BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI
FIRST SEMESTER 2026-27
PHY F312: Statistical Mechanics
Solution of a few problems in Assignment 2 *
“There are three kinds of lies: lies, damned lies, and statistics.” Benjamin Disraeli
Question 3
The length of the mercury column in a certain mercury-in-glass thermometer is 5.0 cm when the
thermometer is in contact with water at its triple point. Consider length of mercury column X as
a thermometric property and let θ be the empirical temperature measured by the thermometer.
(a) Calculate the empirical temperature when the length of mercury is 6cm.
(b) Calculate the length of mercury column at steam point.
(c) If X can be measured with the best possible precision of 0.01cm, can this thermometer
distinguish between the ice point and triple point ? ( Dittman Zemansky 7e. Prob 1.7)
Consider a temperature scale which is defined as
t∗ = aθ2 + b
find the constants a and b, such that t∗ = 0 and t∗ = 100 at ice point and steam point respectively.
Find the value of t∗ when the length of mercury is 7cm. Find the length of the mercury column
when t∗ = 50. (Sears. Salinger 3e. 1.10)
Solution:
(a) We know that we can relate the empirical temperature and the thermometric property using
the following equation:
X
θ = 273.16 K
XT P
6cm
θ = 273.16 K
= 327.79 K
5cm
(b) Let the mercury length at steam-point be XSP .
373.16 = 273.16
XSP
XT P
373.16
= 6.83 cm
273.16
(c) No. Since ice-point and the triple point differ by 0.01 K and the least count of θ is given by:
XSP = 5 cm
=
273.16
0.01 = 0.55 K
5.0
So we wont be able to differentiate between any two temperatures with an accuracy less than
0.55 K.
Given t∗ at θ = 273.16 and 373.16:
0 = a × 273.162 + b
100 = a × 373.162 + b
Solving both equations simultaneously, a = 1.55 × 10−3 and b = −115.5.
If X = 7 cm, then θ = 382.4 K, so t∗ = 111.15.
If t∗ = 50, then θ = 326.8 k, so X = 5.98 cm.
* Compiled by Meet Khatri
Question 6
A certain gas is enclosed in a cylinder with a movable piston. It is observed that if the walls are
adiabatic, a quasi-static increase in volume results in a decrease in pressure according to
P 3 V 5 = constant
(a) Find the quasi-static work done for processes ADB, ACB and the direct linear process AB
as shown in the figure below.
(b) What is the difference of internal energy of the system between states A and B (UB − UA )
?
(c) What is the heat transfer to the system for the three processes ADB, ACB and the direct
linear process AB?
Solution:
(a) For the process ADB:
Z
WADB = −
P dV = −105 × (8 × 10−3 − 10−3 ) = −700J
Similarly, we get WACB = −21.9J and the direct linear process work WAB = −360.9J
(b) Since we are given the equation for an adiabatic process (Q = 0 and ∆U = W) between A and
B, we can write
Z VB
UB − UA = WAB =
Z VB P dV = −PA
VA
VA
VA
V
53
dV
3
PA VA 5/3 (VB 2/3 − VA 2/3 )
2
3
= (25 − 100) = −112.5J
2
=
(c) Continuing from (a) we can write that for the process ADB:
UB − UA = WADB + QADB
QADB = −112.5 + 700 = 587.5J
Similarly, we get QACB = −90.6J and QAB = 248.4J. Do note that here we are able to
calculate Q for ADB and ACB, not for the seperate paths AD, DB, AC, CB as we do not know
UD − UB and UC − UB .
Question 12(a)
Find the efficiency of a Carnot engine operating between two reservoirs at temperatures TH and
TL with ideal gas as the working substance.
Solution:
Consider that the Carnot engine operates on the Carnot cycle given above. Let us denote the work
and the heat associated with each step as Wi and Qi where i = 1, 2, 3, 4. Let us denote the efficiency
of the Carnot engine by η. We know that η is given by:
η=
Total work done by the engine
Heat absorbed by the engine
Process 1 is iso-thermal expansion at temperature TH , and the system does work W1 (positive) and
absorbs heat Q1 . Process 2 is adiabatic expansion, so Q2 = 0 and the system does work W2
(positive). Process 3 is iso-thermal compression at temperature TL and the system does work W3
(negative) and releases heat Q2 . Process 4 is adiabatic compression, so Q4 = 0 with the work
W4 (negative).
So now efficiency is given by:
W1 + W2 + W3 + W4
η=
Q1
We know that for an iso-thermal process Q = W (since U (T ) we have ∆U = 0 for an isothermal
Vf inal
process for an ideal gas) and thus W = nRT ln Vinitial
. While for an adiabatic process
nR
W = γ−1 (Tinitial − Tf inal ). Using this we get the Wi ’s as: W1 = Q1 = nRTH ln VVab ,
nR
nR
W2 = γ−1
(TH − TL ), W3 = nRTL ln VVdc and W4 = −W2 = γ−1
(TL − TH ).Thus
η =1+
nRTL ln VVdc
nRTH ln VVab
η =1−
TL ln VVdc
TH ln VVab
For an adiabatic process pV γ = constant, we can write
p1 V1 V1 γ−1 = p2 V2 V2 γ−1
nRT1 V1 γ−1 = nRT2 V2 γ−1
γ−1
V2
T1
=
T2
V1
So using TTHL =
Vc
Vb
γ−1
=
Vd
Va
γ−1
we get VVab = VVdc , and thus:
η =1−
TL
TH
Question 13
It can be shown that a body in an uniform radiation field (equal intensity of radiation from all
directions) will experience compressive pressure. Thermal radiation of a black body consists of
a photon gas in equilibrium at some temperature T. The following facts are given about the
properties of radiation (a)The energy density u(= U/V ) is a function of T only and (b)Radiation
pressure is related to energy density u as p = 13 u
(a) Prove the relation
∂U
∂p
=T
−p
∂V T
∂T V
Hence find U for radiation as a function of T and V. (You may need to use an integration
constant which can be called σ)
(b) Find the Entropy S for radiation.
(c) Draw a neat diagram of a Carnot cycle in the P-V diagram with radiation as the working
substance between two reservoirs at temperatures at T1 and T2 with T1 > T2 . Label the
adiabatics and the isotherms and write down their slopes.
(d) Find the heat Q1 and Q2 taken and rejected from the two reservoirs.
(e) Find the work done per cycle and hence the efficiency of the cycle.
Solution:
The internal energy of the system will be given by U = uV = σV T 4 and thus p = 13 u = 13 σT 4 is only
a function of temperature. From this we can say that an iso-thermal process for the given system
will also be an iso-baric one. So the Carnot cycle can be depicted as:
Since (p, V, T ) are not all independent, we can take the state variables to be (V, T ) or (p, T ). The
first law of thermodynamics states (for a quasi-static process) that:
dU = d̄Q − pdV
1
4
d̄Q = d(σV T 4 ) + σT 4 dV = σT 4 dV + 4σT 3 V dT
3
3
Thus the condition for an adiabatic process is
4 4
dV
dT
σT V
+3
=0
3
V
T
On integration
log V + 3 log T = constant
V T 3 = constant
Using the equation of state p = 13 σT 4 , we get
4
pV 3 = constant
So Cp can be considered infinite since if the pressure is fixed the temp. can’t be changed, while the
CV is given by:
CV = (∂U/∂T )V = 4σT 3 V
Returning to the Carnot Cycle:
4
Q1→2 = σT14
3
Z V2
dV =
V1
4 4
σT (V2 − V1 )
3 1
Similarly, Q3→4 = 43 σT34 (V4 − V3 ) and Q2→3 = Q4→1 = 0 because they are adiabatic processes. For
the work done we can write:
Z V2
1
p dV = − T14 σ(V2 − V1 )
W1→2 = −
3
V1
1
W3→4 = − T34 σ(V4 − V3 )
3
RV
To calculate W2→3 = − V23 p dV we make us of the fact that:
4
4
pV 3 = p1 V13 =
4
1
W2→3 = − σT14 V23
3
Z V3
1 4 34
σT V
3 1 2
4
4
−1
−1
V − 3 dV = σT14 V23 (V3 3 − V2 3 )
V2
4
−1
= σT14 V23 V3 3 − σT14 V2
From V2 T13 = V3 T33 we can write
W2→3 = σT34 V3 − σT14 V2
Similarly, W4→1 = σT34 V4 − σT14 V1 .
We see that the total heat absorbed is equal to the work done:
Q1→2 + Q3→4 = −(W1→2 + W2→3 + W3→4 + W4→1 )
Further
4
4
Q1→2
= σT13 (V2 − V1 ) = σT33 (V3 − V4 )
T1
3
3
since T13 V2 = T23 V2 = T33 V3 and T13 V1 = T43 V4 = T33 V4
Thus
Q1→2
Q3→4
=−
T1
T3
Or
η =1−
T3
|Q3→4 |
Tcold
=1−
=1−
|Q1→2 |
T1
Thot
Entropy of an Ideal gas
For a simple compressible system, we start from the fundamental thermodynamic relation
T dS = dU + P dV.
For an ideal gas,
P V = nRT,
and the internal energy depends only on temperature:
dU = nCV dT.
Therefore,
T dS = nCV dT + P dV.
Dividing by T ,
dS = nCV
P
dT
+ dV.
T
T
Using the ideal-gas equation,
P
nR
=
,
T
V
we obtain
dS = nCV
dT
dV
+ nR
.
T
V
For a calorically ideal gas, CV is constant. Integrating,
S = nCV ln T + nR ln V + constant.
E
To write the logarithms in dimensionless form, we introduce arbitrary reference values T0 and V0 :
S(T, V ) = S0 + nCV ln
T
T0
V
V0
+ nR ln
where S0 is the entropy at the reference state (T0 , V0 ).
Equivalently, since
PV
T =
,
nR
we have
dT
dP
dV
=
+
.
T
P
V
Hence,
dS = nCV
dP
dV
+ n(CV + R)
.
P
V
Using
CP = CV + R,
this becomes
dS = nCV
dV
dP
+ nCP
.
P
V
Integrating,
S(P, V ) = S0 + nCV ln
P
P0
+ nCP ln
V
V0
.
Alternatively, eliminating V , we obtain
dS = nCP
dP
dT
− nR
.
T
P
Therefore,
S(T, P ) = S0 + nCP ln
T
T0
− nR ln
P
P0
.
Thus, the three equivalent forms for the entropy of an ideal gas are
T
V
S(T, V ) = S0 + nCV ln
+ nR ln
,
T0
V0
T
P
S(T, P ) = S0 + nCP ln
− nR ln
,
T0
P0
P
V
S(P, V ) = S0 + nCV ln
+ nCP ln
.
P0
V0
The additive constant S0 cannot be determined from classical thermodynamics alone. The T dS
relation determines only entropy differences between equilibrium states.