lOMoARcPSD|58847208
CLASSICAL THERMODYNAMICS Classical Laws
0th Law - If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is also in thermal
equilibrium with C.
1st Law - There is a system property called energy which is conserved, but can interconvert to different forms. 𝑑𝐸 =
𝑑𝑄 + 𝑑𝑊
2nd Law - There is a system property called entropy which (if the system is isolated from its environment) either
increases or remains constant during a thermodynamic process.
3rd Law - The entropy of a system is a universal constant at absolute zero temperature, and we set that constant to be
zero.
State variables;
State variables describe the equilibrium condition of a system without reference to any previous events - it only depends
on the current situation, independent of path.
• Extensive state variables - depend on the size of the system, scale up when the system is replicated.
• Intensive state variables - do not change if we replicate the system.
The Ideal Classical Gas;
The energy is entirely kinetic as there are no interactions between the particles.
Equations of state - 𝑝𝑉 = 𝑁𝑘𝑇 (Ideal Gas Law),
Kinetic theory -
(N times energy of each particle, which is just KE formula)
Thermodynamic processing of the Ideal Gas;
We can imagine that the process is conducted very slowly, so that thermal equilibrium between the reservoir and
system is roughly maintained, and this is called a quasistatic process.
1. For a quasistatic compression, in the absence of heat transfer, we can state that 𝑑𝐸 = 𝑑𝑊 = −𝑝𝑑𝑉 as 𝑑𝑄 = 0.
Inserting , we get
Divide by
2. For heat input, with volume held constant so no external work done, the change in energy is now purely due to the
change in heat energy; 𝑑𝐸 = 𝑑𝑄. It is of interest to calculate a heat capacity;
3. If we relax the constraint on volume and allow work to take place, we write 𝑑𝑄 = 𝑑𝐸 − 𝑑𝑊 where we are specifically
talking about the heat and energy given to the system. If the process is quasistatic then 𝑑𝑊 = −𝑝𝑑𝑉 ; if we then divide
𝑑𝑄 by the change in temperature during the process, we get ; this only holds for a quasistatic process.
4. Holding pressure constant instead of volume (as in 2.), becomes and therefore
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. Note that and this ratio is denoted 𝛾 for general systems.
Entropy of the Ideal Gas;
First we transfer heat quasistatically to a system and consider which is the heat transfer modulated by
temperature. Using the first law; , insert and we get
Sum over a complete quasistatic heat transfer from state a to b;
Inserting limits; and we have defined entropy, S.
Entropy is also a state variable as it is just a function of them; . The quantity C in the
denominator is included to make the argument of the logarithm dimensionless. It depends on system properties that do
not change as a result of the process, in this case N so we write 𝐶(𝑁). Note that in order to raise the temperature of
twice the amount of gas, we would need twice the
amount of heat. Because , we require that
5 which gives us 𝐶 = 𝑐̂𝑁2
where 𝑐̂ is now independent
of p, T and N. Inserting the relation between pressure and energy - - gives us more forms for the entropy;
We call the integral of the Clausius Integral and equating it to the change in entropy, 𝑆(𝑏) − 𝑆(𝑎), gives us the
Clausius expression for entropy change.
A system does not contain a quantity of heat Q, it only receives incremental contributions dQ to its energy E during a
thermodynamic process.
By summing the dQs, we obtain something independent of the path, and by dividing by T we produce an increment,
known as an exact differential, in the state variable S.
5
The condition discussed earlier, , is simply stating that 𝑆 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 for the same process.
Quasistatic adiabatic compression of a system (𝑑𝑄 = 0 at every stage) is isentropic. Entropy
Change in Free (Joule) Expansion of an Ideal Gas;
The entropy of an isolated system cannot decrease in spontaneous thermodynamic processes, i.e. releasing a constraint
on a system. In almost all cases, the entropy increases, but in some special cases it can remain the same.
1. Gas held inside a container of volume 𝑉0, pressure 𝑃0 and temperature 𝑇0, contained within a larger volume 𝑉1 that is
otherwise empty and thermally isolated from the environment.
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2. The container bursts and so 𝑉0 expands into 𝑉1 nonquasistatically.
3. Once the new equilibrium has been established, the final entropy is; and the initial
entropy was .
4. Since the gas does no work expanding into a vacuum, and no heat is supplied from the environment, there is no
change in the system energy. This implies that the final temperature is equal to the initial temperature. Entropy
change is therefore; . Note that the constant 𝑐̂ does not appear.
Clearly the system entropy has increased as a consequence of the expansion since 𝑉1 > 𝑉0. However there was no heat
transfer so we must conclude that the entropy doesn’t change solely due to a heat transfer. Previously but
we must modify this for a nonquasistatic process as there was entropy change but no heat transfer. We therefore write;
where 𝑇𝑟 is the temperature of the heat source (reservoir). 𝑑𝑆𝑖 can be considered to be the discrepancy
in the change in entropy not accounted for by . It is called the internal change in entropy. For a free expansion where
𝑑𝑄 = 0, only the 𝑑𝑆𝑖 remains so Δ𝑆 = Δ𝑆𝑖 such that .
Entropy change due to Nonquasistatic Heat Transfer;
Consider a process of heat exchange between a reservoir at temperature 𝑇𝑟 and a monatomic ideal classical gas initially
at temperature 𝑇0. N and V are fixed. No work is done on the system and the heat transfer is nonquasistatic. We assume
the system has acquired the temperature𝑇𝑟 at equilibrium according to the 0th law. Change in system entropy is;
. This can be positive or negative
because the system is not isolated, and it will depend on which temperature is greater. The internal entropy change is;
using . Note this is ALWAYS positive,
regardless of which temperature is larger. For this reason, internal entropy change is also known as entropy generation.
The change in entropy of the universe, here taken to be the combination of the ideal gas system and the reservoir, is
Δ𝑆𝑡𝑜𝑡 = Δ𝑆 + Δ𝑆𝑟. The entropy of the universe tends towards a maximum because of entropy generation in
nonquasistatic processes.
Cyclic Thermodynamic Processes, the Clausius Inequality and Carnot’s Theorem;
We consider taking the ideal gas around a cyclic heating and cooling process driven by a time dependent reservoir
temperature 𝑇𝑟(𝑡), or more properly a set of reservoirs with slightly different temperatures. We assume the initial and
final states are at equilibrium but the process need not be quasistatic. Cyclic means that the conditions at the start and
the end are the same. We integrate to get . For a cycle, 𝑆𝑓𝑖𝑛𝑎𝑙 = 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙 and
; this is called the Clausius inequality. For a quasistatic process, because Δ𝑆𝑖 = 0 and we
can replace 𝑇𝑟 with the system temperature T.
A famous example of a quasistatic cyclic process is the Carnot cycle.
1. Isothermal expansion in contact with a hot reservoir
2. Adiabatic expansion (isentropic)
3. Isothermal compression in contact with a cold reservoir
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4. Adiabatic, isentropic compression returning system to original state.
Since the process is quasistatic, there is no entropy generation during the process, therefore the change in system
entropy of the total cycle should be zero. We can show this explicitly by considering the changes in entropy during the
process, during the isothermal
stages;
2 2
3 3
, and since 𝑇ℎ 𝑉2 = 𝑇𝑐 𝑉3
and
, we can see that . Substituting this into the result above, we get 𝑁𝑘𝑙𝑛(1) = 0 = Δ𝑆.
The cycle is designed to convey heat from the hot to the cold reservoir, with the conversion of some of this flow into
mechanical work, obtained by employing the system expansions and compressions to move a load, e.g.
The hot reservoir passes heat Δ𝑄ℎ into the system from 𝑉1 > 𝑉2. Since Δ𝐸 = 0 for the system during this expansion, this
heat is converted into work; . Similarly the cold reservoir receives heat Δ𝑄𝑐 equal to the work
done on the system during the isothermal compression; . The work done on
the environment per cycle is; . We have an idealized device that converts heat
into work, a type of heat engine known as the Carnot engine. It can operate in both directions, by pumping heat from
the cold to the hot reservoir when mechanical work is supplied, e.g. in a fridge.
The sequence can be used to investigate the efficiency 𝜂𝐶 of a Carnot engine. This efficiency depends solely on the
temperatures of the reservoirs; Carnot’s theorem. . The efficiency is always less than 100% i.e. there
is always waste heat ΔQc transferred to the cold reservoir. The most efficient engines clearly need as large a difference
in temperature as possible to minimize .
Generality of the Clausius Expression for Entropy Change;
for an arbitrary substance undergoing a portion of a Carnot cycle. If any substance receives heat from a
reservoir nonquasistatically, it contributes to the change in its entropy through where we include the
entropy generation 𝑑𝑆𝑖.
Fundamental Relation of Thermodynamics;
We rewrite the 2nd Law, 𝑑𝐸 = 𝑑𝑄 + 𝑑𝑊, substituting to obtain 𝑑𝐸 = 𝑇𝑑𝑆 − 𝑝𝑑𝑉.
This law holds for all processes, despite being derived from a quasistatic one, because at equilibrium the values of state
variables (E, T, S, p and V) do not depend on the process used to reach equilibrium.
This relation is often written as .
We can write an increment in entropy more generally as . Comparing this to the
fundamental relation above, we see that and .
These results provide us with a means to define temperature and pressure if we have a function for S.
We must also consider the change in number of particles, 𝑁. We expect;
CLASSICAL THERMODYNAMICS Classical Laws
0th Law - If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is also in thermal
equilibrium with C.
1st Law - There is a system property called energy which is conserved, but can interconvert to different forms. 𝑑𝐸 =
𝑑𝑄 + 𝑑𝑊
2nd Law - There is a system property called entropy which (if the system is isolated from its environment) either
increases or remains constant during a thermodynamic process.
3rd Law - The entropy of a system is a universal constant at absolute zero temperature, and we set that constant to be
zero.
State variables;
State variables describe the equilibrium condition of a system without reference to any previous events - it only depends
on the current situation, independent of path.
• Extensive state variables - depend on the size of the system, scale up when the system is replicated.
• Intensive state variables - do not change if we replicate the system.
The Ideal Classical Gas;
The energy is entirely kinetic as there are no interactions between the particles.
Equations of state - 𝑝𝑉 = 𝑁𝑘𝑇 (Ideal Gas Law),
Kinetic theory -
(N times energy of each particle, which is just KE formula)
Thermodynamic processing of the Ideal Gas;
We can imagine that the process is conducted very slowly, so that thermal equilibrium between the reservoir and
system is roughly maintained, and this is called a quasistatic process.
1. For a quasistatic compression, in the absence of heat transfer, we can state that 𝑑𝐸 = 𝑑𝑊 = −𝑝𝑑𝑉 as 𝑑𝑄 = 0.
Inserting , we get
Divide by
2. For heat input, with volume held constant so no external work done, the change in energy is now purely due to the
change in heat energy; 𝑑𝐸 = 𝑑𝑄. It is of interest to calculate a heat capacity;
3. If we relax the constraint on volume and allow work to take place, we write 𝑑𝑄 = 𝑑𝐸 − 𝑑𝑊 where we are specifically
talking about the heat and energy given to the system. If the process is quasistatic then 𝑑𝑊 = −𝑝𝑑𝑉 ; if we then divide
𝑑𝑄 by the change in temperature during the process, we get ; this only holds for a quasistatic process.
4. Holding pressure constant instead of volume (as in 2.), becomes and therefore
, lOMoARcPSD|58847208
. Note that and this ratio is denoted 𝛾 for general systems.
Entropy of the Ideal Gas;
First we transfer heat quasistatically to a system and consider which is the heat transfer modulated by
temperature. Using the first law; , insert and we get
Sum over a complete quasistatic heat transfer from state a to b;
Inserting limits; and we have defined entropy, S.
Entropy is also a state variable as it is just a function of them; . The quantity C in the
denominator is included to make the argument of the logarithm dimensionless. It depends on system properties that do
not change as a result of the process, in this case N so we write 𝐶(𝑁). Note that in order to raise the temperature of
twice the amount of gas, we would need twice the
amount of heat. Because , we require that
5 which gives us 𝐶 = 𝑐̂𝑁2
where 𝑐̂ is now independent
of p, T and N. Inserting the relation between pressure and energy - - gives us more forms for the entropy;
We call the integral of the Clausius Integral and equating it to the change in entropy, 𝑆(𝑏) − 𝑆(𝑎), gives us the
Clausius expression for entropy change.
A system does not contain a quantity of heat Q, it only receives incremental contributions dQ to its energy E during a
thermodynamic process.
By summing the dQs, we obtain something independent of the path, and by dividing by T we produce an increment,
known as an exact differential, in the state variable S.
5
The condition discussed earlier, , is simply stating that 𝑆 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 for the same process.
Quasistatic adiabatic compression of a system (𝑑𝑄 = 0 at every stage) is isentropic. Entropy
Change in Free (Joule) Expansion of an Ideal Gas;
The entropy of an isolated system cannot decrease in spontaneous thermodynamic processes, i.e. releasing a constraint
on a system. In almost all cases, the entropy increases, but in some special cases it can remain the same.
1. Gas held inside a container of volume 𝑉0, pressure 𝑃0 and temperature 𝑇0, contained within a larger volume 𝑉1 that is
otherwise empty and thermally isolated from the environment.
, lOMoARcPSD|58847208
2. The container bursts and so 𝑉0 expands into 𝑉1 nonquasistatically.
3. Once the new equilibrium has been established, the final entropy is; and the initial
entropy was .
4. Since the gas does no work expanding into a vacuum, and no heat is supplied from the environment, there is no
change in the system energy. This implies that the final temperature is equal to the initial temperature. Entropy
change is therefore; . Note that the constant 𝑐̂ does not appear.
Clearly the system entropy has increased as a consequence of the expansion since 𝑉1 > 𝑉0. However there was no heat
transfer so we must conclude that the entropy doesn’t change solely due to a heat transfer. Previously but
we must modify this for a nonquasistatic process as there was entropy change but no heat transfer. We therefore write;
where 𝑇𝑟 is the temperature of the heat source (reservoir). 𝑑𝑆𝑖 can be considered to be the discrepancy
in the change in entropy not accounted for by . It is called the internal change in entropy. For a free expansion where
𝑑𝑄 = 0, only the 𝑑𝑆𝑖 remains so Δ𝑆 = Δ𝑆𝑖 such that .
Entropy change due to Nonquasistatic Heat Transfer;
Consider a process of heat exchange between a reservoir at temperature 𝑇𝑟 and a monatomic ideal classical gas initially
at temperature 𝑇0. N and V are fixed. No work is done on the system and the heat transfer is nonquasistatic. We assume
the system has acquired the temperature𝑇𝑟 at equilibrium according to the 0th law. Change in system entropy is;
. This can be positive or negative
because the system is not isolated, and it will depend on which temperature is greater. The internal entropy change is;
using . Note this is ALWAYS positive,
regardless of which temperature is larger. For this reason, internal entropy change is also known as entropy generation.
The change in entropy of the universe, here taken to be the combination of the ideal gas system and the reservoir, is
Δ𝑆𝑡𝑜𝑡 = Δ𝑆 + Δ𝑆𝑟. The entropy of the universe tends towards a maximum because of entropy generation in
nonquasistatic processes.
Cyclic Thermodynamic Processes, the Clausius Inequality and Carnot’s Theorem;
We consider taking the ideal gas around a cyclic heating and cooling process driven by a time dependent reservoir
temperature 𝑇𝑟(𝑡), or more properly a set of reservoirs with slightly different temperatures. We assume the initial and
final states are at equilibrium but the process need not be quasistatic. Cyclic means that the conditions at the start and
the end are the same. We integrate to get . For a cycle, 𝑆𝑓𝑖𝑛𝑎𝑙 = 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙 and
; this is called the Clausius inequality. For a quasistatic process, because Δ𝑆𝑖 = 0 and we
can replace 𝑇𝑟 with the system temperature T.
A famous example of a quasistatic cyclic process is the Carnot cycle.
1. Isothermal expansion in contact with a hot reservoir
2. Adiabatic expansion (isentropic)
3. Isothermal compression in contact with a cold reservoir
, lOMoARcPSD|58847208
4. Adiabatic, isentropic compression returning system to original state.
Since the process is quasistatic, there is no entropy generation during the process, therefore the change in system
entropy of the total cycle should be zero. We can show this explicitly by considering the changes in entropy during the
process, during the isothermal
stages;
2 2
3 3
, and since 𝑇ℎ 𝑉2 = 𝑇𝑐 𝑉3
and
, we can see that . Substituting this into the result above, we get 𝑁𝑘𝑙𝑛(1) = 0 = Δ𝑆.
The cycle is designed to convey heat from the hot to the cold reservoir, with the conversion of some of this flow into
mechanical work, obtained by employing the system expansions and compressions to move a load, e.g.
The hot reservoir passes heat Δ𝑄ℎ into the system from 𝑉1 > 𝑉2. Since Δ𝐸 = 0 for the system during this expansion, this
heat is converted into work; . Similarly the cold reservoir receives heat Δ𝑄𝑐 equal to the work
done on the system during the isothermal compression; . The work done on
the environment per cycle is; . We have an idealized device that converts heat
into work, a type of heat engine known as the Carnot engine. It can operate in both directions, by pumping heat from
the cold to the hot reservoir when mechanical work is supplied, e.g. in a fridge.
The sequence can be used to investigate the efficiency 𝜂𝐶 of a Carnot engine. This efficiency depends solely on the
temperatures of the reservoirs; Carnot’s theorem. . The efficiency is always less than 100% i.e. there
is always waste heat ΔQc transferred to the cold reservoir. The most efficient engines clearly need as large a difference
in temperature as possible to minimize .
Generality of the Clausius Expression for Entropy Change;
for an arbitrary substance undergoing a portion of a Carnot cycle. If any substance receives heat from a
reservoir nonquasistatically, it contributes to the change in its entropy through where we include the
entropy generation 𝑑𝑆𝑖.
Fundamental Relation of Thermodynamics;
We rewrite the 2nd Law, 𝑑𝐸 = 𝑑𝑄 + 𝑑𝑊, substituting to obtain 𝑑𝐸 = 𝑇𝑑𝑆 − 𝑝𝑑𝑉.
This law holds for all processes, despite being derived from a quasistatic one, because at equilibrium the values of state
variables (E, T, S, p and V) do not depend on the process used to reach equilibrium.
This relation is often written as .
We can write an increment in entropy more generally as . Comparing this to the
fundamental relation above, we see that and .
These results provide us with a means to define temperature and pressure if we have a function for S.
We must also consider the change in number of particles, 𝑁. We expect;
