Persamaan Clapeyron dan Persamaan Clausius-
Clapeyron
Dr. M. Masykuri, M.Si.
Natural Science Study Program
Teacher Training and Education Faculty Sebelas Maret Universitfy (UNS)
Website: http://masykuri.staf.fkip.uns.ac.id, email: [email protected]
Transformasi Energi
Outline
1.
Persamaan Clapeyron
2.
Persamaan Clausius-Clapeyron
Gibbs Energies and Phase Diagrams
• Chemical potential (µ) - for one component system µ =Gm
– Measure of the potential of a substance to bring about change
• At equilibrium, µ is the same throughout the system
– If µ1 is chem. potential of phase 1 and µ2 is chem. potential of phase 2, at equilibrium µ1 = µ2
• Consequence of 2nd Law of Thermodynamics
• If dn is transferred from one location
(phase) to another, -µ1dn, is the change in Gibbs energy in that phase and µ2dn is
change in free energy in the second phase.
Gibbs Energies and Phase Diagrams
• dG = -µ1dn + µ2dn = (µ2 - µ1)dn – If µ1 > µ2 dG < 0 and change
spontaneous
– If µ1 = µ2 dG = 0 and system at equilibrium
– Transition temperature is that temperature where µ1 = µ2
• Chemical potentials of phases change with temperature
– At low T and reasonable p, solid has lowest µ and, hence, most stable
– As T raised, µ of other phase may become less than solid(at that temp.) so it
becomes stable phase
Persamaan Clapeyron
Pada perubahan fasa:
X(fasa 1) X(fasa 2)
dG = -S dT + V dP dG1 = -S1 dT + V1 dP dG2 = -S2 dT + V2 dP Sehingga,
-S1 dT + V1 dP = -S2 dT + V2 dP (S2 - S1) dT = (V2 - V1) dP
S dT = V dP = (1)
Karena,
G = H – T S Pada setimbang G = 0
H – T S = 0 T S = H
S = (2)
(2) (1) didapat,
=
Persamaan CLAPEYRON
Pada setimbang, berlaku:
G = 0 G1 = G2 dG1 = dG2
`
Persamaan Clapeyron
Dari persamaan Clapeyron:
=
Pada H dan T konstan, maka:
P2 – P1= ln
•
Persamaan CLAPEYRON
Persamaan Clausius -Clapeyron
Persamaan Clapeyron, = adalah untuk semua perubahan fasa (padat, cair atau gas).
Khusus untuk kesetimbangan yang melibatkan uap, misalnya:
Cair Uap Berlaku,
=
•
Karena Vuap >>> Vcair, maka=
Pada Gas ideal Vuap =RT/P, =
= =
Persamaan CLAUSIUS - CLAPEYRON
Persamaan Clausius -Clapeyron
Persamaan Clausius-Clapeyron, Jika konstan, maka:
lnP2 – lnP1= ln =
•
Persamaan CLAUSIUS - CLAPEYRON
Problem Set Clausius -Clapeyron Eq.
Titik lebur suatu senyawa organik X adalah 80 oC. Jika
tekanan uap cair X = 10 mmHg pada 85,8 oC dan 40 mmHg pada 119,3 oC, serta tekanan uap Xpadat = 1 mmHg pada 52,6 oC. Hitunglah:
a. Panas penguapan cairan dan titik didih (per definisi:
pada 760 mmHg)
b. Tekanan uap pada titik lebunya
Problem Set Clausius -Clapeyron Eq.
Diketahui:
P1 = 10 mmHg T1= (85,8 + 273) K = 358,8 K P2 = 40 mmHg T2= (119,3 + 273) K = 392,3 K Solid Ps = 1 mmHg
Jawab:
ln = R = 0,082 Latm mol-1K-1
R = 8,314 Jmol-1K-1 ln =
•
Jawab:
= 48470 Jmol-1 = 48,47 kJmol-1
a)
Problem Set Clausius -Clapeyron Eq.
Jika T2 adalah Tdidih
(pada P2 =760 mmHg) ln =
ln = = -
Td = 489 K
•
Jawab:
b) Tlebur = 80 + 273 K = 353K Plebur = …?
T1 = 358,8K P1 = 10 mmHg Maka:
ln = ln =
Plebur = 7,635 mmHg
