Physical Chemistry I. practice

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Physical Chemistry I. practice

Gyula Samu

V.: Ideal mixtures

gysamu@mail.bme.hu

http://oktatas.ch.bme.hu/oktatas/konyvek/fizkem /PysChemBSC1/Requirements.pdf

http://oktatas.ch.bme.hu/oktatas/konyvek/fizkem /PysChemBSC1/Important dates.pdf

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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 · p^∗₁, p₂ = x₂ · p^∗₂

p^∗₁, p^∗₂: eq. vapor pressure of comp. 1 and 2

y₂ = ^p_p² = ^x²^·p

∗ 2

p

z₁ + z₂ = x₁ + x₂ = y₁ + y₂ = 1

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Konovalov’s first law, lever rule

y₂ = ^p_p² = ^x²^·p

∗ 2

p

Konovalov’s first law for ideal mixtures:

if p^∗₁ < p^∗₂: p^∗₁ < p < p^∗₂ → ^p_p^∗² > 1, y₂ > x₂

n₁ = (n_g + n_l) · z₁ = n_g · y₁ + n_l · x₁ n_g · (z₁ − y₁) = n_l · (x₁ − z₁)

Lever rule: _nⁿ^l

g = _x^z¹^−y¹

1−z₁

We can also calculate n_g and n_l directly

n₁ = n_g · y₁ + n_l · x₁ = n_g · y₁ + (n − n_g) · x₁ = n · x₁ + n_g · (y₁ − x₁)

→ n_g = ⁿ_y¹^−n·x¹

1−x₁ n_l = ⁿ_x¹^−n·y¹

1−y₁

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Vapor composition

We have an ideal mixture of acetone (Ac) and acetonitrile (An) log(p

^∗_Ac

) = 9, 36457 −

_237,50+T^1279,87

log(p

^∗_An

) = 9, 36789 −

_238,89+T^1397,93

(p in P a, T in

^◦

C ; but no dimensions inside log!) T = 20

^◦

C x

_Ac

= 0, 2

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

p = p

_Ac

+ p

_An

= x

_Ac

· p

^∗_Ac

+ (1 − x

_Ac

) · p

^∗_An

p_Ac x_Ac·p^∗_Ac p_An (1−x_Ac)·p^∗_An

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Vapor composition

log(p

^∗_Ac

) = 9, 36457 −

_237,50+20^1279,87

= 4, 39420

→ p

^∗_Ac

= 10

^4,39420

P a = 24786 P a

log(p

^∗_An

) = 9, 36789 −

_238,89+20^1397,93

= 3, 96818

→ p

^∗_An

= 10

^3,96818

P a = 9294 P a

p = 0, 2 · 24786 P a + 0, 8 · 9294 P a = 12392 P a y

_Ac

=

^0,2·24786₁₂₃₉₂ _{P a}^{P a}

= 0, 4

y

_An

=

^0,8·9294₁₂₃₉₂^{P a P a}_{P a}

= 0, 6

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Solution composition

Reverse example: we know that

y

_Ac

= 0, 4, p

^∗_Ac

= 24786 P a, and p

^∗_An

= 9294 P a What is the composition of the solution?

x_Ac = ^p_p^Ac∗

Ac = ^y_p^Ac∗^·p Ac

→ x_Ac and p unknown:

we need another equation with p and x_Ac! p = x_Ac · p^∗_Ac + (1 − x_Ac) · p^∗_An

Expressing p from the first equation: p = ^x^Ac^·p

∗ Ac

y_Ac

From the two equations for p we have

x_Ac·p^∗_Ac

= x · p^∗ + (1 − x ) · p^∗

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Solution composition

x_Ac·p^∗_Ac

y_Ac

= x

_Ac

· p

^∗_Ac

+ (1 − x

_Ac

) · p

^∗_An

= x

_Ac

· (p

^∗_Ac

− p

^∗_An

) + p

^∗_An

x

_Ac

· [p

^∗_Ac

− y

_Ac

· (p

^∗_Ac

− p

^∗_An

)] = p

^∗_An

· y

_Ac

x

_Ac

=

_[24786 P a−0,4·(24786−9294)]²⁴⁷⁸⁶ ^{P a·0,4}

= 0, 2

x

_An

= 1 − x

_Ac

= 0, 8

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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 = 140

^◦

C ?

p

^∗_cb

= 125, 24 kP a p

^∗_bb

= 66, 10 kP a

Start of boiling: almost all of the mixture is liquid: z

_cb

= x

_cb

p = 100 kP a = x

_cb

· p

^∗_cb

+ (1 − x

_cb

) · p

^∗_bb

= x

_cb

(p

^∗_cb

− p

^∗_bb

) + p

^∗_bb

→ x

_cb

=

^p−p

∗ bb

p^∗_cb−p^∗_bb

=

100 kP a−66,10 kP a

125,24 kP a−66,10 kP a

= 0, 573

x = 1 − x = 0, 427

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Composition of a certain boiling point

What is the composition of the vapor phase?

y

_cb

=

^p_p^cb

=

^x^cb^·p

∗ cb

p

=

0,573·125,24 kP a

100 kP a

= 0, 718

y

_bb

= 1 − y

_cb

= 0, 282

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Amount of substance in phases

Let’s have 3 mol of the previous mixture at p = 95 kP a with z

_cb

= 0, 63. What is the quantity of the solution and the vapor?

n

_l

+ n

_g

= 3 mol

ⁿ_n^g

l

=

^z_y^cb^−x^cb

cb−z_cb

=

₃ _mol−nⁿ^g

g

x

_cb

=

^p−p

∗ bb

p^∗_cb−p^∗_bb

= 0, 4887 y

_cb

=

^x^cb^·p

∗ cb

p

= 0, 6443

n_g

3 mol−n_g

=

0,63−0,4887

0,6443−0,63

= 9, 8811 10, 8811 · n

_g

= 29, 6434 mol

n

_g

= 2, 7243 mol → n

_l

= 0, 2757 mol