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Physical Chemistry I. practice

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Physical Chemistry I. practice

Gyula Samu & Zolt´an Rolik

V.: Ideal mixtures

rolik@mail.bme.hu

(2)

Pressure-composition diagram

2 components, 2 phases

Miscible components, ideal solution, vapor is ideal gas

z2: molar fraction of comp. 2 in the total system

x2: molar fraction of comp. 2 in the liquid phase (solution)

y2: molar fraction of comp. 2 in the

Dalton’s law:

p = p1 + p2, p1 = y1 · p, p2 = y2 · p Raoult’s law:

p1 = x1 · p1, p2 = x2 · p2

p1, p2: eq. vapor pressure of comp. 1 and 2

y2 = pp2 = x2·p

2

p

z1 + z2 = x1 + x2 = y1 + y2 = 1

(3)

Konovalov’s first law, lever rule

y2 = pp2 = x2·p

2

p

Konovalov’s first law for ideal mixtures:

if p1 < p2: p1 < p < p2 pp2 > 1, y2 > x2

n1 = (ng + nl) · z1 = ng · y1 + nl · x1 ng · (z1 y1) = nl · (x1 z1)

Lever rule: nnl

g = xz1−y1

1−z1

We can also calculate ng and nl directly

n1 = ng · y1 + nl · x1 = ng · y1 + (n ng) · x1 = n · x1 + ng · (y1 x1)

ng = ny1−n·x1

1−x1 nl = nx1−n·y1

1−y1

(4)

Vapor composition

We have an ideal mixture of acetone (Ac) and acetonitrile (An) log(pAc) = 9.36457 − 237.50+T1279.87

log(pAn) = 9.36789 − 238.89+T1397.93

(p in P a, T in C; but no dimensions inside log!) T = 20C xAc = 0.2

What is the vapor pressure? What is the composition of the vapor?

p = pAc + pAn = xAc · pAc + (1 − xAc) · pAn

p x ·p p (1−x )·p

(5)

Vapor composition

log(pAc) = 9.36457 − 237.50+201279.87 = 4.39420

→ pAc = 104.39420P a = 24786 P a

log(pAn) = 9.36789 − 238.89+201397.93 = 3.96818

→ pAn = 103.96818P a = 9294 P a

p = 0.2 · 24786 P a + 0.8 · 9294 P a = 12392 P a yAc = 0.2·2478612392 P aP a = 0.4

yAn = 0.8·929412392P a P aP a = 0.6

(6)

Solution composition

Reverse example: we know that

yAc = 0.4, pAc = 24786 P a, and pAn = 9294 P a What is the composition of the solution?

xAc = ppAc

Ac = ypAc·p Ac

xAc and p unknown:

we need another equation with p and xAc! p = xAc · pAc + (1 xAc) · pAn

Expressing p from the first equation: p = xAc·p

Ac

yAc

From the two equations for p we have

xAc·pAc

= x · p + (1 x ) · p

(7)

Solution composition

xAc·pAc

yAc = xAc · pAc + (1 − xAc) · pAn

= xAc · (pAc − pAn) + pAn

xAc · [pAc − yAc · (pAc − pAn)] = pAn · yAc

xAc = [24786 P a−0.4·(24786−9294)]24786 P a·0.4 = 0.2

xAn = 1 − xAc = 0.8

(8)

Composition of a certain boiling point

What is the composition of the ideal chlorobenzene (cb) - bromobenzene (bb) mixture that starts to boil at 100 kP a on T = 140C?

pcb = 125.24 kP a pbb = 66.10 kP a

Start of boiling: almost all of the mixture is liquid: zcb = xcb p = 100 kP a = xcb · pcb + (1 − xcb) · pbb

= xcb(pcb − pbb) + pbb

→ xcb = p−p

bb

pcb−pbb = 100 kP a−66.10 kP a

125.24 kP a−66.10 kP a = 0.573 x = 1 − x = 0.427

(9)

Composition of a certain boiling point

What is the composition of the vapor phase?

ycb = ppcb = xcb·p

cb

p = 0.573·125.24 kP a

100 kP a = 0.718 ybb = 1 − ycb = 0.282

(10)

Amount of substance in phases

Let’s have 3 mol of the previous mixture at p = 95 kP a with zcb = 0.63. What is the quantity of the solution and the vapor?

nl + ng = 3 mol nng

l = zycb−xcb

cb−zcb = 3 mol−nng

g

xcb = p−p

bb

pcb−pbb = 0.4887 ycb = xcb·p

cb

p = 0.6443

ng

3 mol−ng = 0.63−0.4887

0.6443−0.63 = 9.8811 10.8811 · ng = 29.6434 mol

ng = 2.7243 mol → nl = 0.2757 mol

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