Physical Chemistry I. practice

Physical Chemistry I. practice

Gyula Samu & Rolik Zolt´an

IV.: Phase transitions of ideal one-component systems

rolik@mail.bme.hu

p-T diagram

Clapeyron:

dp

dT = _T^∆H_∆V

Clausius-Clapeyron:

dp

dT = _T^λp_2R

→ ln^p_p²

1 = −_R^λ

1

T₂ − _T¹

1

T,∆H, ∆V : Temperature, enthalpy change and volume change during phase transition

λ: Latent heat, T-independent ∆H

∆H, λ, and ∆V can be either molar ( ^J , ^m3 ) or specific ( ^J , ^m3) changes

Clapeyron vs. Clausius-Clapeyron

Calculate the heat of vaporization of benzene at p = 101.3kP a.

T_v(101.3 kP a) = 80.1^◦C ρ_vap. = 2.741 _m^kg₃

dT_v

dp = 0.32_{kP a}^K (around 1 bar) ρ_liq. = 814.4 _m^kg₃

dp

dT_v = _d¹_Tv

dp

= 3.125 ^{kP a}_K

Clapeyron: ∆H = _dT^dp

v · T_v(101.3 kP a) ·

1

ρ_vap. − _ρ¹

liq.

3.125 ^{kP a}_K · 353.25 K ·

1 2.741 ^kg

m3

− ¹

814.4 ^kg

m3

= 401 ^kJ_kg

Clausius-Clapeyron: λ = _dT^dp

v · [T_v(101.3 kP a)]² · R · ¹_p

3.125 ^{kP a}_K · (353.25 K)² · 8.314_{mol K}^J · _{101.3 kP a}¹ = 32 _mol^kJ

= (since 1 kg Benzene = 12.82 mol) 410 ^kJ_kg

Clausius-Clapeyron

n-octane has

p₁ =26660 P a eq. vapor pressure at T₁ =83.52 ^◦C p₂ =39990 P a eq. vapor pressure at T₂ =95.16 ^◦C What is the p₃ eq. vapor pressure at T₃ =90 ^◦C?

Use the Clausius-Clapeyron approximation!

Clausius-Clapeyron

First we need λ: use Clausius-Clapeyron λ = ^−ln

p2 p1·R

1

T2−_T¹

1

= ^−ln

39.9 kP a 26.6 kP a·R

1

368.31 K−_356.67¹ _K = 38044_mol^J

Now we can calculate p₃ with Clausius-Clapeyron:

p₃ = p₁·e⁻

λ R·

1

T3−_T¹

1

= 26.6 kP a·e⁻

38044 J/mol

R ·

1

363.15 K−_356.67¹ _K

= 33.52 kP a

Clapeyron

The melting point of acetic acid as a function of the pressure is given as (p in P a):

T_m(p) = 16.66^◦C + 0.231 · 10⁻⁶_{P a}^◦C · p − 2.25 · 10⁻¹⁶_{P a}^◦C₂ · p²

a) What is ∆H_f at standard pressure (10⁵ P a) if ∆V_f = 0.156 ^dm_kg³?

We need _dT^dp

m and T_m for Clapeyron

dp

dT_m = _dTm¹

dp

= ¹

2.31·10^−7◦_{P a}^C+4.5·10^{−16 ◦C}

P a2·p = 4.328 · 10^{6P a}◦C = 4.328 · 10^{6P a}_K

T_m(10⁵ P a) = 16.683^◦C

∆H_f = _dT^dp

m · T_m · ∆V_f = 4.328 · 10^{6P a}_K · 289.833 K · 1.56 · 10^−4m_kg³

= 196 ^kJ_kg

Clapeyron

The melting point of acetic acid as a function of the pressure is given as (p in P a):

T_m(p) = 16.66^◦C + 0.231 · 10⁻⁶_{P a}^◦C · p − 2.25 · 10⁻¹⁶_{P a}^◦C₂ · p² a) What is ∆H_f at 100 MP a pressure if ∆V_f = 0.115 ^dm_kg³?

Clapeyron

We need _dT^dp

m and T_m for Clapeyron

p = 10⁸P a

T_m(10⁸P a) = 16.66^◦C + 0.231 · 10⁻⁶_{P a}^◦C · 10⁸P a

−2.25 · 10⁻¹⁶_{P a}^◦C₂ · 10¹⁶P a² = 37.51^◦C

dp

dT_m = _dTm¹

dp

= ¹

2.31·10^−7◦_{P a}^C+4.5·10^{−16 ◦C}

P a2·p = 5.376 · 10^{6P a}_K

∆H_f = _dT^dp

m ·T_m·∆V_f = 5.376·10^{6P a}_K ·310.66 K ·1.15·10^−4m_kg³

= 192 ^kJ_kg

Application

From the 10 ^◦C street we go into a room with 20 ^◦C temperature and 60% relative humidity. Will our glasses get steamed?

Eq. vapor pressure of water at 20 ^◦C is 2.3 kP a λ_v = 40.7 kJ/mol, steam is ideal gas

Question: is the partial pressure of water in the room higher than the eq. vapor pressure for 10 ^◦C? If yes, the vapor will conensate on the glasses

p₁ = ^p2

e

−λv R

1 T2− 1

T1

= ^{2.3 kP a}

e

−40700 J/mol R

1

293.15 K− 1 283.15 K

= 1.275 kP a < 0.6 · 2.3 kP a = 1.38 kP a