Exercises for the course
Statistical Physics
Luca Amendola & Nils Strodthoff
University of Heidelberg
l.amendola@thphys.uni-heidelberg.de
2014
http://www.thphys.uni-heidelberg.de/~amendola/teaching.html
Exercises taken from Chandler, Introduction to Modern Statistical Mechanics; Y-K Lim, Problems and Solutions on Thermodynamics and
Statistical Mechanics; Greiner, Neise, Stocker, Thermodynamics and Statistical Mechanics; Amit, Introduction to Statistical Physics; H.
Callen, Thermodynamics and an Introduction to Thermostatistics; Pathria, Statistical Mechanics; Dalvit, Problems on Statistical
Mechanics.
November 20, 2014
Contents
1 Thermodynamics
2
2 Stochastic dynamics
15
3 Statistics
18
4 Statistical mechanics
22
1
Chapter 1
Thermodynamics
1. Consider a monoatomic ideal gas in a reversible cycle which brings the gas in contact with two heat baths
at Th > T` , as in Fig. (1.1). The isothermal processes are connected by isochores. Find the total work
done by the system and the efficiency of the cycle.
Solution. The work done along an isothermal process at temperature Th between V1 , V2 is
ˆ V2
V2
pdV = −N kTh log
δW12 = −
V
1
V1
and analogously for the compression between V3 = V2 and V4 = V1 ,
ˆ V4
V2
δW34 = −
pdV = N kT` log
V
1
V3
(1.1)
(1.2)
Since no work is done during the isochore transformations, the total work done is
δWtot = δW12 + δW34 = −N k(Th log
V2
V2
V2
− T` log ) = −N k∆T log
V1
V1
V1
(1.3)
(∆T = Th − T` ) which is negative since is done by the system.
The efficiency is defined as the absolute value of the total work done by the system divided by the total
heat absorbed by it. For the process 1 − 2 the heat absorbed is equal to minus the work done δW12 , since
in reversible isothermal processes there is no change of internal temperature. For the isochore process
4 − 1 there is no mechanical work, so
3
(1.4)
dU = N kdT = δQ
2
The change in temperature is related to the change in pressure by the ideal gas equation of state pV = N kT ,
so that, for constant volumes
V dp = N kdT
(1.5)
and finally
3
3
V (p1 − p4 ) = N k∆T
2
2
In step 2 − 3 the heat is not absorbed but released. Then we have
∆Q14 =
η
=
=
N k∆T log VV12
|∆W |
=
∆Qabs
N kTh log VV21 + 32 N k∆T
(Th − T` )α
T` α + 23 (r − 1)
=
1
−
,
Th
Th α + 32 (Th − T` )
α + 32 r−1
r
(1.6)
(1.7)
(1.8)
where r = TTh` and α = log(V2 /V1 ). One immediately sees that the efficiency is lower than the efficiency of
Carnot’s cycle since the term in the brackets is always larger than 1.
2
CHAPTER 1. THERMODYNAMICS
3
Figure 1.1: Find the work and efficiency of a cycle.
2. Calculate the dependence of the temperature of the atmosphere on the height above sea level. You may
assume the atmosphere behaves like an ideal monoatomic gas. Moreover, since air is a poor conductor of
heat you may assume that air rises adiabatically in the atmosphere without exchange of heat.
Solution. For a perfect gas undergoing an adiabatic process, we have dU = 3N kdT = −pdV since δQ = 0.
Moreover we must have pV 5/3 = const (monoatomic gas), hence pdV = − 35 V dp and thus dT = 25 NVk dp.
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 acceleration (that we
assume to be constant). Combining our results we find dT = − 52 mg
k dz , where m = V ρ/N is the average
mass of the gas particles. Integrating this relation and calling T0 the temperature at see level (where we
put the origin of the z axis) yields the result:
T (z) = T0 −
2 mg
z
5 k
(1.9)
3. The van der Waals equation of state for N particles connects the pressure p, the volume V = vN and the
temperature T of a gas by
a
(p + 2 )(v − b) = kT
(1.10)
v
where a, b are some positive parameters. Show that one has the relation
∂U
∂p
= T
−p
(1.11)
∂V T
∂T V
valid in general, and the relation
∂U
∂V
=
T
a
v2
(1.12)
for the van der Waals case. The left hand side of (1.12) 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 separated apart, the energy of the system varies. That is, there is
some interaction between particles. For a perfect gas one has a = 0, which implies that the particles are
actually free.
Solution. The first relation is obtained by the chain rule for partial derivatives: from
U (S(V, T ), V )
one has
∂U ∂S
∂U
∂S
∂U
|T =
|V
|T +
|S = T
|T − p
∂V
∂S ∂V
∂V
∂V
(1.13)
CHAPTER 1. THERMODYNAMICS
4
From the Maxwell relations now one has
∂S
∂p
|T =
|V
(1.14)
∂V
∂T
(This can be obtained by taking dF = −SdT −pdV and by considering that since dF is an exact differential,
∂F 2 /∂T ∂V = ∂F 2 /∂V ∂T ). Then we have
∂U
∂p
|T = T
|V − p
∂V
∂T
(1.15)
Applying this to the van der Waals equation, one obtains the final result 1.12.
4. Show explicitly that the entropy change for the transformation of Ex. 1 from p1 , V1 to p3 , V3 is the same
along the path of Ex. 1 and along a path composed uniquely of isochores and isobares (see Fig. 1.2).
Figure 1.2: Find the entropy for two different paths.
Solution. The change in entropy during an isothermal process (where dU = 0) is
ˆ
∆S =
δQ
1
=
T
T
ˆ
δQ =
∆Q
∆W
1
=−
=
T
T
T
ˆ
V2
pdV = N k log
V1
V2
V1
(1.16)
while during an icochoric process (δW = 0) is
ˆ
∆S =
δQ
=
T
ˆ
3
dU
= Nk
T
2
ˆ
Th
Tl
3
T`
dT
= N k log
T
2
Th
Finally during an isobaric process (during which since p = const one has dV =
dU
δW
and therefore
ˆ
∆S =
δQ
=
T
ˆ
(1.17)
Nk
p dT )
one has
3
3
N kdT = pdV
2
2
= −pdV
(1.18)
=
dU − δW
5
= p
T
2
ˆ
dV
5
= Nk
T
2
(1.19)
ˆ
dV
5
V2
= N k log
V
2
V1
(1.20)
The path isothermal+isochore we employed in Ex. 1 gives therefore
∆S = N k log
V2
3
T`
+ N k log
V1
2
Th
(1.21)
CHAPTER 1. THERMODYNAMICS
5
If we go from V1 to V20 = V2 with an isobaric process, we can write p1 V1 = N kTh and p1 V2 = N kTh0 at
the initial and final stage of the isobare, respectively, from which Th0 = Th V2 /V1 . Then we have indeed
5
V2
3
T`
N k log
+ N k log 0
2
V1
2
Th
5
V2
3
T` V1
= N k log
+ N k log
2
V1
2
Th V2
V2
3
T`
= N k log
+ N k log
V1
2
Th
= ∆S
∆S 0 =
(1.22)
5. A system has energy U (S, V ) expressed as a function of entropy and volume given by
U = S 2 f (V )
(1.23)
Construct the free energy as a function of T,V.
Solution. The temperature is obtained as
∂U
∂S
=T
(1.24)
V
from which we obtain
2Sf (V ) = T
(1.25)
F = U − T S = −S 2 f (V )
(1.26)
The (Helmholtz) free energy is defined as
which can be rewritten as F (T, V ) using 1.25 as
F =−
T2
4f (V )
(1.27)
6. Given the Maxwell distribution of speed,
m 3/2
mv 2
4πv 2 e− 2kT
2πkT
f (|v|) =
find the most probable velocity vmax , the mean velocity hvi and the root mean square velocity
Which velocity coincides with velocity estimated from the equipartition theorem?
(1.28)
p
hv 2 i.
Solution The most probable velocity is found
f 0 (vmax ) = 0,
(1.29)
q
which results in vmax = 2kT
m .
The mean velocity is calculated via
ˆ
∞
hvi =
r
dvvf (v) =
0
8kT
πm
(1.30)
and the rms velocity as
p
ˆ
hv 2 i
∞
2
1/2
dvv f (v)
=
0
which coincides with the mean velocity from 21 mv 2 = 32 kT .
r
=
3kT
,
m
(1.31)
CHAPTER 1. THERMODYNAMICS
6
7. Which of these are exact differentials?
(a) df = (x2 + y 2 )dx + (x2 − y 2 )dy
(b) df = (2x + y)dx + xdy
(c) df = 2ex
2
+y 2
dx + 2yex
2
+y 2
dy
Solution For A(x, y)dx + B(x, y)dy check
∂A
∂y
=
x
∂B
∂x
.
(1.32)
y
b and c are exact differentials.
8. Find the work for an ideal monoatomic gas for the four processes (p1 , V1 ) → (p2 , V2 ):
(a) (p1 , V1 ) → (p2 , V1 ) (isochoric), (p2 , V1 ) → (p2 , V2 ) (isobaric)
(b) (p1 , V1 ) → (p1 , V2 ) (isobaric), (p1 , V2 ) → (p2 , V2 ) (isochoric)
(c) (p1 , V1 ) → (p2 , V2 ) (isothermal)
(d) (p1 , V1 ) → (p2 , V2 ) (adiabatic)
Solution
(a) For isochoric processes ∆W = 0 and for isobaric processes ∆W = −p∆V . Hence we have ∆W1 =
−p2 ∗ (V2 − V1 )
(b) ∆W1 = −p1 ∗ (V2 − V1 )
(c) For an isothermal process
ˆ
V2
∆W3 = −
pdV
V1
ˆ
V2
V −1 dV
V1
V1
= p1 V1 log
V2
= −nRT
(d) For an ideal monoatomic gas γ =
Cp
CV
(1.33)
= 35 . Using pV γ = const we find
ˆ
V2
∆W4 = −
pdV
V1
ˆ V2
=−
p1
V1
= −p1 V1γ
V1
V
γ
dV
(1.34)
V21−γ − V11−γ
1−γ
9. From the definitions of the heat capacities CV , Cp , find the relation
∂U
∂V
]
Cp − CV = [p +
∂V T ∂T p
Use this relation to find CV − Cp for a van der Waals gas
a p + 2 (V − b) = N kT.
V
(1.35)
(1.36)
CHAPTER 1. THERMODYNAMICS
7
Show that the ideal gas limit for CP − CV is reached for V → ∞.
Solution. The heat capacities are defined as
Cp =
δQ
|p
dT
CV =
δQ
|V
dT
(1.37)
In a process in which V = const, the change in internal energy dU = δQ, so we have
CV =
∂U
|V
∂T
(1.38)
To find δQ|p we need to introduce the enthalpy H = U + pV . In fact we have
dH(S, p) = dU + pdV + V dp = T dS + V dp
(1.39)
which for p = const gives dH|p = T dS|p = δQ|p . Then we have
Cp =
∂H
|p
∂T
(1.40)
Now since H = U + pV , we can also write
∂H
∂U
∂V
|p =
|p + p
|p
∂T
∂T
∂T
(1.41)
Expressing V as a function of T, p (which is always possible since we are considering N = const and
therefore two state variables are sufficient to determine the system), one can write U = U (T, V ) =
U [T, V (T, p)] and use the chain rule to find
∂U
∂U
∂U ∂V
|p =
|V +
|T
|p
∂T
∂T
∂V ∂T
(1.42)
∂H
∂U
∂U
∂V
∂U
∂V
|p =
|V +
|T + p
|p = CV +
|T + p
|p
∂T
∂T
∂V
∂T
∂V
∂T
(1.43)
∂U
∂V
Cp − CV = p +
|T
|p
∂V
∂T
(1.44)
and inserting in (1.41),
from which finally
From eq. (1.11), this expression can be further simplified to
Cp − CV = T
∂p ∂V
|V
|p
∂T ∂T
(1.45)
Now the two derivatives can be easily computed from the Van der Waals EOS p(V, T ) and we finally
obtain
Nk
Cp − CV =
(1.46)
1 − 2a(1 − b/V )2 /V N kT
When V → ∞, Cp − CV → N k as expected.
10. From the expression
Cp − CV = T
∂p ∂V
|V
|p
∂T ∂T
(1.47)
derived in Ex. 6, obtain the following expression
Cp − CV = V T
α2
βT
(1.48)
CHAPTER 1. THERMODYNAMICS
8
where
α
≡
βT
≡
1 ∂V
|p
V ∂T
1 ∂V
|T
−
V ∂p
(1.49)
(1.50)
are the thermal expansion coefficient and the isothermal compressibility, respectively. What are the
implications for Cp − CV for stable matter?
Solution. We can write Eq. (1.47) as
Cp − CV = αV T
∂p
|V
∂T
(1.51)
Then we consider an EOS V (p, T ) and take the differential
dV =
∂V
∂V
|T dp +
|p dT
∂p
∂T
For a process at V = const, this implies dV = 0 and thus
∂V
∂p
α
∂T p
|V = − =
∂V
∂T
βT
∂p
(1.52)
(1.53)
T
from which finally
Cp − CV = V T
α2
βT
(1.54)
For stable matter βT > 0 and hence Cp > CV .
11. A compressor performs adiabatic compressions on two ideal gases, first a monoatomic one ( i.e. one with
3 d.o.f., translations, for instance helium) and then a diatomic one (with 5 d.o.f, 3 translations plus 2
rotational d.o.f., for instance air, to some approximation), starting with the same pressure and volume
and compressing to the same final volume.
CP
(a) Show γ = C
= d2 + 1 for an ideal gas with d d.o.f.
V
(b) In which case does the compressors heat more?
Solution. The equation of adiabatic processes is
pV γ = p0 V0γ
(1.55)
where γ = Cp /CV . From the previous exercise, we know that
γ=
Nk
+1
CV
(1.56)
For an ideal gas with d d.o.f. one has the internal energy U = d N2 kT , from which CV = (∂U/∂T )V =
dN k/2 so that
2
(1.57)
γ = +1
d
which equals 5/3 ≈ 1.67 for the monoatomic gas and 7/5 = 1.4 for the diatomic one. Since
γ
V0
pfinal = p0
(1.58)
Vfinal
and Vfinal < V0 , the final pressure for the monoatomic gas is larger; from the ideal gas EOS, p ∼ T , so a
larger pressure corresponds to a higher final temperature.
CHAPTER 1. THERMODYNAMICS
9
12. A volume V of ideal gas at pressure p is compressed isothermally to a volume r times smaller and then
expanded adiabatically to the original volume, as in Fig. 1.3. What is the final pressure? What is the
work done on or by the system in the two cases in which the gas is monoatomic or diatomic, as a function
of the initial pressure and volume and of the compression ratio?
Figure 1.3: Isothermal compression and adiabatic expansion.
Solution. During the isothermal compression from state A to B , pV = const so the final pressure is
pB =
VA
pA = rpA
VB
(1.59)
During the subsequent adiabatic expansion from B to C the equation is
pV γ = p0 V0γ
(1.60)
which in this case gives the final pressure
pC = pB
VB
VC
γ
= rγ−1 pA
(1.61)
(VC = VA ) which is therefore larger for the diatomic gas. The work is done on the system, since the
isothermal compression lies above the adiabatic expansion in a p, V plane. For the monoatomic gas the
adiabatic is steeper so the area between the two curves, i.e. the total work, is larger. The work is
ˆ B
ˆ B
VB
dV
= −pA VA log
= pA VA log r
(1.62)
δWAB = −
pdV = −pA VA
V
VA
A
A
ˆ C
ˆ C
δWBC = −
pdV = −pB VBγ
V −γ dV = −(1 − γ)−1 pB VBγ VC−γ+1 − VB−γ+1
(1.63)
B
B
!
1−γ
VB
γ −γ+1
−1
= −(1 − γ) pB VB VA
1−
(1.64)
VA
γ−1
1−γ !
VB
VB
−1
= −(1 − γ) pA
VA 1 −
(1.65)
VA
VA
= −(γ − 1)−1 pA VA 1 − r1−γ
(1.66)
so the total work is
δW = δWAB + δWBC = pA VA [log r − (γ − 1)−1 1 − r1−γ ] > 0
(for r > 1, γ > 1).
(1.67)
CHAPTER 1. THERMODYNAMICS
10
13. The sound speed of longitudinal waves of small amplitude in an ideal gas with pressure p and density ρ is
s
dp
v=
(1.68)
dρ
Find the sound speed in the case in which one can assume that the rarefactions and compressions are a)
adiabatic and b) isothermal.
Solution. Both isothermal and adiabatic EOS can be written as
pV α = const
(1.69)
where α = 1 for isothermal processes and α = γ for adiabatic ones. Then we have
dV
dp
+α
=0
p
V
(1.70)
p
dp
= −α
dV
V
(1.71)
dp
dp dV
pM
p
=
=α
=α
dρ
dV dρ
V ρ2
ρ
(1.72)
from which
Putting ρ = M/V , we obtain
and from the EOS of ideal gases, p = N kT /V = (N/M )ρkT
dp
N kT
kT
=α
=α
dρ
M
m
where m = M/N is the mass per molecule. Then we have
r
kT
v= α
m
(1.73)
(1.74)
In most cases, the fluctuations are adiabatic since they are so fast that heat has no time to escape. For
air (γ ≈ 1.4, m ≈ 5 · 10−26 kg) at room temperature T = 300K, one has (k = 1.38 · 10−23 J/K)
r
kT
≈ 350m/s
(1.75)
v= α
m
14. Derive the Maxwell relation
∂S
∂p
|T =
|V
∂V
∂T
The pressure of isotropic radiation can be written as p = u(T )/3, where u(T ) is the energy density. Show
that u obeys the equation
du
1
T
−u
(1.76)
u=
3
dT
Solve this equation to obtain Stefan’s law that relates u and T .
Solution. From the free energy definition F = U − T S one has dF = −SdT − pdV and therefore
−S =
Since F is a state function
∂F
|V ,
∂T
−p =
∂2F
∂2F
=
∂V ∂T
∂T ∂V
∂F
|T
∂V
(1.77)
(1.78)
CHAPTER 1. THERMODYNAMICS
11
and therefore
∂S
∂p
|T =
|V
∂V
∂T
(1.79)
U (S(T, V ), V ) = u(T )V
(1.80)
∂U ∂S
∂U
∂S
∂U
|T =
|V
|T +
|S = T
|T − p
∂V
∂S ∂V
∂V
∂V
(1.81)
∂U
∂p
|T = T
|V − p
∂V
∂T
(1.82)
We can write now
and one has from the chain rule
and finally, using (1.79)
which gives
u=
1 du
(T
− u)
3 dT
(1.83)
du
d log T
(1.84)
This can written as
4u =
or
d log u = 4d log T
(1.85)
4
which gives u = σT , i.e. Stefan’s law, where σ is an integration constant.
15. A system is stable if for every possible transformation the entropy decreases. Considering two identical
systems, isolated from the environment but in contact through a fixed insulated wall. If we transfer a ∆U
from one to the other (keeping N fixed), we get the stability condition
S(U − ∆U, V ) + S(U + ∆U, V ) ≤ 2S(U, V )
(1.86)
and if we transfer both ∆U and ∆V we get
S(U − ∆U, V − ∆V ) + S(U + ∆U, V + ∆V ) ≤ 2S(U, V )
(1.87)
Show that the two conditions together imply for infinitesimal ∆U, ∆V that
∂2S
|V ≤ 0
∂U 2
and
∂2S ∂2S
−
∂V 2 ∂U 2
∂2S
∂V ∂U
(1.88)
2
≥0
(1.89)
2
∂ S
Solution. Expand both conditions in Taylor series to second order. We get ∂U
2 |V ≤ 0 and, from the
2
∂ S
second equation when ∆U = 0, also ∂V 2 |U ≤ 0 . Then from the second relation we obtain in general
∆T H∆ ≤ 0
(1.90)
where ∆ is a vector with components (∆U, ∆V ) and ∆T its transpose and H is the Hessian matrix formed
by the second derivatives


H =  SU U
SU V
SU V 
SV V
where SXY = ∂ 2 S/∂X∂Y . In other words, we get the conditions
X
∆i Hij ∆j ≤ 0
i,j
(1.91)
(1.92)
CHAPTER 1. THERMODYNAMICS
12
which implies that H is a semi-negative definite matrix (i.e such that multiplied right and left by any
vector ∆ it always gives a negetive or zero value). This is realized only if the eigenvalues of H are all
negative. That is, we should impose the condition that the solutions λ1,2 of the equation
det(H − λI) = 0
(1.93)
λ2 − λ(SU U + SV V ) + (SU U SV V − SU2 V ) = 0
(1.94)
are negative. This gives
This equation will have two negative roots if and only if the two coefficients are both positive (Descartes’
rule). The first one, −(SU U + SV V ) is always positive. Therefore the second must be positive as well to
obtain two negative eigenvalues.
16. Starting from the definitions
CV
=
Cp
=
δQ
∂T V
δQ
∂T p
(1.95)
(1.96)
show that
CV
=
Cp
=
∂S
∂T V
∂S
T
∂T p
T
(1.97)
(1.98)
Then prove that
βS
CV
=
βT
Cp
(1.99)
where
βT
=
βS
=
1
V
1
−
V
−
∂V
|T
∂p
∂V
|S
∂p
(1.100)
(1.101)
are the compressibilities, i.e. the relative change in volume when pressure is applied either isothermally
(βT ) or adiabatically (βS ).
Solution. From the definition of CV , we just need to note that if V = const, dUV = T dSV = δQV to
obtain the result. For Cp , we can note first that dH = T dS + V dp so that dH|p = T dS|p = δQp , and
again we obtain the desired result. Then we need to prove that
∂V
∂V
∂S
∂S
|S /
|T =
|V /
|T
(1.102)
∂p
∂p
∂T
∂p
Since V depends only on two state functions, i.e. p, T or S, p, there must be a relation f (S, p, V ) = 0 (i.e
if V = V (S, p) then V − V (S, p) = f (S, p, V ) = 0). Therefore we can apply the triple product rule
∂V
∂S
∂p
|S
|p
|V = −1
(1.103)
∂p
∂V
∂S
from which
∂V
|S
∂p
=−
∂V
|p
∂S
∂S
|V
∂p
(1.104)
CHAPTER 1. THERMODYNAMICS
13
and analogously from f (V, p, T ) = 0
so that
∂V
|T
∂p
=−
∂V
|p
∂T
∂T
|V
∂p
(1.105)
∂V
∂V
∂V
∂S
∂T
∂p
|S /
|T =
|p
|V
|p
|V
∂p
∂p
∂S
∂p
∂V
∂T
(1.106)
Now by the chain rule
∂V
∂T
|p
|p
∂S
∂V
∂S
∂p
|V
|V
∂p
∂T
from which
=
=
∂T
∂S
∂S
∂T
(1.107)
p
(1.108)
V
∂V
∂T
∂S
∂S
∂S
CV
∂V
|S /
|T =
=
/
=
∂p
∂p
∂S p ∂T V
∂T V
∂T p
Cp
(1.109)
17. An ideal classical gas can be defined as one gas with a fixed number of particles at a fixed temperature
which satisfies the following conditions:
(a) the internal energy does not depend on the volume
(b) the enthalpy does not depend on the pressure
Use this definition to derive the equation of state pV = N kT by employing appropriate thermodynamic
potentials and exploiting Maxwell relations.
Solution. Starting from the differential relations
dU = T dS − pdV
(1.110)
dH = T dS + V dP
(1.111)
one obtains
∂S
∂U
=T
−p
∂V T
∂V T
∂H
∂S
=T
+V
∂p T
∂p T
(1.112)
(1.113)
and finally
∂S
p
=
∂V T
T
∂S
V
=−
∂p T
T
(1.114)
(1.115)
exploiting the definition of the ideal gas from above. The entropy S can now be eliminated using Maxwell
relations. In the first case one employs the Maxwell relation associated to the free energy, (∂S/∂V )T =
(∂P/∂T )V , for the second one the relation associated to the Gibbs free energy, (∂S/∂p)T = −(∂V /∂T )p ).
Therefore one obtains
∂p
p
=
(1.116)
∂T V
T
∂V
V
=
(1.117)
∂T p
T
(1.118)
CHAPTER 1. THERMODYNAMICS
14
These can be integrated, giving
p = f1 (V )T
V = f2 (p)T
(1.119)
with two unknown functions f1 and f2 . However, they can only hold simultaneously if f1 (V ) = α/V and
f2 (p) = α/p for constant α. Therefore the equation of state takes the form
pV = αT
(1.120)
Since p and T are intensive but V extensive, the coefficient α has to scale with the particle number
(which was held fixed so far), α = N k, with an additional constant which we identify with the Boltzmann
constant:
pV = N kT.
(1.121)