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Chem 113A Homework # 4
Solutions
- Use the Maxwell relations to prove the following relations.
๐
๐
๐
Starting with ๐๐ป = ๐๐๐ + ๐๐๐
2
2
๐
๐
๐
๐
By definition ๐ผ and ๐
๐
are given by
๐
๐
๐
Thus,
๐
Starting with ๐๐บ = โ๐๐๐ + ๐๐๐
2
2
๐
๐
๐
๐
๐
๐
- Show that
[
]
๐
- Solve the differential equation derived in Problem 2 by integration from T 1
to T 2
to obtain an
expression that would allow you to relate โ๐ด at the two temperatures. This is similar to what
was done in lecture for โ๐บ.
- It is possible to derive an expression for the chemical potential in terms of the Helmholtz free
energy ๐ด = ๐ โ ๐๐. To accomplish this begin with the expression for ๐๐บ for two species 1
and 2.
1
2
๐
๐
2
๐= 1
The Helmholtz energy is the Legendre transform ๐ด = ๐บ โ ๐๐. First find ๐๐ด, then plug the
above expression for ๐๐บ into your result for ๐๐ด. The new expression for ๐๐ด is a function
of ๐, ๐, ๐
1
, and ๐
2
. Using this expression for ๐๐ด define the quantity ๐
๐
(๐ = 1 , 2 ) in terms of
partial molar quantities of A.
๐๐ด = ๐(๐บ โ ๐๐) = ๐๐บ โ ๐(๐๐) = ๐๐บ โ ๐๐๐ โ ๐๐๐ = โ๐๐๐ + ๐๐๐ + โ ๐
๐
๐๐
๐
2
๐= 1
โ ๐๐๐ โ ๐๐๐
skater of mass 85 kg is balanced on one skate, what pressure is exerted at the
interface of the skate and the ice?
(c) What is the melting point of ice under this pressure?
(d) If the temperature of the ice is - 5.0 ยฐC, do you expect melting of the ice to occur
at the ice-skate interface?
No, the lowering of the melting temperature is less than the temperature of the
ice.
- The change in enthalpy is given by ๐๐ป = ๐ถ ๐
๐๐ + ๐๐๐, which implies that for enthalpy
changes โ๐ป we have the differential ๐(โ๐ป) = โ๐ถ
๐
๐๐ + โ๐๐๐. The Clapeyron equation
relates ๐๐ and ๐๐ at equilibrium, and so in combination the two equations can be used to
find how โ๐ป changes along a phase boundary as the temperature changes and the two
phases remain in equilibrium. Show that ๐
๐
ln ๐
Begin with the Clapeyron Equation
Plug this into the equation for ๐(โ๐ป)
๐
๐
๐
Thus,
๐
Now, we take the equation we are trying to prove and manipulate it into something
more intuitive
๐
๐(ln ๐) =
๐
๐
This is a better representation of what we want to prove. Begin by evaluating the
derivative
2
2
Next we plug our expression for ๐
/๐๐ obtained above
๐
2
๐
2
2
๐
Q.E.D.
- Calculate ๏ r
Gยฐ(375 K) for the reaction 2 ๐ถ๐
2
2
(๐) from the value of
r
Gยฐ( 298 K), ๏
r
Hยฐ( 298 K) and the Gibbs-Helmholtz equation.
๐
ยฐ
๐
ยฐ
2
๐
ยฐ
๐
ยฐ
๐
ยฐ
2
๐
ยฐ
According to the Gibbs-Helmholtz Equation
๐
2
2
๐
1
1
๐
2
1
๐
ยฐ
2
2
1
๐
ยฐ
1
๐
ยฐ
2
1
๐
ยฐ
2
