Problem Set 3
Solutions
3.20 MIT
Professor Gerbrand Ceder
Fall2003
2001
Fall
Problem 1
There are two steps to solving this problem
(1) What is the pressure difference in terms of the height?
(2) What is the height of the hill?
Use the following form of the Clapeyron equation (this is valid for equilibrium between an ideal gas and a solid or
liquid)
H
1
d ln P = d
R
T
Which in this case can be written in terms of the pressure at the top and bottom of the hill, and the respective
boiling points
Ptop
H
1
1
ln
=
Pbottom
R Ttop
Tbottom
Now we write the pressure at the bottom in terms of the pressure at the top
Pbottom = Ptop + air gh
Substituting this,
ln
Ptop
Ptop + gh
= ln
1+
gh
Ptop
1
If we expand ln (1 + x) we get,
gh
H
1
1
=
Ptop
R Ttop
Tbottom
H Ptop
1
1
h=
g
Ttop
Tbottom
Assuming that the air is an ideal gas, we can relate the density and pressure to the temperature using the ideal gas
law
P
PV
nRT
RT
=
=
=
M
M
(M W )
where (MW) is the molecular weight of the air. We now have
H RT
1
1
h=
g(M W ) Ttop
Tbottom
2
Plugging in some numbers (g = 10m/s , M W = 28.8g/mol, and we are dealing with air at 25 C )
1000cal/mol 4.184J/cal 8.314J/molK 298K
1
1
h=
10m/s2 0.0288kg/mol
368K
378K
h = 2.6km
Problem 2
(a) Integrate the Clausius-Clapeyron Equation
d ln P H
=
d(1/T )
R
P2
H 1
1
ln
=
P1
R T2
T1
1
For this example, P1 = 1 atm (at boiling), T1 = 1620K,and T2 = 1000K
1
1
156, 000
ln P2 =
8.314
1000 1620
P2 = 7.6 104 atm
(b) At the melting point of Li, Psolid = Pliquid
19, 314
1
156, 000 1
13.049 =
8.314
1620
T
T
550.5
1.47 =
T
T = 375K
Problem 3
First we can find what the vapor pressure over the liquid would be under these conditions
30671
30671
ln(P ) = + 12.5 = + 12.5 = 7.95
T
1500
P = 3.54 104 atm = 35.8 Pa
If all of the silver was in the vapor phase (1 gram)
1
nRT
=
(8.314) (1500) = 115.61 Pa
P =
107.87
V
Hence, there must be both liquid and vapor present
P = P = 35.8 Pa
Now we can find the moles in the vapor phase
35.8
PV
=
= 2.87 103 moles
n=
8.314 1500
RT
Grams of vapor
2.87 103 moles (107.87 g/mol) = 0.31g
So the fraction in the vapor phase is 0.31
(b) In general the amount of heat necessary would be the sum of the mixing enthalpy and the enthalpy of
vaporization
Htotal = Hmix + Hvap
However, since we are told to assume the solution is ideal, the enthalpy of mixing is zero. (This is related to the
assumption that the mixing species do not interact)
Thus
Q = Htotal = Hvap
Since we were not given the enthalpy of vaporization or the boiling point in the data, we must find these quantities.
To find the boiling point, use the given equation for the vapor pressure and find the temperature where the pressure is
equal to 1 atm.
30671
ln(1) = + 12.5 → Tb = 2454K
T
To find the heat of vaporization, we can use the integrated form of the Clapeyron equation for an ideal gas.
We know the temperature and pressure at two points (from the given equation for vapor pressure) and we can then
calculate Hvap.
P2
Hvap 1
1
ln
=
P1
R
T2
T1
Choosing T= 1500 and 1600 K, the value calculated for Hvap is 255.kJ/mole. (If you look this up you get 251
kJ/mole) Now we can get the enthalpy of vaporization at 1500K, which can be done using a loop, since we know
enthalpy is a state function
2
Hvap(1500K ) =
2454
1500
Cpl dT + Hvap(2454K ) +
1500
2454
Cpg dT
5
Hvap(1500K ) = 33J/molK (2454 1500) + 11, 300J/mol + R(1500 2454)
2
Hvap(1500K ) = 267kJ/mol
Problem 4
At 70 C,
Ptotal = PN2 + PH2 O = 1atm
PH2 O = 0.3atm =⇒ PN2 = 0.7atm
After the isothermal volume reduction:
PH2 O is still equal to 0.3 atm since PH2 O can not increase since it is the saturation vapor pressure! This will
mean that as the volume is being reduced, H2 O will condense, keeping PH2 O at 0.3 atm.
PN2 will double
RT
PN2 = nN2
V
V
=⇒ PN2 → 2PN2
V →
2
So,
Ptotal = 1.4 + 0.3 = 1.7 atm
Problem 5
= 1atm at the regular boiling point.
(a) The boiling point is where P = Patmosphere . So PH
2O
(b) The Clausius-Clapeyron equation allows us to find vapor pressure at some temperature from knowledge of the
vapor pressure at another temperature
H
1
d
d ln P = R
T
We can integrate this between 373K and 288K
40, 000 1
1
P373
ln
=
8.314 373 288
P288
PH
(288K ) = 0.021atm
2O
Relative humidity is defined as
PH2 O
PH
2O
In this problem the relative humidity is 65% so we can find the PH2 O
PH2 O
0.65 =
=⇒ PH2 O = 0.0136atm
0.021
. Again we integrate the Clausius-Clapeyron equation, this
The dew point is the temperature where PH2 O = PH
2O
time solving for T
0.021
40, 000 1
1
ln
=
0.0136
8.314 288 T
R.H. =
T = 281K = 8 C
3
Problem 6
nuc = number of unit cells
V = nuc V uc , where V uc is the volume per unit cell
V Li
nLi = nuc x
∂V
∂ (nuc V uc )
=
=
∂nLi
∂ (nuc x)
And since nuc is a constant,
∂V uc
∂x
This is the slope of the curve given. Thus at x = 1 the partial molar volume of Li is
V Li =
V Li = 1.15
3
atom
Problem 7
The vapor pressure of pure B at 1000 K is 5 104 atm. If B is in an ideal solution with another species at 1000
K, its vapor pressure will be:
Higher than 5 104
Lower than 5 104
Equal to 5 104
Impossible to say
Problem 8
Reaction to go from non-stoichiometric Li1x N i1+x O2 to LiNiO2
x
x
Li1x N i1+x O2 + Li2 O + O2 → LiNiO2 + xN iO
2
4
Principle of LeChatleier indicates that a higher PO2 will push the reaction to the right.
Problem 9
Once again, we start with the Clausius-Clapeyron equation in this form
dP
H 1
H dT
=
→ dP =
dT
V T
V T
Integrating we get
T2
H
P2 P1 =
ln
T1
V
Let P1 be the pressure at the original interface (T1 = 5 C ) and P2 be the pressure at the new interface
(T2 = 5.2C )
We can also relate the densities to the volume change as follows:
s l
1
1
V = l s =
l s
So,
l s
T2
P2 P1 = s
H ln
l
T1
Also, we know
P1 = P2 + s gh or P2 P1 = s gh
Putting this together we get
4
s gh =
Solving for the density of the solid,
l s
H ln
s
l
T2
T1
H
T2
ln
= 1
gh
T1
Putting in the values of the know quantities (being very careful with units)
8371J/kg
267.8K
ln
s = 1000g/m3 1 9.8m/s2 0.4m
268K
s
l
s = 2.6g/cm3
5