Physical Chemistry I. practice
Gyula Samu
IV.: Phase transitions of ideal one-component systems
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
p-T diagram
Clapeyron:
dp
dT = T∆H∆V
Clausius-Clapeyron:
dp
dT = Tλp2R
→ lnpp2
1 = −Rλ
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.
Tv(101,3 kP a) = 80,1◦C ρvap. = 2,741 mkg3
dTv
dp = 0,32kP aK (around 1 bar) ρliq. = 814,4 mkg3
dp
dTv = d1Tv
dp
= 3,125 kP aK Clapeyron: ∆H = dTdp
v · Tv(101,3 kP a) ·
1
ρvap. − ρ1
liq.
3,125 kP aK · 353,25 K ·
1 2,741 kg
m3
− 1
814,4 kg
m3
= 401 kJkg
Clausius-Clapeyron: λ = dTdp
v · [Tv(101,3 kP a)]2 · R · p1
3,125 kP aK · (353,25 K)2 · 8,314mol KJ · 101,3 kP a1 = 32 molkJ
= (since 1 kg Benzene = 12,82 mol) 410 kJkg
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 = 38044molJ
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):
Tm(p) = 16,66◦C + 0,231 · 10−6P a◦C · p − 2,25 · 10−16P a◦C2 · p2
a) What is ∆Hf at standard pressure (105 P a) if ∆Vf = 0,156 dmkg3?
We need dTdp
m and Tm for Clapeyron
dp
dTm = dTm1
dp
= 1
2,31·10−7◦P aC+4,5·10−16 ◦C
P a2·p = 4,328 · 106P a◦C = 4,328 · 106P aK
Tm(105 P a) = 16,683◦C
∆Hf = dTdp
m · Tm · ∆Vf = 4,328 · 106P aK · 289,833 K · 1,56 · 10−4mkg3
= 196 kJkg
Clapeyron
The melting point of acetic acid as a function of the pressure is given as (p in P a):
Tm(p) = 16, 66◦C + 0, 231· 10−6P a◦C ·p− 2, 25 ·10−16P a◦C2 · p2 a) What is ∆Hf at 100 MP a pressure if ∆Vf = 0, 115 dmkg3?
Clapeyron
We need dTdp
m and Tm for Clapeyron
p = 108P a
Tm(108P a) = 16,66◦C + 0,231 · 10−6P a◦C · 108P a
−2,25 · 10−16P a◦C2 · 1016P a2 = 37,51◦C
dp
dTm = dTm1
dp
= 1
2,31·10−7◦P aC+4,5·10−16 ◦C
P a2·p = 5, 376 · 106P aK
∆Hf = dTdp
m·Tm·∆Vf = 5, 376·106P aK ·310, 66 K·1, 15·10−4mkg3
= 192 kJkg
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 λv = 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
