Solutions 1 – 1.13, 1.15, 1.16 and additional problem 2
1.13 You have a thermodynamic potential as the EOS in this problem. This is a rare treat because it
simplifies everything.
You are given
E
(
S, L, N
) so
S, L, N
are the natural independent variables - I hope everyone got that.
Since you have
E
you can get explicit expressions for
T, f,
each as a function of
S, V, N:
dE
=
T dS
f dL
dN.
Differentiating we get:
T
=
2
S L
N
2
,
f
=
S
2
N
2,
=−
2
S
2
L
N
3
.
Gibbs-Duhem says
S dT
Ldf
N d
=
0, but our intensive variables are themselves functions of
S,
V, N.
No matter – we just calculate the differentials d
T, df, d
in terms of
dS, dV, dN
.
dT
=
2
L
N
2
dS
2
S
N
2
dL
−
4
S L
N
3
dN ,
df
=
2
S
N
2
dS
−
2
S
2
N
3
dN ,
du
=−
4
S L
N
3
dS
−
2
S
2
N
3
dL
6
S
2
L
N
4
dN.
Its a good idea when you are deriving such expressions to check the dimensions. Each term on the RHS
of each of these equations should be intensive so as to match the intensive property on the LHS.
The Gibbs-Duhem relation must hold for all variations in
S, V, N
. Once again this means that the
coefficients multiplying each differential must vanish. Substituting for d
T, df, d
in the Gibbs-Duhem
equation, and grouping terms we have:
2
S L
N
2
2
S L
N
2
−
4
S L
N
2
dS
2
S
2
N
2
−
2
S
2
N
2
dL
−
4
S
2
L
N
3
−
2
S
2
L
N
3
6
S
2
L
N
3
dN
=
0.
1.15 Rewrite 1
st
law:
dS
=
dE
p dV
−
dN.
Legendre transformations give the two potentials:
1
=
S
−
E,
and
2
=
S
−
E
N.
The differential relations are:
d
1
=−
Ed
pdV
−
dN ,
and
d
2
=−
Ed
pdV
N d
.
Note that the thermodynamic forces (or conjugate variables) in this case are
,
p,
and
.
Its easier to work with 1/
T
as the independent variable rather than
T
.
Also note that part 1 is zero help in solving part 2 in this case – in fact it will probably lead you astray.
A little joke on Chandler's part.