PHYS 501
Homework II
1. For a monatomic ideal gas, we have
V T 3/2
+ c0 ,
S = kB N ln
N
pV = kB N T ,
3
E = kB N T .
2
where c0 is some constant. You can take c0 = 0 to simplify part (a).
(a) Compute the Helmholtz free energy F , the enthalphy H, the Gibbs free energy
G and the internal energy E as a function of the respective set of independent
parameters (in other words, express H as a function of S, p and N ).
F = F (T, V, N ) = E − T S ,
H = H(S, p, N ) = E + pV ,
G = G(T, p, N ) = E − T S + pV .
(b) Check the following relations by computing the derivatives explicitly
∂F
∂E
=
−p =
∂V S,N
∂V T,N
∂H
∂G
V =
=
∂p S,N
∂p T,N
(c) Check that
∂E
∂V
6=
S,N
∂E
∂V
T,N
by computing the derivatives explicitly. (This is a good example that shows that
the partial derivatives depend on the list of variables that are kept constant.)
(d) Since the Gibbs free energy G depends only on a single extensive variable, show
(argue) that it is proportional to it. Let
G
N
be the Gibbs free energy per particle. Compute µ and find the relation between
µ and g.
(e) Compute E − T S + pV − µN .
g=
2.
(a) Let
f = f (x1 , x2 , · · · , xn ; y1 , y2 , · · · , ym )
be a homogeneous function of y1 , · · · , ym with degree k, i.e., for any scale parameter λ we have
f = f (x1 , · · · , xn ; λy1 , λy2 , · · · , λym ) = λk f (x1 , · · · , xn ; y1 , y2 , · · · , ym ) .
Show that
y1
∂f
∂f
+ · · · + ym
=kf .
∂y1
∂ym
Page 2 of 4
PHYS 501 - HW 2
(b) All thermodynamic functions are homogeneous functions of extensive parameters
and the analogs of the relation in part (a) can be written. For example, for a gas
with one component, the first law reads
dE = T dS − pdV + µdN
Show (argue) that E = T S − pV + µN .
(c) Let G = G(T, p, N ) be the Gibbs free energy of a gas with one component. Show
that G = µN and therefore µ is equal to Gibbs free energy per particle.
(d) Let G = G(T, p, N1 , N2 , · · · , Nm ) be the Gibbs free energy of an m-component
gas. Show that G = µ1 N1 + · · · + µm Nm .
3. Consider a container filled with a one-component gas. The container is mentally divided into two. Let i = 1 and i = 2 denote the divisions, Ti denote the temperature of
division i, and µi denote the chemical potential of division i.
(a) Show that heat will flow from division-1 to division-2 if and only if T1 > T2 .
(b) Show that, when T1 = T2 , particles will flow from division-1 to division-2 if and
only if µ1 > µ2 .
4. Consider a closed vessel containing a mixture of gases of H2 O, H2 and O2 . The following
reaction can take place in either direction (forward and reverse),
H2 O ←→ H2 +
1
O2
2
(a) Use this information to deduce the relation between the chemical potentials µH2 O ,
µH2 and µO2 .
Note: Clearly express your assumptions about the conditions which the vessel is
subjected to (thermally insulated, isothermal, isobaric) and express which quantity needs to be extremum for these conditions.
Hint: The quantity that needs to be extremum is a function of the numbers of
molecules of each component, namely NH2 O , NH2 and NO2 . Suppose that the
reaction above has taken place x times. In such a case, NH2 O will change by
NH2 O −→ NH2 O − x .
How much would NH2 and NO2 change? Equilibrium will be reached for a value
of x which makes your function extremum.
(b) Suppose that at a given moment, the chemical potentials of each component is as
follows:
µH2 O = −0.025 eV ,
µH2 = −0.030 eV ,
µO2 = 0.015 eV .
In which direction will the reaction proceed? (Note: The speed of reaction depends
strongly on temperature, but the reaction will proceed at any temperature no matter
how slow the speed is. Do not be confused by this.)
Page 3 of 4
PHYS 501 - HW 2
5. In a vessel, a movable separator wall with mass m
separates two gases. The separator wall can slide
freely in the vertical direction and the vessel is
in thermal contact with the environment which
has temperature T . Let A be the area of the
separator wall.
(a) Express the Helmholtz free energy of the gases in the vessel and the separator in
terms of the volumes of the compartments V1 , V2 and the height h of the piston.
Note-1: Include only the potential energy of the separator wall. We will be interested in final equilibrium states so that we can take the kinetic energy of the wall
to be 0.
Note-2: You do not know (and do not need to know) what the gases are, so you can
keep the thermodynamical functions of them to be arbitrary, i.e., do not assume
anything specific about the gases except the information given above.
(b) As V1 is a free variable, it will change until the free energy is minimum. Show
that the minimization condition is equivalent to
p1 = p2 +
mg
.
A
(c) Now consider the system shown on the right. In
this case, there is only one compartment of the
vessel. The outside of the vessel is the atmosphere having temperature T and pressure pa . If
we consider the separator wall and the gas inside as the system (i.e., we do not consider the
atmosphere as part of the system), then which
thermodynamic potential has to be extremized
for identifying equilibrium state? Compute that
function and show that
p = pa +
mg
A
at equilibrium.
6. Consider a gas inside a closed vessel. What is the minimum possible value of the work
that you must do to change its macrostate from (T, V, N ) to (T, V /3, N )?
Page 4 of 4
PHYS 501 - HW 2
7. Consider a closed vessel with a movable separator wall as shown. We
will assume that the mass of the separator wall is negligible so that
the pressure inside is equal to the atmospheric pressure. The vessel is
also thermally conducting. Let p and T be the atmospheric pressure
and temperature.
Suppose that the vessel is filled with a pure substance (formed from
single type of molecule, e.g., H2 O) and the substance can be either in
the liquid state (`) or the gaseous state (g). Of course, molecules can
pass back and forth between these phases, i.e., in chemical notation
` ←→ g
is possible in both directions. Let N be the total number of molecules.
(a) Let g` (T, p) be the Gibbs free energy per particle for the liquid state and gg (T, p)
be the corresponding quantity for the gaseous state. Show that,
∗ if g` (T, p) < gg (T, p), then the substance inside the vessel is completely in
liquid state,
∗ if g` (T, p) > gg (T, p), then the substance inside the vessel is completely in
gaseous state, and
∗ if g` (T, p) = gg (T, p), then both the liquid and the gaseous phases co-exist.
Argue that, in this case, the number of molecules N` and Ng in liquid and
gaseous phases can have any value.
Note: When the relation g` (T, p) = gg (T, p) is solved for p, we get a continuous
function of T , i.e., p = f (T ) for some function f . This p–T curve is usually
known as “vaporization curve” or “vapor pressure curve”.
(b) We have kept a possible solid phase out of the discussion only for the sake of
simplicity. In reality, we have to take all three phases into account for the determination of the stable phase. (To complicate matters, there might be more than
one solid phase depending on possible crystal structures, but forget about such
details for the sake of simplicity.) Consider the possibility of three phases. The
molecules can pass from one phase to the other. How do decide which phase is
the most stable? How do you decide if two phases co-exist? Is it possible that
all three phases co-exist? (The values of T, p where all three phases co-exist are
known as a “triple point”.)
(c) In part (a), we have assumed that the volume of the vessel can change in such
a way that the pressure inside has always the same value p? What will happen
if we did not impose this condition? Suppose that this time there is no movable
separator wall and the volume V of the vessel is fixed. As a result, while ` ←→ g
reaction proceeds, the pressure inside can change. Suppose that the walls are still
thermally conducting.
∗ Which thermodynamic potential has to be extremized for finding the equilibrium state?
∗ Is it possible that there can be only a single phase inside? Is it possible that
both phases can co-exist?