Arnold Sommerfeld Center
Ludwig–Maximilians–Universität München
Prof. Dr. Ivo Sachs
Winter term 2013/14
Exercises on thermodynamics and statistical mechanics (T4)
Problem set 3, due November 4 - November 8, 2013
Solutions
T1 – Thermodynamic potentials
In this exercise, we consider that the number of particles N is fixed.
(i) Explain why we can consider that the internal energy U of a system is naturally a function of the
entropy S and the volume V of the system.
Since we have dU = T dS − pdV , the natural variables for U are S and V .
(ii) Using the appropriate transformation of U (S, V ) to eliminate its dependence on S, define a new thermodynamic potential F (T, V ) that is only a function of the temperature T and the volume V . Proof
that we have the relation
∂U
∂F
=
= −p .
(1)
∂V T
∂V S
Since
∂U
∂S
=T,
V
the thermodynamic potential that we seek is given by
F = U − TS .
It follows that dF = dU − d(T S) = −SdT − pdV , and thus
∂F
∂U
= −p =
.
∂V T
∂V S
(iii) Show that for an isothermal process, the potential F measures the amount of internal energy that may
be converted into work. F is therfore called the free energy of the system.
For an isothermal process we have dT = 0 and hence dF = dU − T dS = δQ − δW − T dS. But by
definition of the entropy S, δQ = T dS and thus dF = −δW .
(iv) Using the appropriate transformation of U (S, V ) to eliminate its dependence on V , define a thermodynamic potential H(S, p) that is a function of the entropy S and the pression p. H(S, p) is called the
enthalpy of the system. Proof that we have the relation
∂H
∂U
=
=T.
(2)
∂S p
∂S V
This time we wish to transform U (S, V ) to a function of S and p only; since
∂U
= −p ,
∂V S
the thermodynamic potential H must be given by
H = U + pV .
We have moreover that dH = dU + d(pV ) = T dS + V dp and thus
∂H
∂U
=T =
.
∂S p
∂S V
(v) Using the appropriate transformation of F (T, V ) to eliminate its dependence on V , define a thermodynamic potential G(T, p) that is a function of the entropy T and the pression p. G(T, p) is called the
Gibbs free energy of the system. Show that for a perfect gas undergoing an isothermal process, the
Gibbs free energy G measures the amount of internal energy that may be converted into work.
Since
∂F
= −p ,
∂V T
the thermodynamic potential G(T, p) is given by
G = F + pV .
We have dG = dF + d(pV ). Since for a perfect gaz pV = N kB T , d(pV ) vanishes for an isothermal
process. Using the result of question (iii), we find that for isothermal processes of a perfect gaz,
dG = −δW , hence the result.
T2 – Temperature of the atmosphere
Calculate the dependence of the temperature of the atmosphere on the height above sea level.
Hint 1: You may assume the atmosphere behaves like an ideal gas, with internal energy U = 23 N kB T .
Hint 2: Since air is a poor conductor of heat you may assume that air rises adiabatically in the atmosphere
without exchange of heat.
For a perfect gas undergoing an adiabatic process, we have dU = 32 N kB dT = −pdV since δQ = 0. Moreover
we must have pV 5/3 = cst., hence pdV = − 53 V dp and thus
dT =
2 V
dp .
5 N kB
The variation of pressure dp is related to a variation of height dz by dp = −ρgdz, where z is the height
above sea level, ρ is the mass density of air at height z and g is the gravitational constant (that we assume
to be constant). Combining our results we find
dT = −
2 mg
dz ,
5 kB
where m = V ρ/N is the average mass of a constituent of the gas. Integrating this relation and calling T0 the
temperature at see level (where we choose the origin of the z axis) yields the result:
T (z) = T0 −
2 mg
z.
5 kB
T3 – van der Waals equation of state
The van der Waals equation of state for N particles relates the pressure p, the volume V = vN and the
temperature T of a gas by
a
p + 2 (v − b) = kB T ,
(3)
v
where a and b are some parameters. This equation has been established experimentally, and for all observed
gases the parameters a and b are positive numbers independent of T , N and p. An ideal gas corresponds to
the particular case a = b = 0.
(a) Explain why the parameter b can be interpreted as the proper volume of the gas constituents.
Since the temperature must always be positive, we find the constraint v > b. Since v = V /N represent
the volume per particle, b may be interpreted as the proper volume of the particles.
(b) Show that in general we have the relation
∂U
∂p
=T
− p.
∂V T
∂T V
(4)
Let us derive U (S(T, V ), V ) with respect to V at fixed T . Using the chain rule we obtain
∂U
∂S
∂U
∂U
=
+
.
∂V T
∂S V ∂V T
∂V S
Using the result of the Problem Set 2, question T1, as well as dU = T dS − pdV , we get the result.
(c) Show that in the case of a van der Waals gas, we have
∂U
a
= 2·
∂V T
v
(5)
In the light of this formula, what is the physical interpretation of the parameter a?
Using the equation of state (3) we get
kB
∂p
=
·
∂T V
v−b
Plugging into (4) and using again the equation of state (3), we find the result (5). One may interpret this
result as follows. The left hand side of (5) measures the variation of the internal energy as we vary the
volume (at fixed temperature); if the system is composed of particles, this means that as the particles
are brought together or seperated appart, the energy of the system varies. In other words, they are not
free particles: there is some interaction between them. For a perfect gas, where a = 0, this implies that
the particles are actually free.
Z1 – Internal energy of an ideal gas
The internal energy U is naturally a function of the entropy S and the volume V of the system. The goal
of this exercise is to find this function for an ideal gas. Since we already know that the internal energy U is
given in term of the temperature T of the system by
3
N kB T ,
(6)
2
we must express T as a function of the entropy S and the volume V .
The entropy difference ∆SAB between the state A (that we shall call the state of reference) and any other
state B is defined by
Z B
δQ
∆SAB =
·
(7)
T
A
The value of ∆SAB is independent of the reversible path used to compute the integral in the right hand side
of (7). For definiteness, we choose to caracterize the points A and B by their coordinates on the a (p, V )
diagram, with pA < pB and VA < VB .
U=
(a) Using formula (7), express ∆SAB in terms of pA , VA , pB , VB .
Hint: you may use for example the following path: (i) an isochore transformation from (pA , VA ) to
(pB , VA ) followed by (ii) an isobare transformation from (pB , VA ) to (pB , VB ).
(i) For the isochore tranformation we have dV = 0 and thus δQ = 32 N kB dT . Using the equation of state
pV = N kB T , we find
Z (pB ,VA )
δQ
3V
δQ
3N kB pB
=
dp ⇒
=
ln
·
T
2T
T
2
pA
(pA ,VA )
(ii) For the isobaric transformation we have dp = 0 and thus δQ = 52 pdV . This yields
Z (pB ,VB )
5
dV
δQ
5N kB VB
δQ
= N kB
⇒
=
ln
·
T
2
V
T
2
VA
(pB ,VA )
Adding the two contribution we just found yields the following expression for the total entropy change
∆SAB between state A and B:
5/3
3N kB pB VB
∆SAB =
ln
·
5/3
2
pA V
A
(b) Verify explicitly using the formula derived in (a) that if A and B are on an adiabatic curve, then
∆SAB = 0.
5/3
5/3
If A and B are on an adiabatic curve, then pA VA = pB VB and thus ∆SAB = 0, as it should.
(c) Show that the entropy S of any state caracterized by (p, V ) is given in terms of the internal energy U
and the volume V by
3/2 U V
,
(8)
S(U, V ) = N k ln
c3/2
where c is some constant depending on the choice of the reference point. Give the explicit formula relating
c to the internal energy UA and the volume VA of the reference point A.
Using the equation of state and U = 32 N kB T we can also write
3/2
∆SAB = N kB ln
UB VB
3/2
·
UA VA
3/2
Setting c = UA VA , we directly find that the entropy S(U, V ) of any state of internal energy U and
volume V with respect to the reference point A is given by (8).
(d) Express the internal energy U as a function of S, V and the constant c. Verify that this function U (S, V )
satisfies the relation
∂U
= −p .
(9)
∂V S
Hint: show that for any point (p, V ) on an adiabatic curve, pV 5/3 = 2c/3.
Inverting relation (8) we find
U (S, V ) = cV −2/3 e2S/(3N kB ) .
Let us now explicitly verify the relation (9). We immediately find
2c
∂U
= − V −5/3 e2S/(3N kB ) .
∂V S
3
2/3
5/3
Since c = UA VA , using the equation of state and U = 32 N kB T we find c = 32 pA VA . Moreover, since
the entropy is fixed, the point (p, V ) must be on the same adiabatic curve as the reference point (pA , VA ).
As a result, we must have pV 5/3 = 2c/3 and S = 0. This yields
2c
∂U
= − V −5/3 = −p ,
∂V S
3
proving the result.
Z2 – Legendre transformation
Let f be a smooth function of one variable x. Setting y = f 0 (x), we define a new function f ? by
f ? (y) = f (x) − xy .
(i) Show that f ? is independent of x as suggested by the notation in (10).
We compute the differential of the right hand side of (10). On gets
d(f (x) − xy) = f 0 (x)dx − ydx − xdy ,
and thus using the definition y = f 0 (x) we find the result.
(ii) Show that if f is a convex function of x, then f ? is a concave function of y.
Let us compute the second derivative of f ∗ :
d2 f ∗
d
dx dx
dx
0
=
f
(x)
−
y
−
x
=−
·
2
dy
dy
dy
dy
dy
Using the identity
dx dy
= 1,
dy dx
(10)
we thus find
1
d2 f ∗
= − 00 ·
2
dy
f
Hence the sign of the second derivative of f ∗ is always opposit to the sign of f 00 , proving the result.
In mathematics, −f ? is called the Legendre transform of f .
General information
The lecture takes place on
Tuesday at 12 - 14 c.t. and Friday at 10 - 12 c.t. in H 030 (Schellingstraße 4).
The webpage for the lecture and exercises can be found at
http://www.physik.uni-muenchen.de/lehre/vorlesungen/wise_13_14/T4_statistical_mechanics.