1. No category

Physical Chemistry Quiz 5 solution
1. (5%) Why does the triple point in a P-T diagram become a triple line in a P-V
diagram?
2.
(5%) Is the following statement correct? Because dry ice sublimes, carbon
dioxide has no liquid phase. Explain your answer.
3. (10%) Are the two P-T phase diagrams below likely to be observed for a pure
substance? If not, explain all features of the diagram that will not be
observed.
4. (10%) The normal melting point of H2O is 273.15K, and Hfusion = 6010 J
mol-1. Calculate the change in the normal freezing point at 100. and 500. bar
compared to that at 1 bar assuming that the densities of the liquid and solid
phases remain constant at 997 and 917 kg m-3, respectively. Explain why your
answer is positive (or negative).
Sol:
5. (10%) At 298.15 K,
(HCOOH, g) = -351.0 kJ mol-1 and
(HCOOH,
-1
l) = -361.4 kJ mol . Calculate the vapor pressure of water at this temperature.
Sol:
6. (10%) Consider the transition between two forms of solid tin, Sn(s, gray) →
Sn(s, white). The two phases are in equilibrium at 1 bar and 18 . The
densities for gray and white tin are 5750 and 7280 kg m-3, respectively, and
the molar entropies for gray and white tin are 44.14 and 51.18 J K-1 mol-1,
respectively. Calculate the temperature at which the two phases are in
equilibrium at 350. bar.
Sol:
dP
S

7. (a) (10%) Derive Clapeyron equation ,
dT Vm
equilibrium.
for two phases at
(b) (10%) Derive Causius – Clapeyron equation based on Clapeyron equation.
What are the assumptions you have to make for the derivation?
Sol:
(a)
 g  l
shift to different T,P 
 g'  l'
 d  g  d l
Vm g dp  S g dT  Vm l dp  Sl dT
Vdp  SdT
S dP

V dT
(b)
 vap
RT
take Svap 
and Vmg  b  Vl
Tb
P
into

S dP
dP H



P
V dT
dT RTb
dP H

dT
P RTb
8. (10%) The normal boiling temperature of benzene is 353.24 K, and the vapor
pressure of liquid benzene is 1.9 104 Pa at 293.15K. The enthalpy of fusion
is 9.95kJ mo-1, and the vapor pressure of solid benzene is 137 Pa at 228.9K.
(a) calculate H vaporization
(b) calculate Svaporization
Sol:
H vap
1.01325  105
1
1

(

)
4
1.9 10
8.314 353.24 293.15
1.01325  105
8.314  ln
1.19 104  23.97
H vap  
30.7kJ / mol
1
1
(

)
353.24 293.15
H vap
dH
67.8 J / mol k (Tb  353.24)
dS 
(T, P  constant)  Svap 
 86.9
T
Tb
ln
9. For water at 298K, the vapor pressure ( P0) is 0.0316 bar
 = 0.9967 g/cm3 , and molecular mass is 18.015 amu.
(a) (10%) Please estimate the vapor pressure of water at 298K if 100 bar of
external pressure is applied to the system.
(b) (10%) Following (a), if bulk water at 298K is sprayed on a hydrophobic
surgace to form a series of water droplets with uniform spherical size
(with radius R = 10 nm), estimate the vapor pressure of the system
(  H2O  7.2 102 N / m )
Sol:
(a)
  
l
g
apply external force 
P
  g'

P
P
Vm ((100  0.0316)bar )  RT ln
0.0316
3
M 0.018015kg / mol
5 m
Vm 


1.8

10
996.7 kg / m3
mol

l'     Vm (Pf  Pi )   g  RT ln
l
1.899.97
P  e 8.314298  0.0316  0.03398bar
(b)
Spilled water => external force is aroused from the surface tension.
2
Psurface 
 144bar
r
using the equation form (a)
Pvapor =e
1.8(144  0.0316)
8.314 298
 0.0316  0.035bar
10. (10%) The pressure on the concave side of an interface is always greater than
the pressure on the convex side.The difference in pressure of the two sides is
equal to Ps.
Prove that Ps 
Sol:
2
,where γis surface tension and R is radius
R
method1:
  4 r 2
d  8 rdr
In order to get the information of  ,
assume the bubble is expanding with an infinity small amount.
Then the amount of works can represent by Fdr
 d  Fdr
 Fsurface  8 r
Pinside  Poutside 
Fsurface
4 r 2
2
(for droplet =>only one layer)
r
4

(for bubble =>two layers)
r
Psurface 
Psurface
method 2:
(Here Fsurface means
Fcrossection)
keypoint:pressure is equal any where.
Force apply on any cross section are the same.
Pin  Pout 
Fcrosssection
A
Fcrosssection  2 r
Acrossection   r 2
2 r 2

(single layer)
r
 r2
4 r 4
= 2 
(double layer)
r
r
Psurface 