Question

At 300 K the vapour pressure of a pure liquid A is 553 Torr, of pure liquid B it is 380 Torr. A and B form ideal liquid mixtures.

a) Calculate the vapour pressures of A and B and the total pressure above liquid mixtures which are 25, 50, and 75 mol% B.

b) For a particular liquid mixture, the vapour composition is found to be 0.35 mol% A. Calculate the liquid composition in equilibrium with vapour of this composition. What do you notice about the liquid composition (relative to that of the vapour)?

Answer 1

a)

vapour pressure of pure liquid A = 553 Torr

vapour pressure of pure liquid B = 380 Torr

1) 25 mol% B

mol% = (mole fraction)$\times$100

mole fraction = $\frac{mol percentage}{100}$

mole fraction of B ($\chi _{B}$) = $\frac{mol percentage of B}{100}$

=  $\frac{25}{100}$

= 0.25

mole fraction of A ($\chi _{A}$) = 1- mole fraction B (since there are only two components)

= 1- 0.25

= 0.75

vapour pressure of A in solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)

= (0.75)$\times$( 553 Torr)

= 414.75 Torr

vapour pressure of B in solution = ($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.25)$\times$( 380 Torr)

= 96 Torr

vapour pressure of solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)+($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.75)$\times$( 553 Torr)+ (0.25)$\times$( 380 Torr)

= 414.75 Torr +96 Torr

= 510.75 Torr

2) 50 mol% B

mol% = (mole fraction)$\times$100

mole fraction = $\frac{mol percentage}{100}$

mole fraction of B ($\chi _{B}$) = $\frac{mol percentage of B}{100}$

=  $\frac{50}{100}$

= 0.50

mole fraction of A ($\chi _{A}$) = 1- mole fraction B (since there are only two components)

= 1- 0.50

= 0.50

vapour pressure of A in solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)

= (0.50)$\times$( 553 Torr)

= 276.5 Torr

vapour pressure of B in solution = ($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.50)$\times$( 380 Torr)

= 190 Torr

vapour pressure of solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)+($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.50)$\times$( 553 Torr)+ (0.50)$\times$( 380 Torr)

= 276.5 Torr +190 Torr

= 466.5 Torr

3) 75 mol% B

mol% = (mole fraction)$\times$100

mole fraction = $\frac{mol percentage}{100}$

mole fraction of B ($\chi _{B}$) = $\frac{mol percentage of B}{100}$

=  $\frac{75}{100}$

= 0.75

mole fraction of A ($\chi _{A}$) = 1- mole fraction B (since there are only two components)

= 1- 0.75

= 0.25

vapour pressure of A in solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)

= (0.25)$\times$( 553 Torr)

= 133.25 Torr

vapour pressure of B in solution = ($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.75)$\times$( 380 Torr)

= 285 Torr

vapour pressure of solution = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)+($\chi _{B}$)$\times$(vapour pressure of pure liquid B)

= (0.25)$\times$( 553 Torr)+ (0.75)$\times$( 380 Torr)

= 133.25 Torr +285 Torr

= 418.25 Torr

b)

Let total vapour pressure = P_t

pressure due to A vapour = P_A

pressure due to B vapour = P_B

let mole fraction of A vapour = $\phi _{A}$ = 0.35 % = 0.0035

let mole fraction of B vapour = $\phi _{B}$ = 1-$\phi _{A}$ = 1- 0.0035 = 0.9965

vapour pressure due to A = ($\chi _{A}$)$\times$(vapour pressure of pure liquid A)

= ($\chi _{A}$)$\times$( 553 Torr)

since T and V are constant

pressure due to A vapour =  vapour pressure due to A = (mole fraction of A vapour)$\times$(total vapour pressure)

= $\phi _{A}$$\times$ P_t

₌ 0.0035$\times$ P_t

form above equations

($\chi _{A}$)$\times$( 553 Torr) = 0.0035$\times$ P_t

  P_{t = $\frac{\chi _{A}\times 553}{0.0035}$}

= 1.58 $\times$ 10⁵$\chi _{A}$ (1)

vapour pressure due to B = ($\chi _{B}$)$\times$(vapour pressure of pure liquid A)

= ($\chi _{B}$)$\times$(380)

pressure due to A vapour =  vapour pressure due to A = (mole fraction of B vapour)$\times$(total vapour pressure)

= $\phi _{B}$$\times$ P_t

₌ 0.9965$\times$ P_t

therefore   ($\chi _{B}$)$\times$(380) = 0.9965$\times$ P_t

P_t = _{$\frac{\chi _{B}\times 380}{0.9965}$}

= 381,33 $\chi _{B}$ (2)

from 1 and 2   1.58 $\times$ 10⁵$\chi _{A}$ = 381.33 $\chi _{B}$

$\chi _{A}$ = 414.339 $\chi _{B}$

but since $\chi _{A}$ and $\chi _{B}$ are mole fractions

$\chi _{A}$$\chi _{B}$ = 1

substituting $\chi _{A}$ = 414.339 $\chi _{B}$ in above equation

  414.339 $\chi _{B}$  $\chi _{B}$ = 1

415.339 $\chi _{B}$ = 1

$\chi _{B}$ = $\frac{1}{415.339}$

$\chi _{B}$ = 0.0024

we know that   $\chi _{A}$$\chi _{B}$ = 1

  $\chi _{B}$ = 1- $\chi _{A}$

= 1- 0.0024

= 0.9976

  mole fraction of A ($\chi _{A}$) = 0.0024

  mole fraction of B ($\chi _{B}$) = 0.9976