Problems in thermodynamics and statistical physics
authors: Jakub Fišák and others
1. Let F(x, y) = x · ex2+y2
. Calculate (a) ∂F
∂x
, (b) ∂F
∂y
, (c) ∂2F
∂x2 , (d) ∂2F
∂x∂y
, (e) ∂2F
∂y∂x
, (f)
∂2F
∂y2 .
Solution:
(a)
∂ x·ex2+y2
∂x
= exp (x2
+ y2
) + 2x2
exp (x2
+ y2
) = exp (x2
+ y2
) (1 + 2x2
),
(b)
∂ x·ex2+y2
∂y
= 2xy exp (x2
+ y2
),
(c)
∂2 x·ex2+y2
∂x2 = 4x exp (x2
+ y2
)+(1 + 2x2
) 2x exp (x2
+ y2
) = exp (x2
+ y2
) (4x + 2x + 4x3
) =
exp (x2
+ y2
) (6x + 4x3
) = 2x (3 + 2x2
) exp (x2
+ y2
)
(d)
∂2 x·ex2+y2
∂x∂y
= 2y (1 + 2x2
) exp (x2
+ y2
),
(e)
∂2
(x·e2+y2
)
∂y∂x
= 2y exp (x2
+ y2
) + 4x2
y exp (x2
+ y2
) = 2y exp (x2
+ y2
) (1 + 2x2
),
(f)
∂2 x·ex2+y2
∂y2 = 2x exp (x2
+ y2
) + 4xy2
exp (x2
+ y2
) = 2x exp (x2
+ y2
) (1 + y2
).
2. Let dω = A(x, y)dx + B(x, y)dy be any differential form (Pfaffian). Show that if
dω is an exact differential (there exists a function F(x, y) such that dω = dF), then it
must hold
a)
∂A
∂y
=
∂B
∂x
, b) dω = 0,
(b) for each closed integration path.
Solution:
(a) Assuming satisfied
df =
∂f
∂x
dx +
∂f
∂y
dy = A(x, y)dx + B(x, y)dy
since the interchangeability of the second derivatives holds:
∂
∂x
∂f
∂y
=
∂
∂y
∂f
∂x
=
∂B(x, y)
∂x
=
∂A(x, y)
∂y
because apparently it fullfills
A(x, y) =
∂f
∂x
, B(x, y) =
∂f
∂y
(b) For the calculation, we use Stokes’ theorem, which is of the form
V
dω =
∂V
ω
1
x
y
a
b
c
d
1 2
1
2
Obrázek 1: Representation of the circular process for task 3.
We calculate the left side
C
df =
C
dx
∂f
∂x
+ dy
∂f
∂y
=
S (∂S =C )
d dx
∂f
∂x
+ dy
∂f
∂y
=
S (∂S =C )
d(df) = 0,
because the outer derivative of the outer derivative of any form is always zero.
3. Let dω1 = (x2
− y) dx + x dy. Is it an exact differential, dω2 = dω1/x2
an exact
differential? Calculate the integral dω between the points (1, 1) and (2, 2) along the
lines (1, 1) → (1, 2) → (2, 2) and (1, 1) → (2, 1) → (2, 2).
Solution:
Since the derivative of the term udx with respect to y s equal to -1 and the derivative
of the second term with respect to x is equal to 1, the term is not an exact differential.
Dividing by x2
both derivatives become equal − 1
x2 , so the term is an exact differential.
We calculate the curve integral as follows: we can divide the integral into two parts
C
dω1 =
C1
dω1 +
C2
dω1
Subsequently, we parameterize the integration path, which is shown in Figure 1. For
path a and b, the parameters are as follows:
(a): x = 1, y = t, t ∈ [1, 2],
(b): x = t, y = 2, t ∈ [1, 2].
So we can calculate the relevant integrals:
C1
dω1 =
2
1
dt = 1
since dx = 0. Second part:
C2
dω1 =
2
1
dt t2
− 2 =
t3
3
− 2t
2
1
= −
2
3
2
V
p
a
b
c
d
p1 p2
p1
p2
Obrázek 2: Representation of the circular process for task 4.
So the result is equal
C
dω1 =
1
3
In the case of integration along the blue integration path, we get the result of 10
3
.
Integrating dω1/x2
, we get same result 1 for both cases.
4. Is dω = p dV + V dp an exact differential? If so, determine the function F whose
exact differential is dω. Compute the integral dω between the points (V1, p1) and (V2, p2)
along the lines (V1, p1) → (V1, p2) → (V2, p2) a (V1, p1) → (V2, p1) → (V2, p2).
Solution:
It follows from the integrability conditions that it is an exact differential. The function
ω is easy to find. We know that
∂F
∂V
= p
after that
F = pv + g(p)
the derivative of this function with respect to pressure is equal to V , from here
∂F
∂p
= V = V +
dg(p)
dp
for the derivative of the function g(p) we get that it is zero, and therefore the function
g(p) must be some constant. So the function F is equal to
F(p, V ) = pV + konst
We calculate the curve integral analogously to the previous case. Fig. 2 shows the
integration paths. the curve integral along the blue path comes out as −V1p1 + V2p2
and of course exactly the same along the red path.
5. Is the fom dQ = c dT + RT
V
dV an exact differential? Calculate the integral dω
between the points (V1, T1) and (V2, T2) long the lines (V1, T1) → (V1, T2) → (V2, T2)
3
a (V1, T1) → (V2, T1) → (V2, T2). What function f(V, T) must we multiply dQ by to
make the product f dQ an exact differential? Determine the function S for which
dS = f dQ.c and R are constants.
Solution:
Since the derivative of the first term with respect to volume is zero and the derivative
of the second term with respect to temperature is equal to R
V
, it is not a complete
differential. The integration takes place along similar integration paths as shown in 2
, the integral along the blue integration path is equal to RT2 ln V2
V1
+ c (T2 − T1) and
along the red integration path: RT1 ln V2
V1
+ c (T2 − T1).
Next comes the slightly more complicated part. We need to find a function f(T, V ),
such that the condition is satisfied
∂(f(T, V )c)
∂V
=
∂ f(T, V )RT
¯V
∂T
We calculate the derivative on both sides:
c
∂f
∂V
= R
T
V
∂f
∂T
+ fR
after multiplying by V we get an equation with separated variables
cV
∂f
∂V
= RT
∂f
∂T
+ fR
we can then assume the function f in the form
f(T, V ) = g(T) · h(V )
Substitute and divide by V · g(T) · h(V )
cV
1
h(V )
∂h(V )
∂V
= RT
1
g(T)
∂g(T)
∂T
+ R
The left side now depends only on volume, the right side only on temperature. In
order to satisfy the equality for any V and T, both sides must be constant. Thus, two
differential equations for volume and temperature fall out of this equation:
cV
1
h(V )
∂h(V )
∂V
= konst.
the solution is, after dividing the equation by the volume
c ln(h(V )) = konst · ln(V )
then the function h(V ) is equal to
h(V ) = V
konst.
c .
4
Let’s focus on the right side:
RT
1
g(T)
∂g(T)
∂T
+ R = konst
we also solve by separating variables:
1
g(T)
∂g(T)
∂T
=
k
R
− 1
and the result is
ln(g(T)) =
konst.
R
− 1 ln(T)
after that
g(T) = T
konst.
R
−1
Let us now recall the function f(T, V ) = g(T) · h(V ), into which we will substitute the
result of our calculations
f(T, V ) = V
konst.
c · T
konst
R
−1
where Const. is arbitrary. Therefore, we choose the simplest solution konst.= 0:
f(V, T) =
1
T
6. Let x, y a z be 3 state variables, connected by the state equation f(x, y, z) = 0.
Show the validity of the relations
∂x
∂y z
=
∂y
∂x
−1
z
(1)
∂x
∂y z
= −
∂x
∂z y
∂z
∂y x
(2)
and
∂x
∂y z
=
∂x
∂y w
+
∂x
∂w y
∂w
∂y z
(3)
where the subscript indicates a constant variable and w is another state variable, w =
w(x, y, z).
Solution:
We calculate the partial derivatives with respect to x and y assuming that z is constant:
df
dx
=
∂f
∂x y,z
+
∂y
∂x z
∂f
∂y x,z
= 0
df
dy
=
∂f
∂y x,z
+
∂x
∂y z
∂f
∂x y,z
= 0
5
from each equation we express the term not containing f :
∂y
∂x z
= −
∂f
∂x y,z
∂f
∂y
x,z
,
∂x
∂y z
= −
∂f
∂y
x,z
∂f
∂x y,z
after multiplying both the two expressions we have:
∂y
∂x z
·
∂x
∂y z
=
∂f
∂x y,z
∂f
∂y
x,z
·
∂f
∂y
x,z
∂f
∂x y,z
= 1
which proves (1). Let’s now write the right side of the expression
∂y
∂x z
= −
∂f
∂x y,z
∂f
∂y
x,z
we transform the derivatives on the right-hand side into z z-coordinates
∂y
∂x z
= −
∂z
∂x y
∂f
∂z y,z
∂z
∂y
x
∂f
∂z x,z
= −
∂z
∂x y
∂z
∂y
x
= −
∂z
∂x y
∂y
∂z x
this proves expression (2). Recall the coordinate transformation formula (x, y, z) →
(u, v, w) for the derivative:
∂y
∂x z
=
∂y
∂u v,w
∂u
∂x y,z
+
∂y
∂v u,w
∂v
∂x y,z
+
∂y
∂w u,v
∂w
∂x y,z
we will make use of this relationship assuming that we are only transforming a single
coordinate from z → w
∂y
∂x z
=
∂y
∂x y,w
+
∂y
∂w
7. The equation of state pV = NkT connects the variables p, V and T, where N and
k are constants. Verify by direct calculation that
∂p
∂V T
=
∂V
∂p
−1
T
∂p
∂T V
=
∂T
∂p
−1
V
∂p
∂V T
= −
∂p
∂T V
∂T
∂V p
6
∂T
∂V p
= −
∂T
∂p V
∂p
∂V T
Solution:
So let’s look at the individual equations one by one:
∂p
∂V T
= −
NkT
V 2
= −
p
V
,
∂V
∂p
−1
T
= −
NkT
p2
−1
= −
V
p
−1
= −
p
V
∂p
∂T V
=
Nk
V
=
p
T
,
∂T
∂p
−1
V
=
V
Nk
−1
=
T
p
−1
=
p
T
∂p
∂V T
= −
NkT
V 2
= −
p
V
, −
∂p
∂T V
∂T
∂V p
= −
p
T
p
Nk
= −
p
T
·
T
V
= −
p
V
∂T
∂V p
=
T
V
, −
∂T
∂p V
∂p
∂V T
=
T
p
−
p
V
= −
T
V
8. The equation of state of an ideal gas can be written as
• pV = NkT,
• pV = n1RT,
• p = ρkT
µ
,
• p = nkT,
where p, V, T are pressure, volume and temperature, N is the number of particles, n
is their concentration, k is Boltzmann constant (k = 1, 38 · 10−23
m2
kg s−2
K−1
), R
is the gas constant (R = 8, 31 J mol−1
K−1
), n1 is the amount of substance, ρ is the
gas density and µ is the molecular weight. Verify the units of k and R. What are the
units of n? Show that the individual equations are equivalent rovnice jsou ekvivalentní
NA = 6.022 · 1023
mol−1
.
Solution:
We determine the dimensions of the individual constants from the respective state
equations
[k] =
[p][V ]
[N][T]
=
kg · m−1
· s−2
· m3
K
= kg · m2
· s−2
[R] =
[p][V ]
[n1] [T]
=
kg · m−1
· s−2
· m3
mol · K
= kg · m2
· s−2
· mol−1
n has unit of m−3
. It is the concentration of particles. Now to derive the equivalence
between the individual forms of the equation: We derive the second equation from the
first very easily: we expand the right-hand side of the equation by Avogadro’s constant
pV = N
NA
NA
kT
7
since NAk = R - molar gas constant and N/NA = n, we get the second equation
pV = nRT
To derive the third equation, we start again from the first equation. Assume that we
know the average molecular weight µ, with which we expand the right-hand side of the
equation of state
pV = N
µ
µ
kT
here again Nµ is the mean mass of N particles. We divide the equation by the volume
p =
m
V
1
µ
kT
and we see that we got the desired density in the expression
p =
ρ
µ
kT
The derivation of the last equation of state is simple. It is enough to divide the first
equation by the volume and realize that že N/V = n, i.e. the number density of
particles
p = nkT
That’s all.
9. At a constant temperature of 20◦
C an ideal gas expands quasi-statically from a state
with a pressure of 20 atm to a state with a pressure of 1 atm. What work is done by 1
mole of gas?
Solution:
The work is generally equal to
W =
V2
V1
dV p(T) =
V2(p2)
V1(p1)
dV
nRT
V
= nR ln(V )|
V2(p2)
V1(p1) =
nRT ln
V2 (p2)
V1 (p1)
= nRT ln
nRT
p2
nRT
p1
= nRT ln
p1
p2
≈ 8.31 · 293.15 · ln(10)J ≈ 5.6 kJ (4)
10. During the quasi-static adiabatic expansion of 6 liters of helium at a temperature
of 350 K, the pressure drops from 40 atm to 1 atm. Calculate the resulting volume and
temperature (assume the ideal gas equation of state). Compare the obtained results
with the values that would be obtained for isothermal expansion (κ = 1, 63). Assume
it is an ideal gas.
Solution:
8
It is true for the adiabatic process
pV κ
= konst.
after that
p0V K
0 = p1V K
1
i.e. the new volume is equal to
V1 = κ
p0
p1
V0 =
1.63
√
40 · 61 ≈ 571
To calculate the temperature, we must first find out the mass of the gas. We can do
this easily from the equation of state of an ideal gas, where we use the quantities from
the initial state of the gas
p0V0 = nRT0 → n =
p0V0
RT0
the temperature itself is also calculated from the equation of state, but for the final
state
p1V1 = nRT1 → T1 = p1V1
nR
= p1V1
p0V0
T0 = p1V1
p0V0
T0 = p1
p0
κ p0
p1
T0 = p0
p1
1
κ
−1
T0 ≈
40
1
1.63 − 1 · 350 K = 84.11 K
Now let’s focus on the comparison with isothermal expansion. The temperature is the
same, we don’t have to deal with it. From the equation p0V0 = p1V1 we can easily find
that the volume will be
V1 =
p0
p1
V0 = 40 · 61 = 2401
thus, the final volume is approximately four times larger than in the case of adiabatic
expansion.
11. Calculate the work done by an ideal gas during a quasi-static adiabatic expansion
from a state characterized by p1, V1 to a state p2, V2. Determine the work done by the
gas if it passes from the initial state to the final state first by an isochoric process and
then by an isobaric process, or first by an isobaric process and then by an isochoric
process.
Solution:
For an adiabatic process, pV K
= konst. = p1V K
1 = p2V K
2 . he work can then be easily
W =
V2
V1
dV p =
V2
V1
dV
p1V κ
1
V κ
= p1V κ
1
V 1−κ
1 − κ
V2
V1
=
p1V κ
1
1 − κ
V 1−κ
2 − V 1−κ
1
=
p1V1
1 − κ
V2
V1
1−κ
− 1
9
We can easily calculate the second part of the example by reasoning. The work done
is equal to the area between the curve and the V . axis. If we look at the pV diagram
of individual processes 2, it is clear that no work is done during isochoric events, so
isobaric events will have the main contribution. The calculation of the work is therefore
limited to the calculation of the content of the rectangle. If the process is first isochoric
and then isobaric, the work done is equal
W1 = p2 (V2 − V1)
for the second case
W1 = p1 (V2 − V1) .
12. During the exchange of air between the lower and upper layers of the troposphere,
expansion occurs, or air compression: rising air expands in an area of lower pressure.
Due to the low thermal conductivity of air, the expansion and compression processes
can be considered adiabatic. Calculate the change in temperature with height due to
these processes. (Consider air as an ideal gas.)
Solution:
First, let’s review the basic equations we’ll be using: this is an ideal gas, so its equation
of state will do
pV = nRT
and also from the information that the process can be considered adiabatic
pV κ
= konst.
However, we want to express this equation using temperature instead of volume, so we
use the equation of state and get
p
RT
p
κ
= p1−κ
Rκ
Tκ
= p1−κ
Tκ
= konst.
we differentiate the expression
(1 − κ)p−κ
Rκ
Tκ
dp = p1−κ
Rκ
κTκ−1
dT = 0
something is shortened here and the equation is then simplified to
(1 − κ)dp +
pκ
T
dT = 0 (5)
We will now be interested in where to take the height dependence. Therefore, we will
recall Euler’s fluid flow equation (which will also be useful in some other examples)
dv
dt
+ (v · ∇)v = −
∇p
ρ
− g (6)
10
Assume stationary flow and also neglect the nonlinear term. Then we get the hydrostatic
equilibrium equation in the form
∇p = −ρg
which we can rewrite to
dp = −ρg dh
and substitute in
−ρg
κ − 1
κ
dh
p
=
dT
T
=
1
T
dT
dh
dh
in the equation we are still hindered by the pressure, which we get rid of using the
equation of state of an ideal gas in the form
p =
ρ
µ
kT
after that
dT
dh
= −
µg
k
κ − 1
κ
For g = 9.81 m · s−2
, µ = 29mH a κ = 1.4 will work out
dT
dh
≈ 10 K · km−1
13. Assume that the atmosphere of the planet Venus contains k1 = 96.5% molecules
CO2 and k2 = 3.5% molecules N2. We can neglect the other components. The temperature
of the atmosphere is t = 464◦
C and the atmospheric pressure on the surface of
Venus contains p0 = 9.1MPa. The mass of the planet is M = 4.87 · 1024
kg and the
radius R = 6052 km.
(a) Determine the density ρ0 of the atmosphere and the gravitational acceleration gv
at the surface of Venus.
(b) To research the planet’s atmosphere, we will use an open "hot air balloon"(filled,
of course, with the planet’s atmosphere) with a volume of V = 50 m3
. The weight of
the structure is m = 100 kg. To what temperature t1 must we heat the gas in the
balloon to start rising above the surface of the planet? At what temperature t2 inside
the balloon will we reach a height of 1 km?
Neglect the rotation of Venus and the drop in gravitational acceleration as the
balloon exits. Consider the temperature of the atmosphere up to a height of 1 km as
constant. The molar masses of the two main components of the Venus atmosphere are
Mm (CO2) = 44.0 · 10−3
kg · mol−1
, Mm (N2) = 28.0 · 10−3
kg · mol−1
.
Solution:
(a) We determine the density of the atmosphere using an equation of state of the form
p =
ρkBT
µ
11
where µ is the average molecular weight, which we calculate using the formula
µ =
k1Mm1 + k2Mm2
NA
= 7.21 × 10−24
kg
The density of the atmosphere is then equal
ρ =
µp
kBT
= 64.518 kg · m−3
We calculate the magnitude of the gravitational acceleration by direct substitution in
the formula
ag =
GM
R2
= 8.87 m · s−2
(b) We will use Archimedes’ law to calculate the temperature. The magnitude of
the buoyant force must be greater than or equal to the magnitude of the gravity force,
broken down
ρatm. Vbal g ≥ mg = (mbal + mplyn ) g =
m
V
+
µp
kBT
from this equation we can easily express the temperature in the form
T ≥
Tatm
1 − mkBTatm.
V µp
= 760.73 K
To calculate the temperature required for the balloon to take off to a height of 1 km, we
need to determine the density/pressure of the atmosphere at the appropriate height. We
can derive the barometric equation from the hydrostatic equilibrium equation, which
we write in the form
∂p
∂h
= −ρg
into which we substitute the equation of state of an ideal gas
∂p
∂h
= −
pµ
kBT
g
the solution of this equation is of the form
p(h) = p0 exp −
µ
kBT
h
respectively density (substitution into equation of state)
ρ(h) =
µp0
kBT
exp −
µ
kBT
h = 64.063 kg · m−3
Assuming that the gas in the balloon has the same pressure as the gas in the atmosphere,
then we can use the equation of state to write
p =
ρball
µ
kBT =
ρatm.
µ
kBTatm
12
where do we get
ρbal. = ρatm.
Tatm.
Tbal.
The condition for temperature is then
T2 ≥
T
1 − m
V ρ0
exp µ
kBT
h
= 762 K
14. A gas is described by the equation of state p = p(V, T). Show by direct calculation
that δQ is not an exact differential.
Solution:
We start from the first theorem of thermodynamics dE = δQ − p dV , from which we
express δQ
δQ = dE + p dV
we can break down the energy differential in temperature and volume variables, then
δQ =
∂E
∂T V
dT +
∂E
∂V T
+ p dV
We recall what condition must hold for a complete differential and apply it to the
appropriate formula
∂
∂V
∂E
∂T V
=
∂
∂T
∂E
∂V T
+
∂p
∂T V
due to the interchangeability of partial derivatives, we get
∂p
∂T V
= 0
however, this contradicts the assumption that pressure is a function of volume. Therefore,
this condition is not fulfilled and apparently it is not a complete differential.
15. The energy of a particle enclosed in an infinitely high potential well with dimensions
Lx × Ly × Lz is given by the relation
E =
2
2m
π2 n2
x
L2
x
+
n2
y
L2
y
+
n2
y
L2
y
Assume that Lx = Ly = Lz = L. Assume that the system as a whole is characterized
by the energy of the system. How is the microstate determined, how is the macrostate
determined? Calculate the force that the particle exerts on the walls of the container.
Determine the relationship between the energy of the system and the pressure.
13
Solution:
A macrostate is a characterized bz macroscopic paremeters of a system that we can directly
measure. In this case, the macrostate is characterized by energy E (L, nx, ny, nz).
For a given macrostate, there are several microstates that we cannot distinguish. Here,
the microstate is characterized by the values nx, ny, nz.
Let’s write how the energy of the system changes when it is moved by a bit dx.
Since we assume a really gradual change, we will then perform the expansion only to
the first order
E(L + dx, L, L) = E(L, L, L) −
2
π2
mL3
n2
x dx
After that
dE = −
2
π2
mL3
n2
x dx
Since the relation dE = −F · dris valid, we get a formula for the acting force
Fx =
2
π2
n2
x
mL3
the pressure is equal to p = F/S, and therefore
p =
2
π2
n2
x
mL5
Mean values are as follows: pressure:
px =
2
π2
n2
x
mL5
=
2
π2
mL5
n2
x
and energy
E =
2
π2
mL3
n2
x + n2
y + n2
z =
p
2L3
.
16. Derive from the existence of the equation of state f(p, V, T) = 0 the relation
α = p · β · κ
between the thermal coefficient of expansion α := 1
V
∂V
∂T p
, isochoric expansion coefficeint
β := 1
p
∂p
∂T V
and the coefficient of isothermal compressibility κ := − 1
V
∂V
∂p
T
.
Solution:
We substitute the appropriate expressions on the right side
p · β · κ = p ·
1
p
∂p
∂T V
−
1
V
∂V
∂p T
= −
1
V
∂p
∂T V
∂V
∂p T
=
1
V
∂V
∂T p
= α
14
17. The state equation has the form p = f(V ) · T. Prove::
(a) ∂E
∂V T
= 0
(b) if a holds, then ∂E
∂p
T
= 0.
Solution:
(a) We express the expression δQ/T from the first law of thermodynamics
δQ
T
=
1
T
∂E
∂T V
dT +
1
T
∂E
∂V T
+ p dV
The following formula will be used a lot, so the derivation needs to be done properly!
Since this is an exact differential, the derivative of the first term with respect to the
volume must be equal erivation of the second term according to temperature, i.e. We
will use the second law of thermodynamic for quasi-static processes, which we will
insert into the first thermodynamic theorem
dS =
dE
T
+
p
T
dV (7)
We know that dE is the complete differential, which we will break down into the
variables of temperature and volume
dE =
∂E
∂T V
dT +
∂E
∂V T
dV (8)
we will get
dS =
1
T
∂E
∂T T
dT +
1
T
∂E
∂V T
+ p dV (9)
We know that dS s also an exact differential and therefore must hold
∂
∂V
1
T
∂E
∂T V T
=
∂
∂T
1
T
∂E
∂V T
+ p
V
whence
1
T
∂
∂V
∂E
∂T T V
= −
1
T2
∂E
∂V T
+ p +
1
T
∂
∂T
∂E
∂V V T
+
∂P
∂T V
.
The second derivative disappears and we finally multiply the expression by T2
. We will
get
∂E
∂V T
+ p = T
∂P
∂T V
(10)
Now it is enough to calculate the relevant partial derivatives
∂E
∂V T
= Tf(V ) − f(V )T = 0
15
(b) We rewrite the derivative of energy with respect to volume as follows
∂E
∂p T
=
∂E
∂V T
∂V
∂p T
= 0
18. For an ideal gas, calculate ∂p
∂V ad
, ∂p
∂V T
.
Solution:
19. Show that for a gas described by the equation of state f(p, V, T) = 0 holds
∂p
∂V ad
=
∂p
∂V T
cp
cV
Solution:
We break down the left side using the new variable T
∂P
∂V S
=
∂P
∂V T
+
∂P
∂T V
∂T
∂V S
(11)
Now we can prove three important formulas resulting from the first theorem of thermodynamics,
from which we will express heat
δQ = dE + p dV (12)
First, we derive another important formula describing the difference between cp and
cV . We start again from (12), where we break down dE again
δQ =
∂E
∂T T
dT +
∂E
∂V T
+ p dV
Now we will break down the volume as a function of pressure and temperature
dV =
∂V
∂T p
dT +
∂V
∂p T
dp
Since we want to express the heat capacity at constant pressure, we put dp = 0.
Moreover, it applies
cV =
δQ
∂T V
(13)
We substitute in the relation for δQ
δQ = cV dT +
∂E
∂V T
+ p
∂V
∂T p
dT = cV + T
∂P
∂T V
∂V
∂T p
dT
16
where we used formula (10). It already follows from this formula
cp =
δQ
∂T p
= cV + T
∂P
∂T V
∂V
∂T p
(14)
Before the actual calculation, it will still be necessary to express the adiabatic derivative.
We start again with (12). We are interested in adiabatic processes and therefore
δQ = 0
0 = dE + p dV
dE is classically broken down into volume and temperature variables. We will use the
first derived formula for the derivative of energy with respect to volume. We will get
0 = T
∂p
∂T V
dV + cV dT
where we now write dT as a function of volume and entropy. We are interested in
processes where the entropy is constant, so we can take the sdS term to be zero
0 = T
∂p
∂T V
dV + cV
∂T
∂V S
dV
That’s where we get it from
∂T
∂V S
= −
T
cV
∂p
∂T V
(15)
We have derived all the important identities, which now just need to be combined
correctly. We substitute formula (14) into (15) for the derivation of pressure by tem-
perature
∂T
∂V S
= −
cp − cV
cV
∂T
∂V p
We can substitute this formula in (11), we get
∂P
∂V S
=
∂P
∂V T
−
cp − cV
cV
∂P
∂T V
∂T
∂V p
We can modify the product of partial derivatives
∂P
∂T V
∂T
∂V p
= −
∂P
∂V T
After substitution
∂P
∂V S
=
∂P
∂V T
+
cp − cV
cV
∂P
∂V T
= 1 +
cp − cV
cV
∂P
∂V T
=
cp
cV
∂P
∂V T
and finally we get the search term. When using Maxwell relations, the procedure is
simpler.
17
20. Show the validity of the relation cp − cV = R between the isobaric and isochoric
specific heats of one mole of an ideal gas. The internal energy of an ideal gas does not
depend on its volume.
Solution:
It is sufficient to calculate this equation from the previously proven formula (14). We
substitute the ideal gas equation
cp − cV = T
∂P
∂T V
∂V
∂T p
= T
p
T
V
T
=
pV
T
= nR
For 1 mol of gas this is exactly R, so Mayer’s relation is proved.
21. Calculate the entropy of an ideal gas when cp = konst., cV = konst. Show that δQ
is not an exact differential.
Solution:
We proceed from the formula (9), where we substitute (10), we have
dS =
cV
T
dT +
∂p
∂T V
dV
We substitute the derivation valid for an ideal gas
dS =
cV
T
dT +
nR
V
dV
Obviously, this is a full differential. From the first equation
∂S
∂T V
=
cV
T
we will get
S(V, T) = cV ln(T) + C(V )
we now derive by volume
∂S
∂V T
=
∂C(V )
∂V T
=
nR
V
after that
C(V ) = nR ln(V ) + konst.
The entropy sought is equal to
S(T, V ) = cV ln(T) + nR ln(V ) + konst.
18
22. Prove the validity of the relation pV κ
= konst. ( κ = cp/cV is the adiabatic
exponent) for quasi-static adiabatic process of an ideal gas. Calculate κ assuming that
cV = 3
2
R.
Solution:
We start from the calculated entropy of an ideal gas, which is constant
S(T, V ) = cV ln(T) + nR ln(V ) + konst. = const.
Const. we hide in const. and we will edit further
ln(T) + ln V
nR
cV = ln TV
nR
cV = konst.
from this equation
TV
nR
cV = konst.
Now let’s substitute for temperature from the equation of state of an ideal gas
pV
nR
V
nR
cV = konst.
nR can be included in the const.
pV
1+ nR
cV = konst.
We will use Mayer relation
pV
1+
cp−cV
cV = pV
cp
cV = pV K
= konst.
So we have a proven adiabat equation for an ideal gas. It is good to emphasize that
this really only applies to an ideal gas and not, for example, to a van der Waals gas.
23. For a gas, it has been found experimentally that the product of pressure and volume
is a function of temperature only, teploty, pV = f(T) and that the internal energy also
depends only on temperature. What shape does f(T) have?
Solution:
We calculate the function by substituting the state equation p = f(T)/V into
∂E
∂V T
+ p = T
∂P
∂T V
(16)
after substitution
f(T)
V
= T
∂f(T)
V
∂T
V
=
T
V
∂f(T)
∂T V
and integration
f(T) = konst. · T (17)
19
24. For a photon gas, the energy density is only a function of temperature and the
pressure is given by p = 1
3
u(T), where u(T) = E/V . Calculate
(a) Function u(T),
(b) entropy„
(c) isotherm and adiabat equation.
Solution:
(a) We calculate the function u by substituting the state equation into
∂E
∂V T
+ p = T
∂P
∂T V
(18)
since we know that E = V u(T), then
u(T) +
1
3
u(T) = T
1
3
∂u(T)
∂T
(19)
whose solution is
u(T) = konst. · T4
(20)
(b) Entropy is equal
dS =
cV
T
dT +
∂p
∂T V
dV
we will calculate the heat capacity from the input
cV =
∂E
∂T V
= 4V konst. T3
then for dS we get
dS = 4V konst. T2
dT +
4
3
konst. T3
dV
This is a complete differential, so by simple integration (or two integrations and one
derivative) we get entropy equal to
S(T, V ) =
8
3
V konst. T3
+ konst 2. (21)
(c) The adiabat equation is easily determined from the last calculated expression
for entropy, ta namely, it must be constant in the case of quasi-static events:
S(T, V ) = konst.
from the expression, we can include all the constants by division or subtraction into
the constant on the right-hand side, then
V · T3
= konst (22)
20
r
dϕ
Fdr
Obrázek 3: Twisting the bar.
We find the equation of the isotherm from the equation of state, which says that if the
temperature is constant, the pressure must also be constant. The equation does not
depend on the volume.
25. A rod is twisted by a moment of force M through an angle ϕ. (a) Prove that the
first theorem of thermodynamics in this case is of the form
dE = δQ + M dϕ. (23)
(b) Derive from the definition of heat capacity (and the first theorem of thermodynamics)
expressions for cM a cϕ.
(c) Find the relationship between ∂M
∂ϕ
adiab
a ∂M
∂ϕ
izoterm
.
Solution:
(a) From the expression of work in the form
dW = F · dr = −F · (r × dϕ), (24)
where F is the acting force, dr is the displacement element of the point and dϕ the
rotation vector by the angle dϕ in the axis of rotation. We will show that dW =
−(F × r) · dϕ, , e.g. by means of expression through the antisymmetric tensor εijk. We
express the element of work dW by notation using the antisymmetric epsilon tensor
−F · (r × dϕ) = −Fi · (εijk · rj · dϕk) = −εkij · Fi · rj · dϕk = −(F × r) · dϕ = M · dϕ
where we used the definition of moment of force M = −F×dϕ. With a suitable choice
of coordinate system: the main axis of the Cartesian coordinate system merges with
the axis of symmetry of the bar, then the scalar product is simplified to the product
of two scalars and we can write the first thermodynamic theorem in the form
dE = δQ + M dϕ (25)
(b) we express cϕ analogously cV , for ϕ = konst. we have 1LT:
dE = δQ
21
from where
∂E
∂T ϕ
=
δQ
∂T ϕ
= cϕ (26)
cM is a bit more complicated, we express δQ from 1LT and write dE as a function of
temperature and angle:
δQ =
∂E
∂T ϕ
dT +
∂E
∂ϕ T
dϕ − M dϕ = cϕdT +
∂E
∂ϕ T
− M dϕ
The next task is to calculate the partial derivative with respect to the angle at constant
temperature. This is done analogously to the previous case where we had pressure and
volume variables. We know that δQ/T is a complete differential, so after dividing both
sides of the equation by the temperature we use the condition necessary for a complete
differential, then
∂E
∂ϕ T
− M = −T
∂M
∂T ϕ
(27)
we insert into modified 1LT
δQ = cϕdT − T
∂M
∂T ϕ
dϕ
we are interested in the heat capacity, which is given by the ratio of δQ and the
differential of temperature at constant M, so we rewrite everything in the variables
of temperature and moment of force; members with a moment of force they drop out,
since M is constant
δQ
∂T M
dT = cϕ − T
∂M
∂T ϕ
∂ϕ
∂T M
dT
the partial derivative on the left side of the equation is the definition of cM , since the
equation holds for any dT, we can write
cM = cϕ − T
∂M
∂T ϕ
∂ϕ
∂T M
(28)
(c) First we have to express some partial derivative for adiabatic processes, we get this
from 1LT, adiabatic (quasi-static) processes are for constant Q, i.e.
0 = cϕdT − T
∂M
∂T ϕ
dϕ = −T
∂M
∂T ϕ
∂ϕ
∂T ad.
dT
after that
∂ϕ
∂T ad.
=
cϕ
T
∂T
∂M ϕ
we substitute 28) for ∂T
∂M ϕ
and express ∂T
∂ϕ
a
d.
∂T
∂ϕ ad.
= 1 −
cM
cϕ
∂T
∂ϕ M
22
now we will use the formula (derived at the beginning)
∂M
∂ϕ ad.
=
∂M
∂ϕ T
+
∂M
∂T ϕ
∂T
∂ϕ ad
and we substitute for the last partial derivative
∂M
∂ϕ T
+
∂M
∂T ϕ
1 −
cM
cϕ
∂T
∂ϕ M
=
∂M
∂ϕ T
+
∂M
∂ϕ T
1 −
cM
cϕ
=
cM
cϕ
∂M
∂ϕ T
So the desired formula is
∂M
∂ϕ ad.
=
cM
cϕ
∂M
∂ϕ T
(29)
26. (a) 1 kg of water at 0◦
C is brought into thermal contact with a large reservoir at
100◦
C. Calculate the entropy change of the water, the reservoir, and the entire system
after equilibrium is established.
(b) Calculate the change in entropy of the entire system if the water was first in contact
with a reservoir at a temperature of 50◦
and then with a reservoir at a temperature
100◦
C.
(c) How to ensure that the entropy of the system does not change when the water is
heated?
Solution:
(a) In the calculation, we assume that the body is so large compared to the water that
its temperature changes negligibly. The entropy change is then equal to
∆SR =
∆Q
T
=
mcV (T1 − T2)
T2
= mcV
T1
T2
− 1 = −1125.6 J · K−1
∆ST =
T2
T1
dT
mcV (T2 − T1)
T
= mcV (T2 − T1) ln(T)|T2
T1
= mcV (T2 − T1) ln
T2
T1
= 131022 J · K−1
the total entropy change of the system is therefore
∆S = ∆SR + ∆ST = 1298896.4 J · K−1
(b) Substitute again into the previous formulas, but with different temperatures. The
following entropy change occurs on first heating:
∆SR1 = −649.9 J · K−1
, ∆SR2 = −562.8 J · K−1
and body
∆ST1 = 35300 J · K−1
, ∆ST2 = 30211 J · K−1
23
(c) Consider that we have N +1 reservoirs. The zero reservoir is in thermal equilibrium
with the body. We will gradually add the body to the individual reservoirs gradually
according to the increasing temperature, until it is heated to the temperature T2.
The temperature of the reservoirs always increases by the same temperature, so the
temperature of the i-th reservoir is the same
T(i) =
T2 − T1
N
i − T1 (30)
let’s say that the body now has a temperature of T(i) and we will heat it to a temperature
of T(i + 1). We calculate the change in entropy of the reservoir analogously to
the previous example
∆SR = mcV
T(i) − T(i + 1)
T(i + 1)
(31)
and body
∆ST = mcV ln
T(i + 1)
T(i)
(32)
after substituting specific T(i) we obtain, denote the expression a = T2−T1
N
∆SR = mcV
−a
a(i + 1) − T2
(33)
and body
∆ST = mcV a ln
a(i + 1) − T2
a − T2
(34)
the total entropy change is obtained by summing the entropy changes over all cycles
∆SR = mcV
N
i=1
−a
a(i + 1) − T2
(35)
and body
∆ST = mcV a
N
i=1
ln
a(i + 1) − T2
a − T2
(36)
The sum of logarithms is simple, the middle terms are subtracted and only the extreme
terms remain, i.e.
∆ST = mcV a ln
a(N + 1) − T2
a − T2
(37)
the final sum of the harmonic series cannot be calculated analytically, we know that
the sum can be approximately replaced by an integral and then
∆SR = mcV
N
i=1
−a
a(i + 1) − T2
≈ −
N
1
dim cV a
1
a(i + 1) − T2
= −mcV a ln
a(N + 1) − T2
a − T2
24
for sufficiently high N this is exactly the case. Obviously, this is exactly the opposite
value to the entropy change of the body, so the sum approaches zero as N, increases,
as the summation approximation by integration made in the last step is refined.
27. Show that for small deviations δρ, δp from the equilibrium values of density ρ0 and
pressure p0 the propagation of sound waves can be described by a wave equation
∂2
δp
∂t2
= c2 ∂2
δp
∂x2
, (38)
where the speed of sound is given by the relation c = (∂p/∂ρ)ad if we assume that
events are on- ad fast that there is no heat exchange between the individual elements
of the air. Show that the speed of sound can also be calculated as c = κad/ρ0, where
the adiabatic compressibility κad := −V ∂p
∂V ad
Calculate the speed of sound in air
assuming that air is made up of only N2 molecules and that κ = cp/cV = 7/5.
Solution:
We start from Euler’s equation for fluid motion
∂v
∂t
+ (v · ∇)v = −
∇p
ρ
we simplify the case to 1 D , we assume that the waves move along a straight line and
we point the x-axis in the direction of the group velocity vector of these waves, Euler’s
equation will then be
ρ
∂v
∂t
+ ρv
∂v
∂x
= −
∂p
∂x
(39)
and the continuity equation
∂ρ
∂t
+
∂(ρv)
∂x
= 0 (40)
Subsequently, we will assume that the density changes by a small value compared to
its mean value, which we write as
ρ = ρ0 + δρ = ρ0 +
∂ρ
∂p ad.
δp (41)
on the right side of the equation, we assume that the sound wave is an adiabatic
process. Similarly, we break down the velocity v, for which we can appropriately choose
a coordinate system such that v0 = 0, then
v = δv (42)
we substitute the relevant developments into the individual equations, then:
ρ0
∂δv
∂t
+ δρ∂δv
∂t
+ ρ0δv∂δv
∂x
+ δρδv∂δv
∂x
=
ρ0
∂δv
∂t
+ ∂ρ
∂p
ad.
δp∂δv
∂t
+ ρ0δv∂δv
∂x
+ ∂ρ
∂p
ad.
δv∂δv
∂x
= −∂(p0+δp)
∂x
= ∂δp
∂x
,
in the equation, we will neglect the second, third and fourth terms on the left side,
25
because they are significantly smaller than the first term, on the right side only the
derivative of the pressure fluctuation will remain
ρ0
∂δv
∂t
= −
∂δp
∂x
(43)
we substitute the same in the continuity equation
∂ρ
∂p ad.
∂δp
∂t
+ ρ0
∂δv
∂x
+
∂ρ
∂p ad.
∂(δpδv)
∂x
= 0
we neglect the third term again because the fluctuation multiple is significantly lower
than the first two terms, then
∂ρ
∂p ad.
∂δp
∂t
= −ρ0
∂δv
∂x
(44)
we derive equation (51) according to time, equation (52) again according to the x
-coordinate, the right-hand sides are equal due to the interchangeability of the derivatives,
then we get
∂ρ
∂p ad.
ρ0
∂2
δv
∂t2
= ρ0
∂2
δv
∂x2
(45)
and finally, after a simple modification, we get the equation we are looking for
∂2
δp
∂t2
= c2 ∂2
δp
∂x2
(46)
To rewrite the speed of sound using adiabatic compressibility, we modify the partial
derivative
c =
∂p
∂ρ ad
=
∂p
∂V ad.
∂V
∂ρ ad.
=
∂p
∂V ad.
−
m
ρ2
we rewrite the fraction as
m
ρ2
=
V
ρ
since m = ρV . Eventually
c = −
V
ρ
∂p
∂V ad.
=
κad
ρ
. (47)
To calculate the speed of sound, we will use the knowledge of the ideal gas adiabat
equation
pV κ
= p
µκ
µκ
V κ
=
pµκ
ρκ
= konst. = K
from here we can easily determine the derivative
∂p
∂ρ ad
= κK
ρκ−1
µκ
= κK
ρκ
µκ
ρ−1
= κ
p
ρ
= κ
kBT
µ
26
where we substituted the ideal gas equation of state p = ρkBT/µ. The speed of sound
is then
c = κ
kBT
µ
= 493.67 m · s−1
(48)
28. An ideal gas expands adiabatically from volume V1 into vacuum. Calculate the
increase in entropy if the gas in the final state has volume V2 and prove that the
expansion process is irreversible.
Solution:
First, I will list the formulas that we will use in the calculation. You should find their
derivation in the workbook. The first formula follows from the first law of thermody-
namics
∂E
∂V T
= T
∂p
∂T V
− p
another important formula is the expression of entropy (or complete differential) from
the first theorem of thermodynamics
dS =
cV
T
dT +
∂p
∂T V
dV
Energy is a state quantity. Therefore, its change is determined only by the initial and
final conditions, regardless of the intermediate states through which the gas passed.
Therefore, we can write the energy change in the form
∆E =
∂E
∂T V
∆T +
∂E
∂V T
∆V = cV ∆T + T
∂p
∂T V
− p ∆V
Since this is an isolated system, the total energy must be constant, i.e. ∆E = 0. From
here we easily find that
∆T =
1
cV
p − T
∂p
∂T V
∆V
Entropy is also a state variable, so it can be written for the same reasons
∆S =
∂S
∂T V
∆T +
∂S
∂V T
∆V =
cV
T
∆T +
∂P
∂T V
∆V
where we substitute for ∆T and get
∆S =
p
T
∆V
27
29. The van der Waals equation of state for 1 mole of gas has the form
p +
a
V 2
(V − b) = RT (49)
where a, b are constants. For a given T he curve can have two extremes given by
equation
∂p
∂V T
= 0
In addition, at the critical point determined by the parameters Tc, pc a Vc
∂2
p
∂V 2
T
= 0
Calculate the values of Tc, pc a Vc. Write the state equation using the variables T′
=
T/Tc, p′
= p/pc a V ′
= V/Vc.
Solution:
First, we calculate the first and second derivatives of pressure with respect to volume
at constant temperature:
∂p
∂V T
= −
RT
(V − b)2
+
2a
V 3
= 0 (50)
and
∂2
P
∂V 2
T
=
2RT
(V − b)3
−
6a
V 4
= 0 (51)
We express the expression from both equatio RT
(V −b)2 , that is
3a
V 4
=
2a
V 3(V − b)
hence the critical volume Vc = 3b. By substituting into (40), we obtain the value of the
critical temperature, which is equal to
TC =
8a
27Rb
and we substitute the critical volume and the critical temperature into the equation of
state, from which we get for the critical pressure
pC =
a
27b2
Just put in the equation of state T = T′
· Tc,
p′
=
8T′
V ′ − 1
−
3
V ′2
(52)
28
30. Determine::
(a) the internal energy and entropy of a van der Waals gas,
(b) the work of a van der Waals gas during reversible isothermal expansion,
(c) the temperature change of a van der Waals gas during adiabatic expansion into a
vacuum.
Solution:
(a) We calculate the internal energy by substituting into the formula
dE = cV dT + T
∂p
∂T V
− p dV = cV dT +
a
V 2
dV
from here we easily find that
E = cV T −
a
V
+ konst.
We determine the entropy from the formula
dS =
cV
T
dT +
∂p
∂T V
dV =
cV
T
dT +
R
V − b
dV
After that
S = cV ln(T) + R ln(V − b) + konst.
(b) We determine the work by substituting the equation of state into
W =
V2
V1
dV p(T, V ) =
V2
V1
dV
RT
V − b
−
a
V 2
=
RT ln(V − b) +
a
V
V2
V1
= RT ln
V2 − b
V1 − b
+ a
1
V2
−
1
V1
(c) We work out the entropy equation, the entropy change of the gas itself depends
only on initial and final state, therefore we can write:
∆S =
cV
T
∆T +
∂p
∂T V
∆V (53)
since we are calculating an adiabatic process, ∆S = 0 and after substituting from the
equation of state::
0 =
cV
T
∆T +
R
V − b
∆V (54)
from here
∆T = −
T
cV
R
V − b
∆V (55)
31. The Joule-Thomson coefficient is defined using a parameter
λ = −
∂T
∂p H
29
(a) Show that
dH = T dS + V dp
and
λ =
V
Cp
(1 − Tαp)
αp := 1
V
∂V
∂T p
is the coefficient of isobaric expansion.
(b) Show that
λ =
T ∂p
∂T V
+ V ∂p
∂V T
Cp · ∂p
∂V T
(c) Verify that λ = 0 for a classical ideal gas.
(d) Show that it holds for a van der Waals gas
λ =
bp + 3ab
V 2 − 2a
V
p − a
V 2 + 2ab
V 3 · Cp
(e) Express the equation of the inverse curve that represents the interface between the
region λ > 0 and λ < 0 in the p − V diagram for the case of a van der Waals gas.
Solution:
(a) Since H = H(S, p), it is logical that by Legendre transformation we get dH =
T dS + V dp. From this formula we also know that
∂H
∂S p
= T,
∂H
∂p S
= V (56)
We start editing the expression for λ, into which we substitute the given αp and
cp = T
∂S
∂T p
we will start modifying the right-hand side of the expression step by step, using the
identities for the partial derivatives that we known
V
Cp
(1 − Tαp) =
V
T ∂S
∂T p
1 −
T
V
∂V
∂T p
=
V
T
∂T
∂S p
−
∂T
∂S p
∂V
∂T p
=
∂H
∂p S
∂S
∂H p
∂T
∂S p
−
∂T
∂S p
∂V
∂T p
=
−
∂S
∂p H
∂T
∂S p
−
∂V
∂S p
=
−
∂S
∂p H
∂T
∂S p
−
∂T
∂p S
= −
∂T
∂p H
= λ
30
(b) The procedure is the same as in the previous case: we break down the right-hand
side of a we gradually adjust the partial derivatives:
T ∂p
∂T V
+ V ∂p
∂V T
Cp · ∂p
∂V T
=
T ∂p
∂T V
+ V ∂p
∂V T
T ∂S
∂T p
∂p
∂V T
=
∂T
∂S p
∂V
∂p T
∂p
∂T V
+
V
T
∂T
∂S p
∂V
∂p T
∂p
∂V T
=
−
∂T
∂S p
∂V
∂T p
+
V
T
∂T
∂S p
=
−
∂T
∂S p
∂V
∂T p
+
∂H
∂p S
∂S
∂H p
∂T
∂S p
=
−
∂T
∂S p
∂V
∂T p
−
∂S
∂p H
∂T
∂S p
=
−
∂V
∂S p
−
∂S
∂p H
∂T
∂S p
= −
∂T
∂p S
−
∂S
∂p H
∂T
∂S p
= −
∂p
∂T H
= λ
(c) for an ideal gas we know the equation of state pV = nRT. We will therefore use
the expression for λ from the previous subtask. If λ is to be zero, either the numerator
will be zero or the denominator will go to infinity. Let’s look at the numerator first:
T
∂p
∂T V
+ V
∂p
∂V T
= T
nR
V
− V
nRT
V 2
=
nRT
V
−
nRT
V
= 0 = λ
(d) First, we adjust the denominator to the desired form:
Cp ·
∂p
∂V T
= Cp ·
RT
(V − b)2 − 2a
V 3
=
−Cp
1
V − b
RT
V − b
−
2a(V − b)
V 3
= − Cp
1
V − b
RT
V − b
−
2a
V 2
+
2ab
V 3
=
− Cp
1
V − b
p −
a
V 2
+
2ab
V 3
Analogously, we will adjust the nominator
RT
V − b
−
V
V − b
p −
a
V 2
+
2ab
V 3
=
1
V − b
RT − V p −
a
V 2
+
2ab
V 3
=
1
V − b
p +
a
V 2
(V − b) − pv +
a
V 2
−
2ab
V 3
=
1
V − b
−bp −
3ab
V 2
+
2a
V
=
−
1
V − b
bp +
3ab
V 2
−
2a
V
In the fraction, the factors −1/(V − b), are divided , and thus we get the formula we
are looking for..
31
(e) From the condition that the numerator must be equal to zero, we calculate the
equation
p = a
2
bV
−
3
V 2
(57)
32. Show that the thermal coefficient of expansion
α :=
1
V
∂V
∂T p
meets the relation
∂S
∂p T
= −
∂V
∂T p
= −V α
Solution:
The first equality is the Maxwell relation resulting from the Gibbs potential. If we
expand the middle expression by the volume, the expression for α drops out and we
get the right side we are looking for.
33. Show that the specific heat at constant pressure, cp, and at constant volume, cV ,
satisfy the relation
cp − cV = T
∂p
∂T V
∂V
∂T p
= −T
∂S
∂V T
∂S
∂p T
.
Solution:
The left and middle sides are derived earlier, see (14). Individual partial derivatives
can be rewritten on the right-hand side using Maxwell’s relations.
34. Free energy of the system F(V, T) = −1
3
· const ·V T4
. Determine its pressure,
internal energy, 3 enthalpy and Gibbs potential.
Solution:
• pressure: from the free energy equation
dF = −S dT − p dV
we easily find that
p = −
∂F
∂V T
=
1
3
· const. T4
• entropy: analogously to the previous point:
S = −
∂F
∂T V
=
4
3
· const. V T3
32
• we can easily calculate the internal energy from the free energy:
E = F + TS = konst. V T4
=
4
3
4
3 V
C
1
3
S
4
3
• for enthalpy:
H = E + PV =
4
3
konst. V T4
if we want to express using pressure and entropy, we express V konst. T3
from entropy
and T, from pressure, then
H =
3
4
1
4
Sp
1
4
• We can express the Gibbs potential as
G = F + PV = F +
1
3
konst.V T3
= 0
35. Calculate the efficiency of the Carnot cycle (1. isothermal expansion, T2 = konst,
2. adiabatic expansion„ S = konst, 3. isothermal compression, T1 = konst, 4. adiabatic
compression, S = konst) for an ideal gas using its equations of state.
Solution:
First, we write down the work and heat produced during the individual processes:
(a) Isothermal expansion (p1, V1) → (p2, V2) :
W1 = nRT1 ln
V2
V1
Q1 = nRT1 ln
V2
V1
(b) adiabatic expansion (p2, V2) → (p3, V3) :
W2 = p3V κ
3
V 1−κ
3 − V 1−κ
2
1 − κ
=
p3V3
1 − κ
1 −
V3
V2
κ−1
=
nRT2
1 − κ
1 −
V3
V2
κ−1
Q2 = 0
(c) Isothermic compression (p3, V3) → (p4, V4) :
W3 = nRT2 ln
V4
V3
Q3 = nRT2 ln
V4
V3
33
(d) adiabatic compression (p4, V4) → (p1, V1) :
W4 = p1V κ
1
V 1−κ
1 − V 1−κ
4
1 − κ
=
p1V1
1 − κ
1 −
V1
V4
κ−1
=
nRT1
1 − κ
1 −
V1
V4
κ−1
(58)
Q4 = 0 (59)
t will be most advantageous to express as many quantities as possible in terms of
temperature, for the adiabatic process TV κ−1
, applies, i.e. for adiabatic expansion
T1V K−1
2 = T2V K−1
3
T2V K−1
4 = T1V K−1
1
then we get
T1
T2
=
V3
V2
κ−1
=
V4
V1
κ−1
(60)
then we can rewrite the individual work in adiabatic processes as
W2 =
nRT2
1 − κ
1 −
T1
T2
(61)
and
W4 =
nRT1
1 − κ
1 −
T2
T1
(62)
work during isothermal compression
W3 = nRT1 ln
V1
V2
= Q3 (63)
the efficiency of the machine is then equal
η =
W
Q+
=
nRT1 ln V2
V1
+ nRT2
1−κ
1 − T1
T2
+ nRT2 ln V1
V2
+ nRT1
1−κ
1 − T2
T1
nRT1 ln V2
V1
= 1 −
T2
T1
+
nR
1 − κ
T2 1 −
T1
T2
+ T1 1 −
T2
T1
1 −
T2
T1
+
nR
1 − κ
(T2 − T1 + T1 − T2) = 1 −
T2
T1
36. Calculate the efficiency of the following ideal gas cycle. Can this process be carried
out reversibly?
1. isothermal expansion T2 = konst
2. isochoric cooling V2 = konst
3. isothermal compression T1 = konst
4. isochoric heating V1 = konst.
34
Solution:
Work and heat done in individual processes:
(a) isothermal expansion (p1, V1) → (p2, V2)
W1 =
V2
V1
dV
nRT2
V
= nRT2 ln
V2
V1
= Q1 (64)
(b) isochoric cooling (p2, V2) → (p3, V2)
Q2 = −cV n (T2 − T1) , W2 = 0 (65)
(c) isothermal compression (p3, V2) → (p4, V1)
W3 = −nRT1 ln
V2
V1
= Q3 (66)
(d) isochoric heating (p4, V1) → (p1, V1)
Q4 = cV n (T2 − T1) , W4 = 0 (67)
The total work is then equal
W = nR ln
V2
V1
(T2 − T1) (68)
and the supplied heat Q = Q1 + Q4, then the efficiency is equal
η =
nR ln V2
V1
(T2 − T1)
nRT2 ln V2
V1
= Q1 + cV n (T2 − T1)
=
T2 − T1
T2 + cV
nR ln
V2
V1
(T2 − T1)
< ηCarnot (69)
37. Determine the efficiency coefficient of an (idealized) Otto engine that works with
an ideal gas 5 of specific heat cV = 5
2
R/mol at a compression ratio of 10:1.
1. adiabatic compression„
2. isochoric heating (=burning of fuel),
3. adiabatic expansion (doing work),
4. cooling (=hot gas exhaust, new, cold gas is drawn in).
Solution:
Work and heat done in each process:
(a) Adiabatic compression (p1, V1) → (p2, V2)
W1 =
nRT1
1 − κ
V 1−κ
2 − V 1−κ
1 =
nRT1
1 − κ
ε1−κ
− 1 (70)
where we modify the expression with κ
W1 =
nRT1
1 − cp
cV
ε1−κ
− 1 = −
ncV RT1
cp − cV
ε1−κ
− 1 = −ncV T1 ε1−κ
− 1 (71)
35
Here we used the knowledge of Mayer’s relation for an ideal gas.
Q1 = 0 (72)
(b) Isochoric heating (p2, V2) → (p3, V2)
W2 = 0, Q2 = ncV (T3 − T2) (73)
(c) Adiabatic expansion (p3, V2) → (p4, V1)
W3 =
nRT1
1 − cp
cV
ε1−κ
− 1 = −ncV T4 1 − ε1−κ
(74)
(d) Isochoric cooling (p4, V1) → (p1, V1)
W4 = 0, Q4 = ncV (T1 − T4) (75)
We can then write the efficiency in the form
η =
−ncV T1 (ε1−κ
− 1) − ncV T4 (1 − ε1−κ
)
ncV (T3 − T2)
=
T4 − T1
T3 − T2
ε1−κ
− 1 (76)
We will now use the adiabatic equation in the form of the TV κ−1
= konst., which we
write for both adiabatic process
T1V κ−1
1 = T2V κ−1
2 , T3V κ−1
2 = T4V κ−1
1 (77)
where do we get
T1 = T2
V2
V1
κ−1
= T2εκ−1
(78)
and
T4 = T3
V2
V1
κ−1
= T3εκ−1
(79)
we substitute in the equation for efficiency, we get
η =
T3εκ−1
− T2εκ−1
T3 − T2
ε1−κ
− 1 = 1 − εκ−1
(80)
For the given ε = 1
10
a κ = R+cV
cV
= 7
5
it turns out η = 0.6.
38. The Diesel cycle consists of the following parts:
1. adiabatic compression of atmospheric air,
2. combustion of the injected mixture and isobaric expansion,
3. adiabatic expansion
4. and isochoric cooling.
Determine the cycle efficiency as a function of compression ratio for an ideal gas.
Solution:
36
Work and heat done in each process:
(a) Adiabatic compression
W1 = −ncV T1
V2
V1
1−κ
− 1 = −ncV (T2 − T1) , Q1 = 0 (81)
where we used the adiabatic equation
T1V κ−1
1 = T2V κ−1
2
(b) Isobaric expansion
W2 = p2 (V3 − V2) = p2V2
V3
V2
− 1 (82)
using the equation for the isobar V2
T2
= V3
T3
and the equations of state for an ideal
gas we can T3 rewrite work using temperature to
nRT2
T3
T2
− 1 = nR (T3 − T2) (83)
and heat
Q2 = ncp (T3 − T2) (84)
(c) Adiabatic expansion
Q3 = 0, W3 = −ncV T3
V1
V3
1−κ
− 1 = −ncV (T4 − T3) (85)
(d) Isochoric cooling
W4 = 0, Q4 = ncV (T1 − T4) (86)
We then calculate the efficiency classically
η =
−ncV (T2 − T1) − ncV (T4 − T3) + nR (T3 − T2)
ncp (T3 − T2)
(87)
when using Mayer’s formula, we substitute after R
η =
cV (T1 − T2 + T3 − T4 − T + T2) + cp (T3 − T2)
cp (T3 − T2)
= 1 −
cV (T4 − T1)
cp (T3 − T2)
(88)
Efficiency can still be overridden using compression ratios. Let us recall from the equations
of adiabats and the equation of isobars
T1 =
V2
V1
κ−1
T2, T3 =
V3
V2
T2, T4 =
V3
V1
κ−1
T3 =
V3
V1
κ−1
V3
V2
T2 (89)
37
Substitute for temperature in the equation for efficiency
η = 1 −
V3
V1
κ−1
V3
V2
T2 − V2
V1
κ−1
T2
κ V3
V2
T2 − T2
= 1 −
1
V2
(V κ
3 − V κ
2 ) 1
V κ−1
1
1
V2
κ (V3 − V2)
(90)
Expressions 1
V2
will be shortened. From the numerator, we will write V K
2 and from the
denominator in parenthese V2, then
η = 1 −
V3
V2
κ
− 1
V1
V2
κ−1
κ V3
V2
− 1
(91)
Let us now denote ρ = V3
V2
a ε = V1
V2
, then
η = 1 −
1
κ
1
εκ−1
ρκ
− 1
ρ − 1
(92)
39. What is the total entropy change if we mix 2 kg of water at a temperature of 363
K adiabatically and at constant pressure with 3 kg of water at a temperature of 283
K? (cp = 4184 J/Kkg)
Solution:
First, it will be necessary to calculate the resulting temperature, let’s denote it by T.
The first body must give up the heat Q, which the second body receives, i.e.
m1cp (T1 − T) = m2cp (T − T2)
from here
T =
m1T1 + m2T2
m1 + m2
The entropy change of one body is equal to
∆S1 = dQ
1
T
= m1cp
T
T1
dT
1
T
= m1cp ln
m1 + m2
T2
T1
m1 + m2
analogously, we calculate the change in entropy of the second body
∆S2 = m2cp ln
m1 + m2
m1
T1
T2
+ m2
= m2cp ln
m1 + m2
m1 + m2
T2
T1
+ m2cp ln
T2
T1
The sum is then equal
∆S = ∆S1 + ∆S2 = (m1 − m2) cp ln
m1 + m2
T2
T1
m1 + m2
+ m2cp ln
T2
T1
(93)
38
40. A refrigerator can turn 10 liters of water at 0◦
C into ice at the same temperature
in one hour. For this, the thermal energy Q = 800kcal(= 800 × 1, 163Wh) must be
transferred to the air (27, 3◦
C). What is the minimum power the refrigerator must
have?
Solution:
Assume that this is a Carnot cycle, the efficiency of which is equal to
η = 1 −
T1
T2
(94)
where T1 is the temperature in the refrigerator an T2 is the ambient temperature. Let
us denote Q1 the heat that must be removed from the refrigerator, W the work that the
working machine will do and Q2 the heat that will pass into the external environment,
apparently this heat is equal to
Q2 = W + Q1 (95)
At the same time, we know that the basic definition of the efficiency of a heat engine
is
η =
W
Q2
=
W
W + Q1
(96)
from where we can easily calculate the job sought
W =
ηQ1
1 − η
=
Q1
1
1−
T1
T2
− 1
(97)
Q1 is equal to the quantit Q from the specification, after calculation we get
W = 93 W · h (98)
from which it is clear that the power consumption must be equal to P0 = 93 W.
41. Prove that for T → 0 there is no system described by pV = const ·T.
Solution:
We can write the pressure of the system as
p = konst.
T
V
(99)
where T and V are the independent variables. Free energy is also expressed in these
variables
dF = −S dT − p dV (100)
According to the third law of thermodynamics limT→0+ = 0 regardless of pressure,
volume, density, etc. Therefore, it must also be true that
∂S
∂V T
→ 0 (101)
39
From the free energy we find that
∂S
∂V T
=
∂p
∂T V
=
konst.
V
= 0 (102)
which contradicts the third law of thermodynamics.
42. A closed system consists of two simple subsystems that are separated by a movable
wall that allows
(a) only heat exchange,
(b) both heat and mass exchange,
(c) neither heat nor mass exchange.
What are the corresponding equilibrium conditions?
Solution:
Subsystems can be described by changing individual physical quantities. Since the
sub-systems form a system isolated from the surroundings, so the total energy does
not change, i.e.
dE = dE1 + dE2 (103)
whence apparently dE1 = −dE2.
(a) the same is also true for entropy
dS1 = −dS2 (104)
and for volume change
dV1 = −dV2 (105)
for the total entropy change we can write
dS = 0 =
1
T1
dE1 +
p1
T1
dV1 +
1
T1
dE2 +
p1
T1
dV2 (106)
where we substitute the equations written for the change in volume and energy
0 =
1
T1
dE1 +
p1
T1
dV1 −
1
T1
dE2 −
p1
T1
dV2 =
1
T1
−
1
T2
dE1 +
p1
T1
−
p2
T2
dV1 (107)
for the expression to be equal to zero, the individual parentheses must be equal. The
first parenthesis implies equality of temperatures, the second parenthesis equality of
pressures.
(b) the same conditions apply here as in the previous point, in addition also the condition
for the number of particles dN1 = −dN2, by analogous calculation we arrive at
the equation
0 =
1
T1
−
1
T2
dE1 +
p1
T1
−
p2
T2
dV1 +
µ2
T2
−
µ1
T1
dN1 (108)
40
which also results in the equality of chemical potentials.
(c) In this case, dS1 = dS2 = 0, also for heat and number of particles, only for volume
change we can still write dV1 = −dV2. We describe the change in internal energy
dE(S, V ) = p1 dV1 + p2 dV2 = (p1 − p2) dV1 (109)
from which it follows that only the pressures must be equal.
43. Two equal quantities of ideal gas with the same temperature T and different pressures
p1, p2 are separated from each other by a partition. Determine the change in
entropy resulting from the mixing of the two gases.
Solution:
The entropy of an ideal gas is given by the relation
S = ncV ln T + nR ln V − Rn ln n (110)
The entropy of the separated gases is given by
S1 = n (cV ln T + R ln V1) − Rn ln n, S2 = n (cV ln T + R ln V2) − Rn ln n (111)
the entropy after mixing the gases is equal
S12 = 2n [cV ln T + R ln (V1 + V2)] − 2Rn ln(2n) (112)
the difference of these entropies is then equal
∆S = S12 − S1 − S2 = nR [2 ln (V1 + V2) − ln (V1 · V2)] − 2nR ln 2
= nR 2 ln
nRT
p1
+
nRT
p2
− ln
n2
R2
T2
p1p2
− 2nR ln 2
= nR ln
(p1 + p2)2
p1 · p2
− Rn ln 4 = nR ln
(p1 + p2)2
4p1p2
44. Determine the maximum work that can be obtained by combining equal quantities
of the same ideal gas with the same temperature T0 (and different volumes or pressures).
Solution:
The entropy of the gas before mixing is equal
S1 = n (cV ln T0 + R ln V1) − Rn ln n, S2 = n (cV ln T0 + R ln V2) − Rn ln n (113)
after mixing the gases is entropy
S12 = 2n [cV ln T + R ln (V1 + V2)] − 2Rn ln(2n) (114)
41
We first determine the sum of entropies before mixing
S1 + S2 = 2ncV ln T0 + nR ln (V1 · V2) − 2nR ln n (115)
Again we calculate the entropy difference, so that the work is maximal, the entropy
change must be zero, from here I get the econdition
2cV ln
T
T0
= R ln
V1 · V2
(V1 + V2)2 + 2R ln 2 (116)
by combining the logarithms on the right-hand side and taking the logarithm, we obtain
T
T0
RcV
=
4V1V2
(V1 + V2)2
R
(117)
from where we calculate the resulting temperature
T = T0
4V1 · V2
(V1 + V2)2
R
2CV
= T0
4V1V2
(V1 + V2)2
κ−1
2
(118)
The maximum work is then equal
∆W = 2ncV (T0 − T) (119)
45. Molar volume of water v(2)
= 18 cm3
/mol, the molar volume of ice is 9.1% larger (at
a pressure of 105
Pa ), The molar mass of water is 18 g/mol. The latent heat of melting
of ice is 330 kJ/kg. Calculate the change in melting point as the pressure changes.
Solution:
We will use the Clausius-Capperian equation
dT
dp
=
T∆v
q
(120)
it is enough to substitute the entered values in the right side. Difference in molar
volumes
∆v = v(2)
− v(1)
= 0.091 · 18 · 10−6
= 1.64 · 10−6
m−3
· mol−1
(121)
latent heat is equal
q = ml = 3300000 J · mol−1
· 0.018 mol = 5900 J · mol−1
(122)
After substituting into the CC equation
dT
dp
= 7.5 · 10−8
K · Pa−1
(123)
42
46. When the magnetization M changes by dM the system performs work dW =
−H dM, where H is the intensity of the magnetic field. (This is work done per unit
volume; volume V = konst. = 1.) Determine the difference in heat capacities cH − cM
at a constant field H and at a constant magnetization.
Solution:
The first law of thermodynamics is of the form
dE = δQ + H · dM (124)
according to the condition of constant field H and M we can choose a coordinate
system by rewriting the scalar product as H · dM. We can now proceed analogously as
in the case with volume and pressure: for cM for M = konst.:
dE = δQ
from where
∂E
∂T M
=
δQ
∂T M
= cM (125)
to calculate cH we express δQ term from 1LT and rewrite dE = dE(T, M)
δQ =
∂E
∂T M
dT +
∂E
∂M T
− H dM = cM dT − T
∂H
∂T M
dM
we convert to the variables of temperature and magnetic field intensity
δQ
∂T H
dT = cM dT − T
∂H
∂T M
∂M
∂T H
dT
whence apparently it is true that
δQ
∂T H
= cH = cM − T
∂H
∂T M
∂M
∂T H
(126)
then the desired result is of the form
cH − cM = −T
∂H
∂T M
∂M
∂T H
(127)
47. Determine the equation of the adiabat of an isotropic magnetic.
Solution:
from the previous example we can easily determine δQ from dS
dS =
cM
T
dT −
∂H
∂T M
dM
43
or the derivative at dM we substitute the result cH − cM , and dT = dT(M, H), then
dS =
cM
T
∂T
∂M H
dM +
∂T
∂H M
dH −
cH − cM
−T
∂T
∂M H
dM = 0
after editing
cH
∂T
∂M H
dM + cM
∂T
∂H M
dH = 0
in the end we multiply
1
cM
∂H
∂T M
and we get (using the formula for the relationship between partial derivatives)
cH
cM
dM −
∂M
∂H T
dH = 0 (128)
48. Show that it holds
cH
cM
=
χT
χS
,
where
χT =
∂M
∂H T
and
χS =
∂M
∂H S
Solution:
In the expression derived in the previous example
cH
cM
dM −
∂M
∂H T
dH = 0
we rewrite dH = dH(M, S), then
cH
cM
dM −
∂M
∂H T
∂H
∂M S
dM =
cH
cM
dM −
χT
χS
dM = 0
apparently it applies from there
cH
cM
=
χT
χS
(129)
49. The gamma function is defined by an integral
Γ(n) :=
∞
0
dt exp(−t)tn−1
.
44
1. Prove the relationship
Γ(n + 1) = nΓ(n),
2. calculate Γ(n), n ∈ N,
3. calculate
Γ n +
1
2
, n ∈ N.
Solution:
We will prove 1, first, as we will use this formula in the following points as well. So
let’s write the relation for Γ(n+1) from the definition and adjust this expression using
per partes
Γ(n + 1) =
∞
0
dt exp(−t)tn
= −tn
exp(−t)|∞
0
0
+n
∞
0
dt tn−1
exp(−t) = nΓ(n).
In part 2 we first determine Γ(1):
Γ(1) =
∞
0
dt exp(−t) = − exp(−t)|∞
0 = 1.
We will now use the formula from 1 to calculate the values of gamma functions for
other natural numbers
Γ(2) = Γ(1 + 1) = 1 · Γ(1) = 1,
Γ(3) = 2 · Γ(2) = 1 · 2,
Γ(4) = 3 · Γ(3) = 1 · 2 · 3 = 6,
Γ(5) = 4 · Γ(4) = 1 · 2 · 3 · 4 = 24,
we can therefore determine a formula for general n:
Γ(n + 1) = n · (n − 1) · · · · · 3 · 2 · 1 = n!.
Analogously, we calculate 3: first Γ(1/2):
Γ
1
2
=
∞
0
dt exp(−t)t− 1
2 = 2
∞
0
ds exp −s2
=
√
π,
where in the last integral we substituted t = s2
. Now we will calculate the values of
the gamma function for the next n:
Γ
3
2
= Γ
1
2
+ 1 =
1
2
Γ
1
2
=
1
2
√
π,
45
Γ
5
2
=
3
2
Γ
3
2
=
1
2
·
3
2
√
π =
3
4
√
π,
Γ
7
2
=
5
2
Γ
5
2
=
1
2
·
3
2
·
5
2
√
π =
15
8
√
π,
Γ
9
2
=
7
2
Γ
7
2
=
1
2
·
3
2
·
5
2
·
7
2
√
π =
105
16
√
π,
generally for n
Γ n +
1
2
=
n − 1
2
·
n − 3
2
· · · · ·
7
2
·
5
2
·
3
2
·
1
2
√
π =
(n − 1)!!
2n
√
π.
50. Using the Gamma function, calculate the approximate expression ln(n!) for large
values of n (Stirling’s formula).
Solution:
From the previous example, we know that Γ(n + 1) = n!. We will use this fact in the
calculation. So let’s write the Gamma function and adjust the argument accordingly
Γ(n + 1) =
∞
0
dt exp(−t)tn
=
∞
0
dt exp(−t) · exp[n ln(t)] =
∞
0
dt exp[n ln(t) − t].
We make the substitution t = n + x
∞
0
dt exp[n ln(t) − t] =
∞
−n
dx exp[n ln(n + x) − n − x]
≈
∞
−n
dt exp n ln(n) − n −
x2
2n
= exp[n ln(n) − n]
∞
−n
dt exp −
x2
2n
.
Since the Gaussian function is non-zero only on a certain interval, the width of which,
however, is much smaller than n, we can send the lower limit of the integral to −∞.
Then we have a Gaussian function that we can integrate
exp[n ln(n) − n]
∞
−n
dt exp −
x2
2n
=
+∞
−∞
dt exp −
x2
2n
=
√
2πn exp[n ln(n) − n].
51. A hydrogen atom is in the level n = 3. Assuming that the occupation of the energy
levels is given by the microcanonical distribution, calculate the probability that the
atom is in states with the same secondary quantum number l.
46
Solution:
First, it is necessary to determine the number of possible states for each quantum state
described by l. For n = 3 the possible quantum numbers are l = {0, 1, 2}. Each state
with the orbital quantum number l is further divided according to magnetic quantum
numbers, namely m = {−l, −l + 1, . . . , l − 1, l}. Each such state is further divided
according to the spin quantum number s = ±1/2. Now we count the number of all
possible states for each state with a given l:
• l = 0: m = {0}, s = {±1/2}, and thus two states are possible,
• l = 1: m = {−1, 0, +1}, s = {±1/2}, six possible states,
• l = 2: m = {−2, −1, 0, +1, +2}, s = {±1/2}, ten possible states.
We can easily calculate the probability using a formula
wi =
# states with quantum number l = i
# all possible states
,
from here we easily find that
w0 =
1
9
, w1 =
1
3
, w2 =
5
9
.
52. Entropy for an isolated system is given by the relation S = kB ln Γ, where Γ is
the number of microstates. For a closed system S = −kB n wn ln wn. Show that both
relations are not contradictory.
Solution:
The system is divided into (body) and A′
The energy of the isolated system is the sum
of the energies of these two systems, whose energy is equal to E0 = E +E′
= konst. Let
us now calculate the entropy in individual cases. In an isolated system A each state n
is associated with Γ(E0 − En) states of the system A′
. We can modify this expression,
let’s take the logarithm of the number of states, which we expand according to the
Taylor formula to the first order according to En
kB ln[Γ(E0−En)] ≈ kB ln Γ(E0)−
∂
∂E
kB ln Γ(E)En = kB ln Γ(E0)−
∂S
∂E
En = kB ln Γ(E0)−
En
T
,
(130)
after subtracting the logarithm and dividing kB in this relation, we get
Γ(E0 − En) ≈ Γ(E0) exp −
En
kBT
. (131)
The total entropy of the isolated system is then equal to
S0 = kB ln
n
Γ(E0) exp −
En
kBT
= kB ln Γ(E0)+kB ln
n
exp −
En
kBT
= kB ln Γ(E0)+kB ln Z.
(132)
47
In the case of a closed system, it is necessary to add the entropy in both subsystems
S0 = S + S′
= −kB
n
wn ln wn + kB ln Γ(E0 − E) =
− kB
n
1
Z
exp −
En
kBT
− ln Z −
En
kBT
+ kB ln Γ(E0) −
∂kB ln Γ(E0)
∂E0
E =
kB ln Z +
1
T n
1
Z
exp −
En
kBT
En +kB ln Γ(E0)−
E
T
= kB ln Z
E
T
+kB ln Γ(E0)−
E
T
=
kB ln Γ(E0) + kB ln Z. (133)
In both cases, the entropy is given by the same expression.
53. Show that the heat capacity cV is given by the energy fluctuation, i.e.
cV =
1
kBT2
∆E2
.
Solution:
The mean energy of the system is equal to
E =
n
wnEn,
for V = konst. En does not change, then we can write for the heat capacity
cV =
∂E
∂T V
=
n
En
∂wn
∂T V
.
The probability of occupying a given state is equal
wn = exp
F − En
kBT
,
from which we easily obtain the derivative with respect to temperature
∂wn
∂T V
=
∂
∂T
F − En
kBT V
exp
F − En
kBT
=
T ∂F
∂T V
− F + En
kBT2
exp
F − En
kBT
.
Now we will use the knowledge from thermodynamics, from which we know the definition
of free energy
F = E − TS, dF = −pdV − SdT,
after that
∂F
∂T V
= −S,
and
E = F + TS,
48
which we substitute in the relation for the derivative wn with respect to temperature
∂wn
∂T V
=
En − E
kBT2
exp
F − En
kBT
=
En − E
kBT2
wn.
The heat capacity is then in the form
cV =
n
Enwn
En − E
kBT2
=
1
kBT2
n
E2
nwn − E
n
Enwn =
1
kBT2
E2
− E2
,
from there we get the relationship we are looking for
cV =
1
kBT2
∆E2
.
54. Calculate the thermodynamic properties of the system of N distinguishable classical
harmonic oscillators with frequency ω.
Solution:
The energy of the system of harmonic oscillators is equal
E =
N
n=1
p2
n
2m
+
1
2
mω2
q2
n . (134)
We first calculate the statistical sum
Z =
1
hN
R2N
dN
q · dN
p · exp −
N
n=1
p2
n
2mkBT
+
mω2
q2
n
2kBT
=
1
hN
R2N
dN
p · dN
q ·
N
n=1
exp
p2
n
2mkBT
+
mω2
q2
n
2kBT
=
1
hN
N
n=1 R
dpn · dqn · exp
p2
n
2mkBT
+
mω2
q2
n
2kBT
=
1
hN
∞
0
dp · exp
p2
2mkBT
N


∞
0
dq exp
mω2
q2
2kBT


N
=
1
hN
2mkBTπ ·
2πkBT
mω2
N
=
2πkBT
hω
N
=
kBT
ω
N
. (135)
We can easily calculate the free energy from the relation
F = −kBT ln(Z) = −kBT ln
kBT
ω
N
= −NkBT ln
kBT
ω
. (136)
49
We will use Maxwell’s relations to calculate pressure and entropy
p = −
∂F
∂V T
= 0, (137)
S = −
∂F
∂T V
= NkB ln
kBT
ω
+ 1 (138)
55. Consider a gas with diatomic molecules. Calculate the molar heat capacity of the
given gas. Consider only the vibrational motion of molecules, when the energy is given
by the relation
En = ω n +
1
2
. (139)
First, calculate the statistical sum from which you will determine the free energy and
from the free energy you can already determine the required heat capacity. You can
write the resulting heat capacity in the approximation of low and high temperatures.
Solution:
We calculate the statistical sum (partition function)
Z =
∞
n=0
exp −
−En
kBT
=
∞
n=0
exp −
ω n + 1
2
kBT
= exp −
ω
2kBT
∞
n=0
exp −
ω
kBT
n .
(140)
An infinite series is a geometric series for which we can find the sum. Then we can
write the statistical sum in the form
Z =
exp − ω
2kBT
1 − exp − ω
kBT
=
2
2 exp ω
2kBT
− exp − ω
2kBT
=
1
2 sinh ω
2kBT
. (141)
The free energy can be easily found using the statistical sum
F = −kBT ln(Z) = kBT ln 2 sinh
ω
2kBT
.
From a computational point of view, the most advantageous procedure seems to be to
first calculate the internal energy using a formula
E = F + TS = F − T
∂F
∂T V
.
After substitution and derivating it comes out
E = kBT ln 2 sinh
ω
2kBT
− kBT ln 2 sinh
ω
2kBT
+
ω
2
cotanh
ω
2kBT
=
ω
2
cotanh
ω
2kBT
.
50
We can easily calculate the required heat capacity from the energy
cV =
∂E
∂T V
=
ω
T
2
1
kB cosh2 ω
kBT
. (142)
Finally, we will determine the heat capacity for high and for low temperatures:
• approximation of low temperatures: in the equation for heat capacity, we break
down the cosh function using exponentials
cV =
∂E
∂T V
= 4
ω
T
2
1
kB exp ω
kBT
− exp − ω
kBT
2 .
The first exponential acquires significantly larger values than the second exponential,
so we can write the heat capacity in the form
cV ≈
4
kB
ω
T
2
exp −
2 ω
kBT
. (143)
• approximation of high temperatures: in this case, sinh acquires very small values,
therefore we develop according to the Taylor formula to the first order, then
cV ≈ 4
ω
T
2
1
kB
2 ω
kBT
2 = kb. (144)
56. Derive the form of the Maxwell-Boltzmann distribution of the speeds of gas molecules.
Proceed only from the assumption that the space is isotropic and that the
movement of gas molecules in individual directions is independent.
Solution:
Let us denote the desired function f(p) = f (px, py, pz). Equations follow from the
conditions of the assignment: independence of direction
f (px, py, pz) = fx (px) · fy (py) · fz (pz) (145)
isotropy of the space
f (px, py, pz) = f( p ) (146)
from here we get the equation
fx (px) · fy (py) · fz (pz) = f( p ) (147)
which we logarithmize (denote fx (px) = fx a f( p ) = f
ln fx + ln fy + ln fz = ln f (148)
51
we calculate the derivative with respect to vx, only the first term remains from the left
side, the right side is derived using the knowledge of the norm p = p2
x + p2
y + p2
z
f′
x
fx
=
∂f
∂px
1
f
=
px
p
f′
f
(149)
In this equation, the equations can be separated
f′
x
px · fx
=
f′
f p
(150)
Both sides must be equal to a constant. Let’s take the right side, which we multiply
by the speed norm
f′
f
= konst. p (151)
we can find the solution easily
f = C exp
k p 2
2
(152)
for a physically acceptable result, k must be negative.
57. Derive the Maxwell-Boltzmann distribution of atomic momentum using the canonical
distribution.
Solution:
Normovací konstantu nalezneme zintegrováním funkce přes celý hybnostní prostor, takový
integrál musí být roven jedné
1 =
+∞
−∞ −∞
C exp −
p2
2mkT
(153)
Převedeme na sférický integrál
+∞
−∞ −∞
C exp −
p2
2mkT
= C
2π
0
dϕ
π
0
dθ sin(θ)
∞
0
dpp2
exp −
p2
2mkT
(154)
Úhlová část je rovna 4π. U radiální části provedeme substituci t = p2
2mkBT
∞
0
dpp2
exp −
p2
2mkT
=
∞
0
dt
mkBT
2t
2mkBTt exp(−t) =
√
2 (mkBT)
3
2
∞
0
dt
√
t exp(−t)
(155)
integrál je roven gamma funkci Γ 3
2
= 1
2
Γ 1
2
=
√
π/2 Následně
1 = C
√
2 (mkBT)
3
2
√
π
2
= (2πmkBT)
3
2 (156)
52
normovací konstanta je tedy rovna
C =
1
(2πmkBT)
3
2
(157)
58. Assuming the validity of the Maxwell-Boltzmann distribution of the speeds of
gas molecules, calculate (a) px , (e)
√
∆E2, (b) p , (c) p2
, (f) the most probable
magnitude of momentum, (d) v2
, (g) probability that pz > 0.
Solution:
At the beginning, we calculate the general integral
+∞
−∞ d
ppn
· exp −ap2
Again we convert to spherical coordinates. We will perform the integration over the
angular part in exactly the same way as in the previous example. The radial part
throws out the Jacobian p2
, then
+∞
−∞ −∞
dppn
· exp −ap2
= 4π
∞
0
dppn+2
exp −ap2
(158)
by substituting t = ap2
we arrive at the integral
π
2an+ 1
2
∞
0
dtt
n
2
+ 1
2 exp(−t) (159)
from where we get the desired value of the integral
+∞
−∞
dppn
· exp −ap2
=
2π
a
n
2
+ 3
2
Γ
n
2
+
3
2
(160)
(a) We don’t even need to calculate this integral. If we write the triple integral, it
is easy we convert to three integrals. The only integral over the x coordinate is of the
form +∞
−∞
dpxpx exp −
p2
x
2mkBT
(161)
at first sight it is obvious that we are integrating an odd function over a symmetric
interval. The integral is apparently equal to zero.
(b) Using the formula for the integral for n = 1, then
p =
2πC
a2
Γ(2) =
2π (2mkBT)
3
2
(2πmkBT)
3
2
= 2
2mkBT
π
(162)
where C is the normalization constant, which we also need for the calculation.
(c) We apply exactly the same formula as in the previous case, only for n = 2.
p2
= 3mkBT (163)
53
(d) Assuming the classical relationship between momentum and velocity p = mv we
can write
v2
=
p2
m2
=
1
m2
p2
=
3kBT
m
(164)
(e) First we need to modify the fluctuation equation
∆E = (E − E )2
= E2
− 2E E + E 2
= E2
− 2 E + E 2
= E2
− E 2
(165)
we assume the classical dispersion relation E = p2
2m
, which we substitute in the fluctuation
equation
∆E =
p4
4m2
−
p2
2m
2
=
1
4m2
p4
− p2
(166)
the calculation was thus reduced to the calculation of mean values. In addition, we
have already calculated the mean squared value, so all that remains is to calculate the
fourth power, then
∆E =
1
4m2
15 (mkBT)2
− 9 (mkBT)2
=
6 (mkBT)2
4m2
=
6
4
(kBT)2
(167)
after square root we get
∆E =
3
2
kBT (168)
(f) We must determine the most probable magnitude of momentum from the definition
of probability density
dW = f (px, py, pz) dpx · dpy · dpz = 4πp2
f(p)dp = 4π2
p2
exp −
p2
2mkBT
dp (169)
so the expression in front of dp is enough to differentiate to find the extremum of the
function
2p exp −
p2
2mkBT
− p2 2p
2mkBT
exp −
p2
2mkBT
= 0 (170)
after a simple modification, we obtain p2
= 2mkBT. Then p =
√
2mkBT.
(g) this case, we solve the triple integral with the only difference that the integration
over z takes place only in the interval (0, ∞). Considering that we are integrating over a
symmetric interval an even function, the contribution of both the positive and negative
parts must be equal. Then the probability is equal to one half.
59. Calculate the density distribution in a column of gas with base A under the
influence of a homogeneous gravitational field (in the atmosphere). Assume that the
gas is made up of indistinguishable particles, each of mass m.
Solution:
54
Again, we start by calculating the statistical sum. We write this for a system of
indistinguishable particles in the form
Z =
1
N!(2π )3
R3N ×Ω
d3N
p · d3N
q · exp −
n
p2
n
2mkBT
+
mgq
kBT
,
where Ω = {[x, y, z]; x ∈ [−L, L], y ∈ [−L, L], z ∈ [0, ∞), L ∈ R+
}. We calculate the
statistical sum for one particle
Z =
1
(2π )3


R
dp · exp −
p2
2mkBT
Ω
dq · exp −
mgq
kBT

 ,
the first integral is from a rescaled Gaussian, therefore it is equal
R
dp · exp −
p2
2mkBT
= (2πmkBT)3.
Second integral
Ω
dq · exp −
mgq
kBT
=
L
−L
dx
L
−L
dy
A
∞
0
dz exp −
mgz
kBT
kBT
mg
= A
kBT
mg
.
After that
Z =
1
(2π )3
(2πmkBT)
3
2 A
kBT
mg
. (171)
The probability density that the particle occurs in the phase volume d3
p · d3
q is given
dwn =
1
Z
d3
p · d3
q
(2π )3
exp −
p2
2mkBT
exp −
mgq
kBT
=
mg
(2πmkBT)
3
2 AkBT
exp −
p2
2mkBT
exp −
mgq
kBT
d3
p · d3
q.
We will only be interested in the probability of finding a particle in a given position
q, therefore we will integrate wn over the momentum space q, where wn denotes the
probability density of the given particle in the element
p
dwn =
mg
AkBT
exp −
mgq3
kBT
P
·d3
q,
where P denotes the probability density of the given particle in the element d3
q. Then
we get the number density
n =
Nmg
AkBT
exp −
mgq3
kBT
. (172)
55
The density is equal
̺(z) = ̺(0) · exp −
mgq3
kBT
. (173)
60. Show that pressure and energy density have the same unit.
Solution:
ověříme jednotky z definičních vztahů:
[p] =
[F]
[S]
=
kg · m · s−2
m2
= kg · m−1
· s−2
,
[e] =
[E]
[V ]
=
kg · m2
· s−2
·
m3
= kg · m−1
· s−2
.
61. Calculate the density of states for relativistic particles and find the limit relations
for classical and ultra relativistic particles.
Solution:
The relation holds for the density of states
ρ(E) =
gV
πd
1
2d
Vol Sd−1 [k(E)]d−1
dE
dk
, (174)
where g is the degeneracy of energy levels, d he dimension of space, and Vol Sd−1
surface
of d−1 dimensional sphere. In our case, we have d = 3. The dispersion relationship
E(k) is given by the relation
E =
√
m2c4 + 2k2c2,
from here
k =
√
E2 − m2c4
c
.
Substituting the appropriate quantities into (174) we get
ρ(E) =
4πgV
(2π )3
1
c3
E
√
E2 − m2c4. (175)
• Classical limit – Energy in this case is equal to:
Ekl. ≈ E − mc2
=
√
m2c4 + 2k2c2 − mc2
= mc2
1 +
2k2
m2c2
− mc2
≈
mc2
1 +
1
2
2
k2
m2c2
− mc2
=
2
k2
2m
. (176)
56
• Ultrarelativistic limit – in this case E ≫ mc2
, applies , then we can neglect the
second term in the square root
ρ(E) =
4πgV
(2π )3
1
c3
E2
.
62. Show that in the classical case it is possible to derive the Maxwell-Boltzmann
velocity distribution law from the grand canonical distribution of a single particle.
Solution:
It holds for the number of boson states in the energy interval (E, E + dE)
dN =
ρ(E)dE
exp E−µ
kBT
+ 1
.
Since in the classic case it is
exp −
E − µ
kBT
≪ 1,
we can neglect the one in the denominator. Then, after substituting for ρ(E) we get
dN =
4πgV
(2π )3
√
2m3E exp
E − µ
kBT
dE.
We transform the expression from energy coordinates to velocity coordinates
dE =
∂E
∂v
dv = mv,
because we consider the classical formula for kinetic energy E = 0.5 · mv2
dw =
4πgV
(2π )3
√
2m3
m
2
exp
0.5mv2
− µ
kBT
mdv =
4πgV
(2π )3
m3
v2
exp −
v2
2mkBT
exp −
µ
kBT
dv.
To calculate µ we start from the relation for one particle
1 = N =
gV
(2π )3
(2πmkBT)
3
2 F3
2
µ
kBT
=
gV
(2π )3
(2πmkBT)
3
2
1
Γ 3
2
∞
0
dx
x
1
2
exp x − µ
kBT
+ 1
.
In the integral, the exponential acquires significantly larger values than one, so we
neglect the number one, we get
∞
0
dx x
1
2 exp −x +
µ
kBT
≈ exp
µ
kBT
∞
0
dx x
1
2 exp (−x) = exp
µ
kBT
Γ
3
2
.
57
We substitute in the previous formula
gV
(2π )3
exp
µ
kBT
= (2πmkBT)− 3
2 .
Where do we receive
dw = 4π
m
2πkBT
v2
exp
v2
2kBT
dv. (177)
63. Let’s define functions
Bn(y) =
1
Γ(n)
∞
0
dx
xn−1
exp(x − y) − 1
, (178)
and
Fn(y) =
1
Γ(n)
∞
0
dx
xn−1
exp(x − y) + 1
. (179)
For these functions prove
dBn+1(y)
dy
= Bn(y),
and
dFn+1(y)
dy
= Fn(y).
Solution:
Let’s do the math first
dBn+1(y)
dy
=
d
dy

 1
Γ(n + 1)
∞
0
dx
xn
exp(x − y) − 1

 =
1
Γ(n + 1)
∞
0
dx xn d
dy
1
exp(x − y) − 1
.
In this phase, we will use the knowledge of the derivative of the function in the argument
of the integral. As
d
dy
1
exp(x − y) − 1
= −
d
dx
1
exp(x − y) − 1
,
we can substitute the expression on the right-hand side into the integral, in which we
get integrating by parts
−
1
Γ(n + 1)
∞
0
dx xn d
dx
1
exp(x − y) − 1
=
1
Γ(n + 1)



−
xn
exp(x − y) − 1
∞
0
0
+
∞
0
dx
n · xn−1
exp(x − y) − 1



.
58
Furthermore, we know that Γ(n + 1) = nΓ(n). Then we get
dBn+1(y)
dy
d
dy
=
1
Γ(n)
∞
0
dx
xn−1
exp(x − y) − 1
= Bn(y). (180)
The relation for fermions can be proved analogously.
64. From a relationship
Ω = −kBT
gV
(2π )3
(2πmkBT)
3
2 B5
2
µ
kBT
, (181)
valid for a non-relativistic ideal boson gas, count the number of particles N and derive
the relation for the chemical potential within the classical limit.
Solution:
The number of particles is given by the relation
N = −
∂Ω
∂µ T,V
,
where we substitute relation (181), then we get
N = −kBT
gV
(2π )3
(2πmkBT)
3
2
∂
∂µ
B5
2
µ
kBT
.
From the previous example we know that
dBn+1(y)
dy
= Bn(y),
from where
dBn+1
µ
kBT
dµ
=
1
kBT
Bn
µ
kBT
.
The number of particles will then work out for us
N =
gV
(2π )3
(2πmkBT)
3
2 B3
2
µ
kBT
. (182)
65. Calculate cV of a non-relativistic fermion gas and verify the validity of the classical
limit for cV /N.
Solution:
In the calculation, we will use the knowledge of the internal energy of the fermion gas
in the classical case
E =
3
2
gV
λ3
T
kBTF5
2
µ
kBT
, (183)
59
and the relation for the number of particles
N =
gV
λ3
T
F3
2
µ
kBT
, (184)
where
λT =
2π 2
mkBT
. (185)
We expand the energy using the expression indicating the number of particl N/N(T, µ),
we get
E =
3
2
NkBT
F5
2
µ
kBT
F3
2
µ
kBT
,
We will introduce
Fn
µ
kBT
= Fn.
We calculate the heat capacity according to the well-known formula
cV =
∂E
∂T N,V
=
3
2
NkB
F5
2
F3
2
+
3
2
NkBT
∂F 5
2
∂T
F3
2
− F5
2
∂F 3
2
∂T
F3
2
2 ,
the derivative of the function Fn with respect to T looks as follows
∂Fn
∂T V,T
=
∂
∂T
µ
kBT
F′
n =
∂µ
∂T V,N
1
kBT
−
µ
kBT2
Fn−1.
We put in the relationship for cV and receive
cV =
3
2
NkB
F5
2
F3
2
+
3
2
N
∂µ
∂T V,N
−
µ
T


1 −
F5
2
F1
2
F3
2
2


 .
For the next calculation, we will use the fact that the total number of particles does
not depend on the temperature, i.e
∂N
∂T V,N
=
3
2
F3
2
+
1
kB
∂µ
∂T V,N
−
µ
kBT
F1
2
= 0.
From there we express the partial derivative of the chemical potential with respect to
temperature
∂µ
∂T V,N
=
µ
T
−
3
2
kB
F3
2
F1
2
. (186)
60
After replacing it in cV we get
cV =
3
2
N


1 −
F5
2
F1
2
F3
2
2


 −
3
2
kB
F3
2
F1
2
+
3
2
NkB
F5
2
F3
2
=
3
2
NkB



F5
2
F3
2
−
3
2
N
F3
2
F1
2


1 −
F5
2
F1
2
F3
2
2






.
In the classic case, we can approximate the function Fn by
Fn ≈ exp
µ
kBT
, (187)
then all the ratios of the function F are equal to one and we get the well-known relation
for the heat capacity
cV ≈
3
2
NkB. (188)
Another way of solving: First, using thermodynamics, we calculate what cV,N . is equal
to. We know that
cV,µ = T
∂S
∂T V,µ
,
we introduce entropy as S = S (N(µ, V, T), V, T), then
T
∂S
∂T V,µ
= T
∂S (N(µ, V, T), V, T)
∂T V,µ
= T
∂S
∂N V,µ
∂N
∂T V,µ
+ T
∂S
∂T V,N
.
We will use Maxwell’s relations arising from the free energy relation
dF = −SdT − pdV + µdN,
from where
∂S
∂N V,T
= −
∂µ
∂T N,V
.
Assuming that at constant volume the number of particles does not change, i.e.
dN =
∂N
∂µ T,V
+
∂N
∂T µ,V
= 0,
we get a relationship
∂µ
∂T N,V
= −
∂µ
∂N T,V
∂N
∂T µ,V
.
We substitute the calculated partial derivatives into the expression for cV,µ
cV,µ = T
∂N
∂µ
−1
T,V
∂N
∂T
2
µ,V
+ T
∂S
∂T V,N
cV,N
.
61
For the heat capacity we then get the relation
cV,N = cV,µ − T
∂N
∂T
2
µ,V
∂N
∂µ
T,V
. (189)
In this step, we will finish with thermodynamics and use the results of statistical
physics. It is enough to just substitute the relevant quantities for the classical fermion
gas. for a non-relativistic fermion gas. The thermodynamic potential of such a system
of particles is equal to
Ω = −
gV
λ3
T
kBTF5
2
µ
kBT
. (190)
From (190) we find the entropy of the gas
S = −
∂Ω
∂T V,µ
=
−gV kB
λ3
T
5
2
F5
2
−
µ
kBT
F3
2
.
From the entropy, we can easily find the heat capacity cV,µ
cV,µ = T
∂S
∂T V,µ
= −
gV kB
λ3
T
15
4
F5
2
− 3
µ
kBT
F3
2
+
µ2
k2
BT2
F1
2
.
Z (184) We calculate the partial derivative from
∂N
∂T V,µ
=
gV
λ3
T
3
2T
F3
2
−
µ
kBT2
F1
2
.
And finally
∂N
∂µ V,T
=
gV
λ3
T
1
kBT
F1
2
.
Substituting everything back into (189), we obtain
cV,N =
gV kB
λ3
T
15
4
F5
2
− 3
µ
kBT
F3
2
+
µ2
k2
BT2
F1
2
− T2 9
4T2
F2
3
2
F1
2
−
3µ
kBT3
F3
2
+
µ2
k2
BT4
F1
2
,
We divide the whole expression by kBN, we substitute (184) on the right-hand side
cV,N
kBN
=
15
4
F5
2
F3
2
− 3
µ
kBT
+
µ2
k2
BT2
F1
2
F3
2
−
9
4
F3
2
F1
2
+
3µ
kbT
−
µ2
k2
BT2
F1
2
F3
2
.
In the classical case, we can again approximate the function Fn by expression (187),
then
cV,N
kBN
=
15
4
− 3
µ
kBT
+
µ2
k2
BT2
−
9
4
+ 3
µ
kBT
−
µ2
k2
BT2
.
We get to the best part of the calculation, the only non-zero term will be 3/2, the
others disappear, then we have the result
cV,N
kBN
=
3
2
, (191)
which is the same result as in the previous case.
62