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Chemical potential

The chemical potential of a species is the change in Gibbs energy with respect to the change in the number of moles of the species, with temperature, pressure and the amount of other substances held constant. It is a measure of the potential of a species for change.

In the previous article, we derived eq143 and eq148 for a closed system that does only pV work. These equations are not applicable to open or closed systems with composition changes or open systems that interchange matter with their surroundings. We now derive equations that apply to such systems.

Consider a one-phase system containing a single chemical species (i.e. a pure substance) in a container that is in thermal equilibrium. The container may be open or closed. The Gibbs energy of the system is represented by the function:

The total differential of for this system is:

where is defined as the chemical potential of the pure substance.

Substituting the definition of molar Gibbs energy  in :

Since is an intensive quantity for a pure substance,

Hence, for a pure substance, its chemical potential is equal to its molar Gibbs energy and we can write eq149 as

Comparing eq150 with the total differential of for a closed system, is the chemical potential due to the exchange of material between the system and its surroundings for an open system.

Question

Does eq150 only apply to a single-component (pure substance), single-phase system?

Answer

Yes. If the pure substance undergoes molecular rearrangment (e.g. cis-trans isomerisation), dissociation (e.g. ) or melting, eq150 does not apply. Systems involving cis-trans isomerisation and dissociation reactions are considered multi-component, whereas a pure substance undergoing melting is treated as a single-component, two-phase system.

 

If the one-phase system contains different chemical species (e.g. cis-trans isomerisation or dissociation reaction), its Gibbs energy is represented by the function:

where  is the number of moles of the -th species in the system.

The corresponding total differential of for a multi-component, single-phase system is:


where the subscripts refers to maintaining the composition of all species constant, while means keeping the composition of all species constant except .

Substituting eq148a and eq148b into eq151 yields

where is the chemical potential of the -th species in the system.

We now have a Gibbs energy equation that is applicable to a ‘one-phase, multi-species’ system that is either open or closed. If the system is closed and involves a reaction,  in eq152 refers to composition changes within the system. For an open system,  may be attributed to either composition changes within the system or material exchange with the surroundings or both, depending on the specified conditions. Eq152 reduces to eq149 if there is only one component in the summation. Note that  in a ‘one-phase, multi-species’ system may not be equal to of the pure species .

Consider a one-phase closed system containing two species that react reversibly (note that reversibility here refers to the reaction having a forward component and backward component, and not in terms of a reversible thermodynamic process) as follows:

If the reaction is occurring at constant pressure and constant temperature, eq152 becomes

Suppose the forward reaction is spontaneous. From eq145,  and hence, , i.e.

Let  and ,

As , we have or

Similarly, in the event that the reverse reaction is spontaneous,  and ,  and , which gives

The spontaneity of a reaction due to the difference in the chemical potentials of the reaction species is analogous to the tendency of the gravitational potential energy of a body to change from a higher value to a lower value. In other words, the term ‘chemical potential’ relates to the potential energy of the system to effect a change.

With reference to eq154 and eq155, the remaining scenario is

When both substances have the same chemical potential, they have equal tendency to change. This means that the reaction is at equilibrium.

We can also express the change in chemical potential of the reaction in a closed system is a different way. Let’s rewrite the equation as and define

where is the stoichiometric number of the -th species in the reversible reaction and is the amount of substance that is being changed in the reaction, i.e. a proportionality constant that measures how much reaction has occurred. is dimensionless, while is called the extent of a reaction and has units of amount in moles. This proportionality constant is the same for all species. By convention, the stoichiometric number for reactants and products in a reversible reaction are negative and positive respectively with respect to the forward reaction.

From eq157,

Since the Gibbs energy of a reaction system decreases for a spontaneous reaction until equilibrium is achieved,

For example, for the reaction  with no initially, mole, with moles and mole, assuming all is consumed. If the reaction is at equilibrium when mole, we have mole and mole. At this point, there are 1 mole of and 0.5 mole of in the system. When the extent of the reaction reaches mole (point R), there are 0.8 mole of and 1.2 moles of in the system. Let’s now analyse the reverse reaction from point R to the equilibrium point. , with and . Therefore, in a reversible reaction is always positive for the forward reaction and negative for the reverse reaction.

Substituting eq157 in eq153, we have

For a spontaneous forward reaction,  and , which makes . When the reaction attains equilibrium, and hence . For a spontaneous reverse reaction, and , which makes . By denoting , we have a new indicator of the nature of a reversible chemical reaction, where

We call the indicator , the reaction Gibbs energy. A reaction where is known as an exergonic reaction (Greek for work producing), while a reaction where is an endergonic reaction (Greek for work consuming). The change in Gibbs energy with respect to the change in extent of a reaction for a reaction at constant and is shown in the diagram below.

We can also derive the first law of thermodynamics for a one-phase open system that involves pV work only, by substituting the definition of enthalpy in eq143 to give , whose differential form is:

Substituting eq152 in eq159

Similarly, the change in enthalpy for an open system is derived as follows:

Substitute eq159 and eq160 in the above equation,

Finally, the total differentials of for a single-component, multi-phase system (e.g. melting of ice), and a multi-component, multi-phase system (e.g. seawater in contact with air) are

and

respectively, where

Eq162 and eq163 can also be expressed, by combining with eq148, as:

and

respectively.

Consider a single component, two-phase system at constant pressure and temperature. At equilibrium, eq162 becomes . Since the net change of each component is zero at equilibrium, . This implies that , i.e. the chemical potentials a component in two coexisting phases are equal at equilibrium. It follows that the chemical potentials a component in three coexisting phases are also equal at equilibrium, because and , so and therefore . This has important applications in phase equilibria.

 

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Gibbs energy

Gibbs energy is a thermodynamic state function that is used to measure the spontaneity of a process at constant temperature and pressure.

As explained in an earlier article, the change in total entropy can help determine if a process in an isolated adiabatic system is spontaneous. However, most reactions proceed at constant pressure and temperature, which typically happens when the system is contained in a vessel with a movable piston and is in thermal contact with a temperature-controlled bath. In such cases, the system’s temperature is kept constant through the transfer of thermal energy from the bath, making the system no longer adiabatic.

Therefore, we need another state function to predict if such reactions are spontaneous or not. Consider a closed system that is in thermal equilibrium with a constant temperature bath. From eq139, . Let and substitute in the differential form of eq24 to give

If the system only does pV work,

Since and are constant and the change in enthalpy at constant pressure is defined as

Since , and are all state functions, we can define a new state function:

where is called the Gibbs energy.

Substituting eq143 in eq142,

Since we have derived eq144 using eq139, where the  equality identifies with irreversible processes and the relation, with reversible processes, the equality and relation in eq144 must also identify with the respective processes, where

 

Question

Why is an irreversible process a spontaneous process?

Answer

See this article for explanation.

 

As mentioned in the article on reversible and irreversible processes, a reversible process is an idealisation, which makes all real processes irreversible or spontaneous. Hence, the Gibbs energy of a system undergoing a real process at constant and decreases until such time when the process attains equilibrium.

Although all real processes are spontaneous, they may not occur within a finite timeframe. Whether a real process occurs is also dependent on chemical kinetics.

The integrated form of eq144 is:

where

 

Question

Is a state function?

Answer

Yes. , known as the Helmholtz energy, is a state function because , and are all state functions. The relation also indicates whether a process is spontaneous, but under conditions of constant temperature and volume.

 

A useful equation can be derived from eq143 as follows:

If the closed system only involves pV work, we can substitute eq121 in the above equation to give:

Eq148 becomes for a reversible process carried out at constant pressure and temperature, which is consistent with eq144. Eq148 also shows that the measure of the change in Gibbs energy of a process in general is not restricted to the process being carried out at constant pressure and constant temperature. However,  becomes a reflection of whether a process is reversible or irreversible if the process occurs at constant pressure and constant temperature.

Finally, Gibbs energy is previously called Gibbs free energy, where the term ‘free’ refers to energy that is “free to do non-pV work”. However in 1988, IUPAC dropped the term to avoid confusion.

 

Question

Show that

Answer

At constant , eq148 becomes . We can further impose the condition that the components of the system remain constant without any loss of generality, i.e.

Similarly, at constant , we have , and hence

 

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Clausius inequality

The Clausius inequality, mathematically expressed as ,  is an entropy change inequality applicable to all  heat transfer processes.

The motivation behind the derivation of the inequality can be perceived as the search for a less restrictive inequality given by eq130, which states that for any adiabatic process in an isolated system.

As we know, a system and its surroundings form the universe, with their entropies related by the following equations:

Since , and from eq133  for any process,

Let’s assume that the surroundings is an infinite heat reservoir with uniform heat capacity and temperature. In other words, equilibrium is always attained in the infinite heat reservoir and therefore, any process occuring in the surroundings is reversible. If so, we can write,

Since any transfer of energy to the surroundings  must come from the system, . So,

or

Eq139 is called the Clausius inequality. We have shown in eq119 that  for a reversible process, where . So the component of eq139 must be for a reversible process, which leaves the remaining component for an irreversible process:

We have therefore developed a change in entropy inequality (eq139) for a system that encompasses all processes.

 

Question

If entropy is a state function, why is there a difference in the change of entropy for reversible versus irreversible processes in eq139?

Answer

Entropy is a state function and the change in entropy of a system undergoing a reversible process from state A to B must be the same as that for an irreversible process between the same two states, i.e. .

With reference to eq132 and eq136, . Similarly, from eq131 and eq136, . Since , the change in entropy of the surroundings associated with a reversible process occurring in the system from A to B must be different from the change in entropy of the surroundings associated with an irreversible process occurring in the system between the same states. This is what the Clausius inequality is trying to convey.

 

 

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Entropy of the universe

The entropy of the universe tends to increase with time. To understand this statement, we need to study the change in entropy of adiabatic processes.

For a reversible adiabatic process in an isolated system, and from eq119, . Therefore,

For an irreversible adiabatic process AB with well-defined start and end states (see diagram above), we can set up two reversible paths, BC and CA, to create a cyclic process ABC, where BC is a reversible isochoric process and CA is a reversible adiabatic process. The change in entropy for the cycle is:

Since entropy is a state function, and because CA is a reversible adiabatic process, . So,

Since BC is a reversible process, we can substitute eq119 in  to give

From eq42, since BC is an isochoric process,

As  and , and therefore

Combining eq128 and eq129

If we regard our physical universe as an isolated system with any process occurring within it as adiabatic,

Since an irreversible process is a spontaneous process,

We can therefore also define reversibility and irreversibility (or spontaneity) in terms of entropy of the universe, where a process is reversible if there is no change in entropy of the universe, and where a process is irreversible (or spontaneous) if the change in entropy of the universe is positive. As mentioned in the article on reversible and irreversible processes, a reversible process is an idealisation, which makes all real processes irreversible (or spontaneous). Hence, the consequence of eq131 is that the entropy of the universe tends to increase with time. If there comes a time when no more irreversible process in the universe occurs, equilibrium is attained and . This hypothesis is known as the heat death of the universe.

 

Question

Show that for an isochoric process.

Answer

The change in internal energy for any process that involves only pV work is given by

For an isochoric process, , so or . Hence, .

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Spontaneous process (thermodynamics)

A Spontaneous process is an irreversible process that occurs naturally under certain conditions in one direction, but requires continual external input in the reverse direction. An example is the flow of heat from a hot body to a cold body.

Consider a closed system with 1 mole of a monoatomic gas at that is brought into contact with another compartment, which contains the same amount of the same gas at (see diagram above). The green boundary is thermally insulating, while the blue divider is thermally conducting. If , heat will flow from the system to the compartment on the right. Such a process is an irreversible process because the thermodynamic properties of the initial and final states of the system, but not the intermediate states, are well-defined. For example, the temperature of the system near the divider will be different from the temperature of the system further away from the divider when the system is first brought into contact with the right compartment. When both compartments attain thermal equilibrium, the system can only be reversed to its initial state via a heat pump (continual external input).

Question

Why does heat always flow from a high temperature body to a low temperature body?

Answer

The heat content of a substance is the total energy of motion of particles making up that substance, while the temperature of a substance is a measure of the average energy of particle motion in that substance. An easy way to understand the flow of heat is to study the motion of a monoatomic perfect gas, which consists entirely of kinetic energy. Consider two compartments that are separated by a thermally insulating valve (see diagram above). The left compartment contains 1 mole of a monoatomic gas at , while the right compartment houses the same amount of the same gas at , with  i.e. . If the separator is removed, atoms move randomly in an enlarged volume, with those on the left moving to the right faster than those on the right moving to the left. Over time, an even distribution of atoms is achieved in the container, where the average kinetic energies per unit volume sampled from different parts of the container () are the same. Since , heat must have flowed from the left to the right.

If we consider energy transfer during collisions, i.e. we no longer have a perfect gas, we end up with the same conclusion. In the case of elastic collisions, the conservation of momentum and kinetic energy lead to:

Combining the two equations and letting , we have and . This means that any two colliding gas atoms exchange their initial velocities, resulting in an equal distribution of atomic velocities and the same average velocity in any sampled volume.

 

Another example of a spontaneous process is the irreversible expansion of a system consisting of a gas in an isolated vertical cylinder (see diagram below).

If the force exerted by the gas on the bottom surface of the frictionless piston is greater than the weight of the piston, the gas expands and pushes the piston up when the catches are removed. The piston moves over a distance  until the force exerted by the expanded gas on the bottom surface of the piston equals to the weight of the piston. This process is an irreversible one since there is only sufficient time for the gas to equilibrate at the start and at the end of the process. To return the system to its initial state, we need to push the piston down over the same distance (continual external input).

 

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Entropy

Entropy is a state function , whose differential is expressed as .

With reference to eq98, for a reversible cycle. Dividing eq98 by gives:

Substituting the definition of from the first law of thermodynamics in eq116 yields:

We have shown in an earlier article that  is an exact differential for an ideal gas. Since the line integral of any exact differential involved in a cyclic process is zero,

The function appears to be the differential of a state function. If it is, we must prove that eq118 applies to any working substance, not just an ideal gas. The proof follows the same logic as the proof that eq98 is applicable to any working substance in a Carnot engine. This is accomplished by replacing  with  and repeating the steps from eq99 through eq105. We call this new state function, entropy, :

 

Question

Is eq119 applicable to irreversible processes? If not, is that in conflict with the fact that is a state function, in which its change is path-independent?

Answer

Eq119 is only applicable to reversible processes. It is often written as , or its integral form .  In other words, can only be calculated by integrating for a reversible process, even though is a state function. Furthermore, is only applicable to a reversible process in a closed system, as is poorly defined for an open system. However, we can still calculate for an irreversible process using eq119 if we can find a reversible process or a combination of reversible processes from point A to point B.

 

Note that in eq119 is the same for both the system and the reservoir (surroundings) for a reversible process. For a closed system undergoing reversible processes, we can substitute eq119 in the differential form of eq24 to give:

If only pV work is involved,

Similarly, we can compute for an irreversible process if we can find a reversible process or a combination of reversible processes from point A to point B. Eq121, which is the combination of the first and second laws of thermodynamics, is called the fundamental equation of thermodynamics.

 

Question

Under what circumstances is ?

Answer

From eq119, for a process occurring in the surroundings. If the surroundings is a constant pressure reservoir, , where is a constant. Since both and are state functions, we can express the integral form of as .

 

Question

Show that .

Answer

Rearranging eq121 gives:

Substituting the total differential into eq122 yields:

Comparing eq123 with the total differential results in:

Since is a state function, the order of differentiation does not matter:

Substituting eq124 into eq125 gives:

Carrying out the derivatives yields:

Multiplying eq126 throughout by and rearranging results in:

 

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Carnot’s theorem

Carnot’s theorem states that no engine operating between two heat reservoirs of differing constant temperatures can be more efficient than a Carnot engine operating between the same reservoirs.

We have mentioned in an earlier article that an irreversible heat engine (real heat engine) is less efficient than a Carnot engine due to the occurrence of dissipative processes during the cycle, e.g. friction, with some of the work converted to heat. Therefore, we just need to compare the efficiencies of reversible engines to prove Carnot’s theorem.

Consider a system of two reversible engines (see diagram above), where a reversible engine X is more efficient than the Carnot engine Y. This hypothetical greater efficiency for engine X can be brought about by a difference in design, a difference in working substance (e.g. different gas used) or both. From eq96 of a previous article,

In other words, for the same amount of thermal energy transferred to each engine (), the net work done by engine X is greater than that of engine Y, .

Suppose we couple the engines, with engine Y running in reverse as a Carnot heat pump (see diagram above), where some amount of work from engine X () is used to run engine Y. Since , the amount of heat transferred to engine Y from the cold reservoir is greater than the amount of heat deposited by engine X into the cold reservoir, i.e.

Engine Y is then adjusted to run at a rate such that it extracts heat from the cold reservoir at the same rate as engine X deposits heat to the cold reservoir (see diagram below).

For example, if engine Y extracts 1.5 times as much thermal energy per cycle from the cold reservoir as engine X deposits to the cold reservoir, engine Y is adjusted to run at two-third of its normal rate so that:

where  is the factor of adjustment to engine Y.

Substitute eq107 in eq106, and noting that ,

From eq108, the work done by engine X is greater than the work required to run engine Y at the adjusted rate. If an equivalent work of  from engine X is used to run engine Y, a net work of is produced by the system. The amount of heat deposited in the hot reservoir by engine Y is , which after substituting eq107 in it, becomes , resulting in a net energy change of the hot reservoir of . This means that the net energy extracted from the hot reservoir is completely used to produce a net work of with zero net energy deposited at the cold reservoir, which violates the Kelvin-Planck statement of the second law of thermodynamics.

The assumption that the reversible engine X is more efficient than the Carnot engine Y is therefore false. Any reversible engine operating between two constant temperature reservoirs is either of equal efficiency or lower efficiency than a Carnot engine operating between the same reservoirs, i.e.

This dispels the notion that a reversible heat engine can be more efficient than a Carnot engine by a difference in design, a difference in working substance or both.

Next, we need to prove that any reversible engine has the same efficiency as a Carnot engine. We now run engine X in reverse and the Carnot engine Y in forward mode (see diagram above). If the Carnot engine Y is more efficient than the reversible engine X, we again violate the Kelvin-Planck statement. This means that

Eq109 and eq110 can only be reconciled if . Therefore, a Carnot engine has the same efficiency as any reversible heat engine working between the same constant temperature reservoirs.

Combining the two proofs, a Carnot engine, like any reversible heat engine operating between the same constant temperature reservoirs, is the most efficient heat engine.

For a Carnot cycle with an ideal gas as the working substance (see diagram above), the work done during each of the processes is:

1) reversible isothermal expansion:
2) reversible adiabatic expansion:
3) reversible isothermal compression:
4) reversible adiabatic compression:

Even though the above work done equations pertain to an ideal gas, they are applicable to the calculation of the efficiencies of reversible engines with all types of working substances, as we have proven that the efficiency of a Carnot engine is independent of its working substance.

Net work done by the cycle is:

We have shown in an earlier article that . So,

According to the definition of a Carnot heat engine, is the transfer of heat from the constant temperature hot reservoir during step 1 of the Carnot cycle. This process is a reversible isothermal expansion with no change in internal energy of the system. So, from the first law of thermodynamics,  and

Substituting eq112 and eq113 in eq96,

Combining eq114 and eq97,

 

Question

What is the significance of eq114?

Answer

Prior to the derivation of eq114, temperatures are commonly measured on a mercury thermometer scale or a gas thermometer scale. In developing the mercury thermometer scale, the volume of mercury in a narrow column of a mercury thermometer is calibrated at the freezing point of water and the boiling point of water. The scale is then obtained by dividing the length of the mercury column into regular intervals between the two points. A similar procedure is employed to calibrate the volume of gas in a gas thermometer to form the gas thermometer scale. In each case, the calibrated temperature scale is dependent on the thermal expansion property of the working substance, which may not be completely linear. With eq114, a new temperature scale called the thermodynamic temperature scale is defined, where

Theoretically, the temperature of an object (cold reservoir) can be measured against a source (hot reservoir) at the triple point of water, 273.16 K by finding the efficiency of the engine used, which is independent of the property of the working substance as explained in the description of the Carnot engine. The triple point of water is chosen as a source temperature because it is a stable state that is easily reproduced. By making a temperature interval equal to a Kelvin, we have the thermodynamic temperature scale, which is equivalent to the absolute temperature scale of the ideal gas law. The equation above also suggests that the absolute zero of a system cannot be achieved. Otherwise, , which violates the Kelvin-Planck statement of the second law of thermodynamics.

In practice, the thermodynamic temperature scale is based on the interpolation of a few easily reproduced temperature points, e.g. triple point of water (273.16 K), triple point of hydrogen (13.8033 K), triple point of oxygen (54.3584 K), etc., all of which can be accurately measured by sensitive instruments.

 

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Carnot heat pump

A Carnot heat pump is a reverse Carnot engine. It is a theoretical construct that transfers heat from a cold reservoir to a hot reservoir, as a result of net work done on the system.

               

Consider a gas in a piston-cylinder device (see diagram above) that repeatedly undergoes a cycle that consists of four reversible processes as follows:

  1. (D to C) The pressure and temperature of the gas at D is equal to and respectively. At this point, the device is in thermal contact with a cold reservoir at a constant temperature . Heat is transferred from the cold reservoir to the gas with  being reduced infinitesimally, causing the gas to expand and push the piston up to do work on the surroundings. This process is a reversible isothermal expansion of the gas.
  1. (C to B) The contact with the cold reservoir is removed and the device is thermally insulated. Work is done on the system, with increasing infinitesimally. The gas is consequently compressed and its internal energy increases, leading to a rise in temperature to . This process is a reversible adiabatic compression of the gas.
  1. (B to A) The device is placed in contact with a hot reservoir at a constant temperature and the insulation is removed. Work done on the gas continues, with  increasing infinitesimally, resulting in the transfer of heat from the gas to the reservoir. This process is a reversible isothermal compression of the gas.
  1. (A to D) The contact with the hot reservoir is removed and the device is thermally insulated again. is now decreased infinitesimally, resulting in a decrease in the internal energy of the gas, with its temperature dropping to . This process is a reversible adiabatic expansion of the gas.

It is important to note that steps 2 and 3 of the cycle are accomplished with energy from the surroundings (external source) in the form of work. The Carnot heat pump, together with the Carnot heat engine, is used to explain Carnot’s theorem.

 

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Carnot cycle

The Carnot cycle, developed by the French scientist Sadi Carnot in 1824, is a theoretical construct that sets the maximum efficiency of a thermodynamic engine.

               

Consider a gas in a piston-cylinder device (see diagram above) that repeatedly undergoes a cycle consisting of four reversible processes as follows:

  1. (A to B) The pressure and temperature of the gas at A is equal to and respectively. At this point, the device is in thermal contact with a hot reservoir at a constant temperature . Heat is transferred from the hot reservoir to the gas with being reduced infinitesimally, causing the gas to expand and push the piston up to do work on the surroundings. This process is a reversible isothermal expansion of the gas.
  1. (B to C) The contact with the hot reservoir is removed and the device is thermally insulated. is again allowed to decrease infinitesimally and the gas continues to expand to do work on the surroundings with its internal energy, resulting in a drop in temperature to . This process is a reversible adiabatic expansion of the gas.
  1. (C to D) The device is placed in contact with a cold reservoir at a constant temperature and the insulation is removed. Work is now done on the gas with increased infinitesimally and heat is transferred from the gas to the reservoir. This process is a reversible isothermal compression of the gas.
  1. (D to A) The contact with the cold reservoir is removed and the device is thermally insulated again. Work done on the gas continues with increasing infinitesimally, resulting in an increase in its internal energy and a hence an increase in temperature back to . This process is a reversible adiabatic compression of the gas.

The theoretical piston-cylinder device that operates via the Carnot cycle is called a Carnot heat engine, i.e. one that operates by transferring energy from a body at a higher constant temperature to a body at a lower constant temperature and, through a cyclic process, converting some of that energy to mechanical work (see diagram below).

By the law on conservation of energy:

where  is the transfer of heat into the system from the hot reservoir (), is the net work produced by the cycle () and  is the transfer of thermal energy out of the system into the cold reservoir ().

The efficiency, , of a heat engine is defined as the fraction of heat transferred to the system from the hot body that is converted to work:

The modulus sign ensures that the value of is positive. Substituting eq95 in eq96,

The greater the work produced by the engine, the smaller the value of and hence the closer the efficiency of the engine to one.

For a reversible heat engine to function, it can operate via different processes (e.g. reversible isochoric processes in place of reversible adiabatic processes). However, regardless of the processes involved, it must undergo a cycle where thermal energy is transferred from a hot reservoir to a cold reservoir with some of the energy converted to work, which is how a heat engine is defined. Therefore, it is impossible to construct an engine that undergoes a cyclic process where thermal energy extracted from the hot reservoir is completely converted to work without any energy deposited at the cold reservoir (see diagram above). This is the Kelvin-Planck statement of the second law of thermodynamics.

 

Question

Can an engine cycle consist of just the first and third processes i.e. a reversible isothermal expansion followed by a reversible isothermal compression?

Answer

If so, the net work done by the cycle is zero (net area under the PV curve is zero), which is not much of an engine.

 

An irreversible heat engine (real heat engine) is less efficient than a Carnot engine due to the occurrence of dissipative processes during the cycle, e.g. friction. Some of the work is converted to heat, which is transferred to the cold reservoir as (see diagram above). So,

and according to eq96,

Even though we have proven that an irreversible heat engine is less efficient than a Carnot engine, we cannot conclude that the Carnot cycle sets the maximum efficiency of a thermodynamic engine. To do so, we need to understand the Carnot heat pump and Carnot’s theorem.

Question

For an ideal gas,

Show that eq98 is applicable to any reversible cycle for a system containing any substance.

Answer

Let’s assume that Carnot cycles of different working substances may have different shapes. Therefore, we need to prove that applies to any cycle.

Consider a cyclic process AEFGBCDA (see diagram above). for AEFGBCDA is:


This cycle is actually made up of two Carnot cycles ABCDA and FGBEF with for ABCDA given by

and  for FGBEF being:

Since  and ,

Substitute eq102 in eq100

Adding eq101 and eq103 and comparing with eq99,

From eq98, and . Therefore, . This means that eq98 holds true for a cycle that is composed of two different sized Carnot cycles. The same logic can be applied to cycles that are made up of more than two Carnot cycles.

Let’s now look at an arbitrary reversible cycle that can represent a system containing any substance (see diagram above). The arbitrary cycle is exactly composed of an infinite number of Carnot cycles as the sizes of the Carnot cycles approach zero. Hence, for any arbitrary reversible cycle,

Therefore, eq98 applies to any reversible cycle for a system containing any substance.

 

 

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Second law of thermodynamics (overview)

The second law of thermodynamics states that the entropy of an isolated system increases during a spontaneous process. Another commonly used definition is the Kelvin-Planck statement, which we shall introduce in the next article.

The first law of thermodynamics states that the total energy of a system and its surroundings is constant. The state functions and , though very insightful and useful, are not able to indicate whether a process can proceed spontaneously. To determine the spontaneity of a process, scientists sought another state function. Rudolf Clausius, a German scientist, analysed the Carnot cycle, which had been developed earlier by Sadi Carnot, a French engineer, and eventually conceptualised a new thermodynamic state function , called entropy, which predicts the spontaneity of certain processes.

 

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