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132 5 SIMPLEMIXTURES
�erefore the pressure are pB = 4.8 kPa , pA = 21 kPa , and ptot = 26 kPa .
�e partial pressure of the gas is given by pA = yAptot, where yA is the mole
fraction in the vapour
yA =
pA
ptot
= 21.4... kPa
26.3... kPa
= 0.82
yB =
pB
ptot
= 4.81... kPa
26.3... kPa
= 0.18
E5B.10(b) Let 1,2-dimethylbenzene be A and 1,3-dimethylbenzene be B. If the solution
is ideal the vapour pressure obeys Raoult’s law, [5A.22–151], pJ = p∗J xJ. �e
mixture will boil when the sum of the partial vapour pressures of A and B equal
the external pressure, here 19 kPa.
pext = pA + pB = xAp∗A + xBp∗B = xAp∗A + (1 − xA)p∗B
hence xA =
pext − p∗B
p∗A − p∗B
and by analogy xB =
pext − p∗A
p∗B − p∗A
xA =
(19 kPa) − (18 kPa)
(20 kPa) − (18 kPa)
= 0.5 xB =
(19 kPa) − (20 kPa)
(18 kPa) − (20 kPa)
= 0.5
�e partial pressure of the gas is given by pJ = yJpext, where yJ is the mole
fraction in the gas, and pJ is given by pJ = p∗J xJ, hence yJ = xJp∗J /pext
yA =
xAp∗A
pext
= (0.50) × (20 kPa)
(19 kPa)
= 0.53
yB =
xBp∗B
pext
= (0.50...) × (18 kPa)
(19 kPa)
= 0.47
E5B.11(b) �e vapour pressure of component J in the solution obeys Raoult’s law, [5A.22–
151], pJ = p∗J xJ, where xJ is the mole fraction in the solution. Is the gas the
partial pressure is pJ = yJptot, where yJ is the mole fraction in the vapour.
�ese relationships give rise to four equations
pA = p∗AxA pB = p∗B(1 − xA) pA = ptot yA pB = ptot(1 − yA)
where xA + xB = 1 is used and likewise for the gas. In these equations xA and
ptot are the unknowns to be found.�e expressions for pA are set equal, as are
those for pB, to give
p∗AxA = ptot yA hence ptot =
p∗AxA
yA
p∗B(1 − xA) = ptot(1 − yA) hence ptot =
p∗B(1 − xA)
1 − yA