciprocal of the equation for $Y_A$: $\frac{1}{Y_A} = \frac{P_B^0 + X_A (P_A^0 - P_B^0)}{X_A P_A^0}$ Separate the terms: $\frac{1}{Y_A} = \frac{P_B^0}{X_A P_A^0} + \frac{X_A (P_A^0 - P_B^0)}{X_A P_A^0}$ Simplify each term: $\frac{1}{Y_A} = \left(\frac{P_B^0}{P_A^0}\right) \left(\frac{1}{X_A}\right) + \left(\frac{P_A^0 - P_B^0}{P_A^0}\right)$ This can be rewritten as: $\frac{1}{Y_A} = \left(\frac{P_B^0}{P_A^0}\right) \left(\frac{1}{X_A}\right) + \left(1 - \frac{P_B^0}{P_A^0}\right)$

6. Identifying the slope: The equation is in the form of a straight line, $y = mx + c$, where:

   • $y = \frac{1}{Y_A}$ (plotted along the y-axis)
   • $x = \frac{1}{X_A}$ (plotted along the x-axis)
   • $m = \text{slope}$
   • $c = \text{y-intercept}$

   Comparing the derived equation with $y = mx + c$: Slope ($m$) = $\frac{P_B^0}{P_A^0}$

The slope of the straight line is $\frac{P_B^0}{P_A^0}$.