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)100

mole fraction =

mole fraction of B () =

=  

= 0.25

mole fraction of A () = 1- mole fraction B (since there are only two components)

= 1- 0.25

= 0.75

vapour pressure of A in solution = ()(vapour pressure of pure liquid A)

= (0.75)( 553 Torr)

= 414.75 Torr

vapour pressure of B in solution = ()(vapour pressure of pure liquid B)

= (0.25)( 380 Torr)

= 96 Torr

vapour pressure of solution = ()(vapour pressure of pure liquid A)+()(vapour pressure of pure liquid B)

= (0.75)( 553 Torr)+ (0.25)( 380 Torr)

= 414.75 Torr +96 Torr

= 510.75 Torr

2) 50 mol% B

mol% = (mole fraction)100

mole fraction =

mole fraction of B () =

=  

= 0.50

mole fraction of A () = 1- mole fraction B (since there are only two components)

= 1- 0.50

= 0.50

vapour pressure of A in solution = ()(vapour pressure of pure liquid A)

= (0.50)( 553 Torr)

= 276.5 Torr

vapour pressure of B in solution = ()(vapour pressure of pure liquid B)

= (0.50)( 380 Torr)

= 190 Torr

vapour pressure of solution = ()(vapour pressure of pure liquid A)+()(vapour pressure of pure liquid B)

= (0.50)( 553 Torr)+ (0.50)( 380 Torr)

= 276.5 Torr +190 Torr

= 466.5 Torr

3) 75 mol% B

mol% = (mole fraction)100

mole fraction =

mole fraction of B () =

=  

= 0.75

mole fraction of A () = 1- mole fraction B (since there are only two components)

= 1- 0.75

= 0.25

vapour pressure of A in solution = ()(vapour pressure of pure liquid A)

= (0.25)( 553 Torr)

= 133.25 Torr

vapour pressure of B in solution = ()(vapour pressure of pure liquid B)

= (0.75)( 380 Torr)

= 285 Torr

vapour pressure of solution = ()(vapour pressure of pure liquid A)+()(vapour pressure of pure liquid B)

= (0.25)( 553 Torr)+ (0.75)( 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 = = 0.35 % = 0.0035

let mole fraction of B vapour = = 1- = 1- 0.0035 = 0.9965

vapour pressure due to A = ()(vapour pressure of pure liquid A)

= ()( 553 Torr)

since T and V are constant

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

= P_t

₌ 0.0035 P_t

form above equations

()( 553 Torr) = 0.0035 P_t

  P_{t =}

= 1.58 10⁵ (1)

vapour pressure due to B = ()(vapour pressure of pure liquid A)

= ()(380)

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

= P_t

₌ 0.9965 P_t

therefore   ()(380) = 0.9965 P_t

P_t =

= 381,33 (2)

from 1 and 2   1.58 10⁵ = 381.33

= 414.339

but since and are mole fractions

= 1

substituting = 414.339 in above equation

  414.339    = 1

415.339 = 1

=

= 0.0024

we know that   = 1

   = 1-

= 1- 0.0024

= 0.9976

  mole fraction of A () = 0.0024

  mole fraction of B () = 0.9976