1. 12. Thermodynamics 12.1. Temperature
The essential concepts here are heat and temperature. Heat is a form of energy transferred between a system and an environment due to the temperature difference existing between them.
12.1. Temperature
Temperature is one of SI base quantities and is measured in kelvins (K).
Zeroth law of thermodynamics
If body A is in thermal equilibrium with body T (a thermoscope) and body B is in
thermal equilibrium with body T, then bodies A and B
are in thermal equilirium with each other.
We say that both bodies have the same temperature.
Measurements of temperature
A temperature scale must be defined. The most fundamental is the Kelvin scale but
in practice some others (Celsius, Fahrenheit) are used. Selecting a certain
thermometric body, the thermometric feature and assuming a relationship between this
feature and temperature, a thermometer is defined. The constant volume gas
thermometer is a standard against which all other thermometers are calibrated. A

2. Constant volume gas thermometer
The temperature is determined by measuring the pressure p of a gas in a fixed volume. Introducing a linear temperature scale
(12.1)
and taking the temperature of the triple point of water as a
reference temperature Ttr = K, one can eliminate constant A
(12.2)
Dividing eq.(12.1) by (12.2) one obtains
(12.3)
As the indications of such a thermometer depend on the kind of
a gas used, one has to use smaller and smaller gas quantities
and calculate the extrapolated value
(12.4)
The extrapolated value
gives the ideal gas
temperature.
Figure from HRW,2
A gas thermometer measuring the temperature of a liquid
A gas thermometer measuring T of a boiling water. The different
gases give only one extrapolated
value K.

3. 12.2. Heat
Heat is a form of energy transferred between bodies of different temperatures (it is not a feature of a system). Similarly to work, it is measured in joules (J). Previously the unit calorie (cal) was used.
The specific heat is defined as a heat needed to warm a unit of mass of a given body by 1 K
(12.5)
Above definition is more strict in a differential form, moreover determining c a body has to be warmed in well defined conditions, under certain parameter constant, e.g. pressure or volume
x – a constant parameter (p, V,…)
Some specific heats are listed below
C[J/g·K]
Al 0.90
Cu 0.39
Pb 0.13
Water 4.19
Mercury 0.14

4. 12.3. Work done by a gas
The work done on a gas by acting a force F on a piston with surface
area A is
(12.6)
where dV is the differential change in the volume.
The total work between volumes V1 and V2 is
(12.7)
and depends on the path of changing the volume. Lets calculate the
work between the initial state A and final state E done on different paths.
The work done along path ABE is
along ADE is
and along ACE is
In each case one obtains a different result. Work is a path-dependent quantity.
A gas with pressure
p in the container A

5. 12.4. Special processes for ideal gas
Starting from ideal gas law n – number of moles, m – mass of a gas,
μ – mass of a mole, R –gas constant,
it is easy to introduce some processes for ideal gas.
1. Isothermal process (T = const)
2. Adiabatic process (Q = 0), no exchange of heat with
the ambient
κ = Cp/CV
Cp – specific heat at constant pressure
CV – specific heat at constant volume
3. Isochoric process (V = const)
4. Isobaric prrocess (p = const)
Comparison of isothermal and adiabatic processes on p-V diagram A
Isochoric 1-3, isobaric 1-2 and isothermic 2-3 processes on p-V diagram

6. 12.5.The first law of thermodynamics
The variation in internal energy ΔU of the system is equal to the heat Q added to it,
lowered by work W done by the system
(12.8)
If a work is done on a system, the sign „-” in formula (12.8) is changed for „+”.
For the processes shown in the figure from
1 to 4 the work done by the gas decreases, heat heat delivered to the gas also changes but
the change of internal energy is always the
same
The internal energy is a function of state and its variation is independent of
the process path. It depends only on the positions of initial and final states
(the variations of heat and work are path dependend) .
For the ideal gas the variation in internal energy is only a function of temperature
cV – specific heat at a constant volume

7. The first law of thermodynamics, cont.
Example
What is a change in internal energy in adiabatic, constant-volume (isochoric) and isothermal processes?
Adiabatic process means Q = 0
In this case ΔU = Q – W = -W = W’, where W’ – work done on the system
Isochoric process means V = const
In this case W = 0 and then ΔU = Q
Isothermal process means T = const
In this case ΔU = 0 and Q = W

8. 12.6. Entropy and the second law of thermodynamics
The processes which obey the first law of thermodynamics are not always possible.
For some processes the energy is always conserved but they are irreversible, i.e.
cannot be reversed by only small changes in the environment. The direction of
processes is indicated by the changes of entropy; for isolated system the entropy
for irreversible processes always increases.
The change in entropy is given by the integral
(12.9)
where i and f mean the initial anf final states, Q is the energy transferred to or from
the system at temperature T. Integral (12.9) can be calculated only for reversible
processes.
Problem 1
Calculate the change in entropy for 1 mol of nitrogen when it expands freely in the isolated
system to the double volume.
The process of free expansion is irreversible. For a free expansion
we have: In this case we can
replace this process by the reversible isothermal expansion between
the same initial and final states and calculate ΔS with eq.(12.9).

9. Changes in entropy The change in entropy in an isothermal process is
Q – heat transferred to the gas
The work done by a gas during expansion in this process
R – gas constant
n – number of moles
For isothermal process
because
Finally one obtains
As ΔS is positive, the entropy increases what could be expected for an irreversible
process of a free expansion in a closed system.
Analysing the reversible process of isothermal expansion one has to take into
account a heat that is exchanged between the gas and the heat reservoir (in contrast
to a free expansion for which Q=0) . For the whole closed system gas – heat
reservoir the total change in entropy is equal to zero, as the change in entropy of a
reservoir is of reverese sign vs. that of a gas.
Isothermal expansion process for ideal gas

10. Second law of thermodynamics
There are several equivalent formulations of the second law. One of them is:
In a process that occurs in an isolated system, the change in entropy is greater than zero for irreversible processes and equal to zero for reversible processes.
(12.10)
In a real world almost all processes are irreversible and reversibility is an
idealization. In living organisms many processes are irreversible but the entropy of
the organism does not increase because the processes involving the surroundings
cause an increase of the entropy of the environment.
The Carnot cycle
It is a thermodynamic cycle in which the conversion of heat into work is done with a
maximum efficiency. Heat Q1
is absorbed during isothermal
expansion AB and Q2 is
discharged during isothermal
compression CD. Processes
AB and CD are reversible. A
The full cycle consists of two isothermal processes AB, CD and two adiabatic processes BC, DA. Positive work is done by the gas in processes AB and BC. In processes CD and DA the environment does work on the gas. The net work W is the area enclosed by the cycle.

11. The Carnot cycle, cont. The changes in entropy in a Carnot cycle are
process AB: (12.11)
process BC: (12.12)
process CD: (12.13)
process DA: (12.14)
The net change in entropy in a complete cycle is zero (entropy is a state function).
From (12.11) – (12.14) one obtains (12.15)
The efficiency of any thermal engine is defined as the ratio of work done per cycle
to the thermal energy absorbed per cycle
(12.16)
For the Carnot engine with the use of (12.15) we thus obtain
No series of processes is possible whose sole result is the tranfer of heat from a thermal reservoir and the complete conversion of this heat to work – second law of thermod.

12. The Carnot refrigerator
Using work an engine can transfer heat from a low temperature reservoir to a high
temperature reservoir. Examples: houshold refrigerator, air conditioner, heat pump.
This can be done by reversing the Carnot cycle
Carnot refrigerator
From (12.15) one obtains
and
what can be written as
The coefficient of performance of a Carnot refrigerator is then
(12.17)
Example:
What is the coefficient of performance of a Carnot refrigerator working between
temperatures t2 = -130C and t1 = 270C.
For real fridges A

13. 12.7. The statistical interpretation of entropy
To calculate the changes of entropy for the free expansion of an ideal gas, the
statistical mechanics will be used. We analyse the distribution of gas particles in
the two identical halves of a box. As an example we take 3 gas molecules and
calculate the number of microstates corresponding to a given configuration
Configurations Multiplicity W
(number of microstates)
First configuration
Second configuration
Third configuration
Fourth configuration
Multiplicity W of
configuration is
calculated from
where
(n! – „n factorial”)
For the 2nd configuration we have

14. The statistical interpretation of entropy, cont.
Different configurations have different number of microstates, then the
configurations are not equally probable.
In the example discussed configurations 2 and 3 are more probable than 1 and 4.
The probability for configuration 2 is 3/8 and for configuration 1 is 1/8.
Example 2
We take 100 molecules. What is the number of microstates for configurations:
n1 = 100, n2 = 0
n1 = 50, n2 = 50
a) b)
The most probable are therefore these configurations where the molecules distribute
equally between the halves of the box.
L. Boltzmann formulated a relationship between the entropy of a given
configuration and the multiplicity W of that configuration
where the Boltzmann constant
As all states are equally probable, then configurations with a large number of
microstates are the most frequent and related entropy takes the highest values.
for large N one can
use the Stirling’s
approximation

15. The statistical interpretation of entropy, cont.
Problem
We solve again the problem of change in entropy for 1 mol of nitrogen when it expands freely in the isolated system to the double volume but using the statistical theory.
In n moles we have N molecules.
Initially (the molecules occupy the left side of a container) the muliplicity of configuration (N,0) is 1.
When nitrogen occupies the full volume, i.e. the configuration is (N/2, N/2) one
obtains
The entropy for initial state:
The final entropy:
Because NkB = nR one gets
We obtained the same result as when the change in entropy was calculated in terms
of temperature and heat transfer. A