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)?
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 = Pt
pressure due to A vapour = PA
pressure due to B vapour = PB
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)
=
Pt
= 0.0035
Pt
form above equations
()
( 553 Torr) =
0.0035
Pt
Pt =
= 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)
=
Pt
= 0.9965
Pt
therefore ()
(380) =
0.9965
Pt
Pt =
= 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
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