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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 ,

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

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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Polytropic processes

A polytropic process is described by the formula , where . The table below shows the association of different pV curves for ideal gases with the corresponding thermodynamic processes:

Question

Why is the projection an adiabat when and why is a constant when ?

Answer

From eq44 and eq46, and for an ideal gas. For adiabatic processes, and hence, . Furthermore, . Since , . So,


which rearranges to , where .

For the second why, . If , .

 

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Line integral

A line integral is the integral of a multi-variable function along a curve. Line integrals can be categorised into scalar line integrals and vector line integrals. They are usually evaluated using parametric equations, which are then reduced to one or more scalar integrals.

Scalar line integrals

Let’s begin with scalar line integrals. A scalar function of two variables is represented geometrically by a surface on a three-dimensional graph (see diagram above). Consider two points, A and B, on the surface. There are infinite ways to move along the surface between these two points. For simplicity, we have indicated two of these infinite paths using the red and pink curves, which can be projected onto the xy plane for better visualisation (see diagram below).

Due to the many paths available between the two points, the integral of between the points is not well defined. In other words, the integral is path-dependent. For a particular path, the integral of along this path, which is specified by one of the curves on the plane, is called a line integral.

Geometrically, the result is the area of the ‘curtain’ extending from that path, between and on the xy plane, to the curve on the surface (see diagram above), i.e.:

where denotes ‘curve’, which is the specified path linking the two points on the xy plane, and is the infinitesimal change in arc length along the curve.

Since the specific path is also a function of and , we convert and into parametric equations consisting of a single parameter and carry out the integration analytically (see 3rd Q&A below for an example).

 

Question

What if the two points, A and B, are the same?

Answer

In the scenario where the two points are the same, the infinite paths from to  are infinite loops. Similarly, integrating the function via different loops may give different results. Each line integral in this case is represented by:

where the circle on the integral sign denotes a loop or a cyclic process.

 

Vector line integrals

Let’s now look at vector line integrals, an example of which is the work done by a vector field (see diagram above). For instance, the work done by a variable force on a charged particle moving along some path in an electric field is

As mentioned in the opening paragraph, such a line integral is evaluated by converting it into one or more scalar integrals. We write and in terms of their two-dimensional Cartesian components:

which can then be evaluated when is specified.

 

Question

What is the difference between a scalar field and a vector field?

Answer

A scalar field, expressed by a scalar function , associates a scalar with each point in some region of space, while a vector field, expressed by a vector function  associates a vector with each point.

 

Line integral with respect to coordinates

Line integrals can also be carried out with respect to one of the function variables instead of with respect to the arc length, e.g. the scalar integral . The geometric interpretation of is the projection of on the plane (see above diagram). An example of such a line integral is the work done on an ideal gas in a reversible process:

If the path of the above integral is defined by introducing a constraint, e.g. when , where is a constant, we get a plane that intersects with the surface at a particular contour (see diagram below). A single curve, an isothermal curve, is then projected on the plane.

Consequently, the integral reduces to one involving a single variable:

Similarly, if the intersecting plane is or , the projection on the plane is a horizontal line (isobaric process) or a vertical line (isochoric process) respectively.

 

Line integral of differential forms

Line integrals are sometimes written in differential form. Consider the work done by a vector field on a particle moving along some path . We can write  and in terms of their components: and . So, eq92 becomes

Compared to the general differential equation of two independent variables of the form , the RHS of the third equality of the above equation is the line integral of a differential equation, which in this case is an inexact differential. In other words, .

In the case of an exact differential , its line integral is equal to the difference in values of the function at the final point and at the starting point:

This is known as the fundamental theorem of line integral. We call such a function, whose output is independent of the path taken to reach it, a state function.

 

Question

How do we proof the fundamental theorem of line integral?

Answer

Consider a function  with an exact differential of the form or

where  and .

To carry out the line integral between two points and , we convert the function into its parametric form, which is to let and  be some function of , i.e. . Correspondingly, we have and hence . So,

Using the definition of the chain rule for a multivariable function (see eq14b),

 

Finally, we have shown in a previous article that the change in internal energy of a system is path-independent for a system containing a perfect gas. We will show in the article on entropy that  is path-independent for any system. This makes an exact differential and , a state function. Thus, we can write

If the two points are the same, ,

Therefore, the change in internal energy in a cyclic process is zero: . In general, the line integral of any exact differential involved in a cyclic process is zero.

 

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Applications of the Joule-Thomson effect

Two significant applications of the Joule-Thomson effect involve the Hampson-Linde cycle and the vapour-compression cycle.

The Hampson-Linde cycle, also known as the ‘Linde refrigerator’ (see diagram below), employs the Joule-Thomson effect to liquefy gases.

The cycle is as follows:

    1. The gas is compressed to allow it to circulate throughout the system. This raises the temperature of the gas because the kinetic energy of the gas molecules increases as they collide with the moving wall.
    2. The gas cools as it passes through the heat exchanger and cools further when it expands through the Joule-Thomson orifice. The system is designed with specific pressures before and after the orifice to ensure the gas cools upon expansion, as described by the Joule-Thomson effect.
    3. The low pressure gas is routed to the heat exchanger, cooling the warmer incoming gas, before returning to the compression chamber.
    4. Warm replenishment gas for replacing any liquefied gas is mixed with the returning gas. The net temperature of the gas after mixing is lower than its temperature in step1. Steps 1 to 4 is repeated until the air is cool enough to condense into the compartment below the orifice.

The vapour-compression cycle is used in household refrigerators and air conditioning units.

The cycle is as follows:

    1. Compression: The refrigerant gas is compressed by the compressor, allowing it to circulate throughout the system. This is not an isenthalpic process because work is done on the gas, which increases both the pressure and temperature of the refrigerant.
    2. Condensation: The hot, high-pressure refrigerant gas then passes through the condenser coils, usually located on the back or bottom of the appliance. As it cools, it condenses into a high-pressure liquid.
    3. Expansion: The flow of the high-pressure liquid refrigerant through an expansion valve, known as the Joule-Thomson orifice, can be modelled using the Joule-Thomson experiment. This causes the refrigerant’s pressure to drop, resulting in a mixture of liquid and vapour. The refrigerant is carefully selected based on its inversion temperature to ensure it cools upon expansion, as described by the Joule-Thomson effect.
    4. Evaporation: The low-pressure refrigerant then enters the evaporator coils (inside the refrigerator or air handler of an air conditioning unit). Here, it absorbs heat from the surrounding environment, which allows it to fully evaporate into a gas. This process cools down the air inside the refrigerator or the room.
    5. Cycle repeats: The refrigerant gas is then drawn back into the compressor to repeat the cycle. The desired temperature of the room or the interior of the refrigerator is monitored by a thermostat, which then signals the appropriate compression pressure to apply for subsequent cycles.

The desired temperature of the room or the interior of the refrigerator is monitored by a thermostat, which then signals the appropriate compression pressure to apply for subsequent cycles.

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The Joule-Thomson experiment

The Joule-Thomson experiment, an improved version of the Joule experiment, was conducted by James Joule and William Thomson in 1852 to study the thermodynamic properties of a gas expanding into a vacuum.

The experiment involves a system that is adiabatically insulated from its surroundings (see diagram above). The system consists of a double-piston cylinder that is separated into two compartments by a rigid porous plug. Constant pressures and , where , are applied to the left piston and the right piston respectively. As a result, the gas in the left compartment is slowly throttled irreversibly to the right compartment. Assuming that pressure is well defined in each compartment,  and . The total work done on the system is

According to the first law of thermodynamics for the entire system,

So,  or . If there is a change in temperature of the system, the experiment measures the change in gas temperature with the change in gas pressure at constant enthalpy, i.e. , which is defined as the Joule-Thomson coefficient :

The experiment is conducted multiple times by lowering  for a constant  and . A temperature-pressure plot of the results gives an isenthalpic curve (see diagram above). The gradient at a particular point of the curve is the Joule-Thomson coefficient with respect to and at that point. Since the change in pressure of the system is always negative, points on the curve that are associated with negative gradients correspond to the gas warming on expansion. Conversely, points on the curve that are associated with positive gradients correspond to the gas cooling on expansion. The point at which the gradient is zero is called the inversion point. Repeating the experiment by holding at new constant values and varying , we have multiple isenthapic curves:

The line connecting all inversion points is the inversion curve. The area on the left of the inversion curve is where cooling occurs for a gas when it expands into a vacuum, while the area on the right is where heating occurs. The temperature corresponding to the inversion curve at a given pressure is the inversion temperature of the gas at that pressure. If a vertical line is extended across the inversion curve at a particular pressure, it will intersect the curve at two points: an upper inversion temperature and a lower inversion temperature. Cooling upon expansion usually occurs when the temperature of the gas is between its upper and lower inversion temperatures prior to expansion. In other words, whether the expansion results in heating or cooling of the gas depends on the initial conditions and the properties of the gas. This phenomenon, known as the Joule-Thomson effect, is exploited in refrigeration processes.

Modern methods of measuring involve evaluating , which is known as the isothermal Joule-Thomson coefficient. To determine the relation between and , we refer to eq16, where . Using eq15, . Substituting eq46 and eq90 in this expression,

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The Joule experiment

The Joule experiment was an attempt by James Joule in 1843 to determine for a gas expanding into a vacuum.

The experiment involves a system that consists of a gas-filled compartment (A) and an evacuated compartment (B), both of which are immersed in a water bath (see diagram above). When the valve between the compartments is opened, the gas expands irreversibly from A to B, doing no work in the process. Assuming that no heat transfer occurs between the system and its surroundings, the first law of thermodynamics states that . If there is a change in temperature of the water bath, the experiment measures the change in gas temperature with the change in gas volume at constant internal energy, i.e. , which is defined as the Joule coefficient :

To determine , we refer to eq16, where . Using eq15, . Substituting eq44 and eq88 in this expression,

For an ideal gas,  is independent of the volume of the gas, i.e. , because there is no intermolecular forces between the molecules (see eq43a). We would therefore expect a non-zero  for a real gas. However, Joule’s experiment setup was too crude to measure a non-zero . An improved version of Joule’s experiment is the Joule-Thomson experiment.

 

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