Reversible isochoric process

A reversible isochoric process is a reversible thermodynamic process that occurs at constant volume. Consider an ideal gas in a piston-cylinder device immersed in a water bath. The piston is soldered to the cylinder walls and is immovable.

       

AD and BC in the above diagram are reversible isochoric processes. According to the ideal gas law, a system undergoing a reversible isochoric process from A to D requires the ratio of to be constant as the pressure of the system decreases. This is, in practice, carried out by continuously decreasing the temperature of the water bath by infinitesimal amounts. Since , work done for a reversible isochoric process is zero (see eq5). Many chemical reactions that take place in a bomb calorimeter are carried out under isochoric conditions.

 

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Reversible isobaric process

A reversible isobaric process is a reversible thermodynamic process that occurs at constant pressure.

       

Specifically, a reversible isobaric process that follows the path of either A to B or D to C is a reversible isobaric expansion, while one that goes from either B to A or C to D is a reversible isobaric compression. We shall use the same setup of a piston-cylinder device immersed in a water bath as the one described in the previous article to illustrate reversible isobaric processes.

With reference to the ideal gas law, an isobaric expansion of a gas requires the ratio of to be constant as the volume of the system increases. Theoretically, for a reversible isobaric expansion to occur, the system has to be in contact with an infinite sequence of temperature baths, with one bath having a temperature that is infinitesimally higher than the next one. In practice, to achieve a result that is close to reversibility, the temperature of the water bath is continuously increased by infinitesimal amounts to maintain a temperature gradient across the cylinder walls so as to supply the system with energy for the expansion at constant pressure.

Since the pressure of the system remains constant and equals to that of the external pressure throughout the process, eq5 becomes

For a reversible isobaric compression, the temperature of the water bath is continuously decreased by infinitesimal amounts to maintain a constant pressure in the system. Many chemical reactions that involve phase changes are conducted under isobaric conditions, e.g. .

 

Question

The change in work done is sometimes represented by . How does this relate to eq6 and eq7?

Answer

The expression or by convention is the general differential equation for work. If the change in pressure is infinitesimal, , then the expression reduces to . If the process is a reversible isothermal process, we substitute the ideal gas law in to give , which is the differential form of eq6. If the process is isobaric, , and again reduces to , which is the differential form of eq7.

 

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Reversible isothermal process

A reversible isothermal process is a reversible thermodynamic process that occurs at constant temperature.

       

A reversible isothermal expansion process for an ideal gas follows the path from A to C, while a reversible isothermal compression moves from C to A (see diagram above). The curve that describes an isothermal process is called an isotherm. We can express work done for a reversible isothermal process, which involves the change in pressure and volume of the system, by substituting the ideal gas law in eq5 to give:

To determine the work done on a system undergoing reversible isothermal expansion or compression, we immerse the piston-cylinder device in a constant temperature water bath. For a reversible isothermal compression, the external force balances the force exerted by the gas on the underside of the piston before the system is compressed. At the start of the compression, the external force is increased infinitesimally causing the piston to move down over an infinitesimal distance.

Consequently, infinitesimal energy is transferred via work to the system, leading to an infinitesimal increase in the kinetic energy of the gas and hence, an increase in the system’s temperature. The infinitesimal difference in temperature between the system and the water bath results in the flow of energy from the system through the thermally conducting cylinder walls to the temperature controlled water bath. Hence, constant temperature of the system is maintained throughout the process. This compression process is repeated stepwise with further infinitesimal increases in the external force until the desired final volume of the system is achieved. Work done is then calculated using eq6. If the external force is now decreased infinitesimally in a stepwise manner, we have a reversible isothermal expansion of the gas. Many chemical reactions involving the determination of rates of reactions are carried out isothermally.

 

Question

Does the temperature of the water bath increase if energy is continuously transferred from the system to the water bath in the compression process?

Answer

Theoretically, the constant temperature water bath is infinitely large. This means that transfer of energy from the system to the layer of water in the vicinity of the system (due to a temperature difference between the two) is the first of infinite steps of transfer, with the second step being the transfer of energy from the body of water in the vicinity of the system to a layer of water further away from the system and so on. In practice, the temperature of the water bath is controlled by a computer and adjusted throughout the process to maintain constant temperature for the system.

 

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Reversible work in thermodynamics

In classical mechanics, the amount of work done on an object, , is the product of the external force, , acting on the object and the distance moved in the direction of the force, :

Pressure-volume work, the most common form of work in chemical thermodynamics, uses the same concept and is defined as the transfer of energy between a system and its surroundings due to a force differential between the two. Another type of work usually encountered in chemical thermodynamics, which is discussed in the article on electrochemistry, is electrical work.

Consider a system containing an ideal gas in a frictionless, closed piston-cylinder with a weightless piston of area (see diagram above). Initially, the piston, which is part of the surroundings, is stationary and the gas molecules are evenly distributed in the system. This implies that the force exerted by the atmosphere on the piston balances the internal force on the piston.

When an additional external force is exerted on the piston, the gas is compressed. Substituting in eq1, the work done on the system is:

However, there are two ways to move the piston to compress the gas. If the piston moves rapidly over a relatively great distance that results in a large change in volume, , there may be a momentary uneven distribution of molecules in the system, where the density of molecules adjacent to the piston wall is greater than that further away from the piston. In this case, we cannot equate the external pressure on the piston with the internal pressure. Instead, if the piston moves very slowly over an infinitesimal distance that results in an infinitesimal change in volume, , such that the molecules in the system have enough time to equilibrate and redistribute themselves evenly throughout the infinitesimally smaller volume, the external pressure is equal to the internal pressure for the infinitesimal compression. As the internal pressure is the pressure of the gas :

Substituting eq3 in eq2,

The total work done for a series of infinitesimal compressions on the system from an initial volume to a final volume is:

As explained in the previous section, we call such a process of a series of infinitesimal changes that occur very slowly so that the system is always in equilibrium, a reversible process. According to IUPAC convention, the negative sign is added so that work done on the system by the surroundings is positive. In other words, if , , we have reversible compression with work done by the surroundings on the system, and if , , we have reversible expansion with work done by the system on the surroundings.

Reversible processes can occur under the following conditions:

  1. Isothermal (constant temperature)
  2. Isobaric (constant pressure)
  3. Isochoric (constant volume)
  4. Adiabatic

We shall discuss the four reversible processes in subsequent articles.

 

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Reversible and irreversible thermodynamic processes

A reversible process is an idealised thermodynamic process that occurs infinitesimally slow in a system to the extent that the thermodynamic properties of the system are well defined at all times throughout the process.

Consider a thermally-conducting, frictionless, piston-cylinder device immersed in a water bath whose temperature can be adjusted (see above diagram). The device is regarded as the system and contains a gas at an initial state of equilibrium A, which is characterised by thermodynamic properties like volume, pressure, temperature and entropy. The external pressure is set equal to the pressure of the gas , and the bath, which is defined as the surroundings, has the same temperature as the device.

The change of state A to another equilibrium state B can be achieved through different paths or processes (see diagram below, which depicts the paths linking the two states of the system).

For example, we can successively reduce  in infinitesimal steps at constant temperature such that each infinitesimal change is slow enough for the molecules to have sufficient time to equilibrate and spread evenly throughout the cylinder, resulting in uniform pressure and temperature throughout the system. This ensures that the thermodynamic properties of the system are well defined at every infinitesimal step and that we can determine and plot the change of values of these properties on a graph.

We can reverse the process from B to A by infinitesimally increasing , with the path taken being exactly the same as the forward process, except in the reverse direction. The infinitesimal increase or decrease in pressure implies that a reversible process requires infinite time to accomplish.

Another way to alter the state of the gas in the piston-cylinder device above is to instantaneously lower to the extent that the system only has enough time to equilibrate at the initial and final states but not at the intermediate states. We called such a path, an irreversible process.

As a reversible process is an idealised process, all real processes occurring within a finite timescale are considered irreversible. An irreversible path is shown as a dotted line on the graph, with only two points – the initial state and the final state – having defined properties and hence defined coordinates. The dotted line is drawn arbitrarily because there isn’t sufficient time for the system to equilibrate at the intermediate states, and hence no exact coordinates between A and B. Since the forward path of the irreversible process is poorly defined to begin with, we cannot explicitly describe the reverse path (other than another arbitrarily drawn dotted line) when we reverse the reaction irreversibly from B to A by instantaneously compressing the gas and transferring some heat from the system to its surroundings.

Although a reversible process is an idealised construct, it is used to derive most of the thermodynamic equations because equations require variables to be well defined to work.

 

Question

How do we apply thermodynamic equations that are derived from an idealised process to real processes?

Answer

A real process, when carried out slowly within a finite timeframe, approximates an ideal process. The graph below shows that the greater the number of well-defined intermediate points, the closer the real process is to an ideal one. Even if a real process cannot be carried out slowly, we can use thermodynamic equations to approximate results by making certain assumptions, e.g. uniform pressure and temperature of the system is maintained.

 


The concepts of reversibility and irreversibility are further elucidated in subsequent sections (see sections on entropy of an isolated system and Clausius inequality).

 

Question

Why is an irreversible process a spontaneous process?

Answer

This is due to the definition of a spontaneous process, which is defined as an irreversible process that occurs naturally under certain conditions in one direction, but requires continual external input in the reverse direction. See this article for details.

 

Lastly, reversible and irreversible processes can also be defined as follows:

A reversible process involves the passage of a system from its initial state to its final state and then back to the initial state without any change to either the system or the surroundings. However, the system and its surroundings cannot be simultaneously restored to their original states for an irreversible process.

In the above example involving the device in a water bath, both the system and the surroundings are precisely restored to their initial states for the reversible process (including no change in entropy of either the system and surroundings). As for the irreversible process, the system and its surroundings cannot be simultaneously restored to their initial states. If the system is restored to its initial state, a finite amount of heat must be transferred to the surroundings, leading to a positive change in entropy of the surroundings.

 

Question

What is the difference between a reversible reaction and a reversible process?

Answer

A reversible chemical reaction, e.g. , is where X forms Y, and at the same time, Y forms X in a particular thermodynamic state. An irreversible reaction, denoted by , is when the reverse reaction cannot proceed in any state under reasonable conditions. On the other hand, a reversible process describes a path linking two thermodynamic states of a system, where thermodynamic properties of the system are well defined at all times.

For example, the above diagram shows four different states of the reaction , each with a set of well defined p, V and T.  Since the reaction is reversible, X forms Y and Y forms X at any point on the graph. Let’s assume that the equilibrium amounts of X and Y at the four states are in the ratio of:

State

X:Y

A

4:1

B

3:2

C

1:1

D

1:4

Clearly, there are different paths linking any two states. For example, if we want to convert an initial equilibrium amount of X and Y in the ratio 4:1 to a final equilibrium amount in the ratio of 1:1, we can accomplish that by infinitesimally increasing the volume of the system from to , followed by infinitesimally reducing of the pressure of the system from to . Alternatively, we can proceed from A to D to C or directly from A to C.

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Thermodynamic property, state, equilibrium and process

A thermodynamic property is a measurable or computable characteristic of a system, such as pressure, volume and temperature. Two systems with all thermodynamic properties being quantitatively equivalent are said to be in the same thermodynamic state. The state of a system is therefore defined by the values of the system’s thermodynamic properties.

A system is in a state of thermodynamic equilibrium when there is no net transfer of energy or matter with its surroundings, resulting in its thermodynamic properties being constant over time.

We are mainly concerned with systems in thermodynamic equilibrium in our discussion of chemical thermodynamics. This is because the properties of the system are only well defined and no longer time-dependent when the system is at equilibrium. For example, when we subject a system to a change (e.g. by increasing the temperature of the system), we are interested in analysing the initial state of the system before the change when the system is in a state of thermodynamic equilibrium at an initial temperature, and the final state of the system a certain time after the change when all regions of the system have the same final temperature (i.e. the system is again in a state of thermodynamic equilibrium).

Finally, the path taken by the system from its initial state to its final state when a change is initiated is known as a thermodynamic process. A process can also be viewed as an integrated sequence of thermodynamic states at equilibrium, starting with an initial state and ending with a final state.

 

Question

Consider two systems that are initially at different temperatures. They are brought in contact via an immovable thermally conducting boundary. Do the two systems necessarily have the same state when they are in thermal equilibrium?

Answer

No. Even though the two systems have the same temperature over time, they may have different volumes and pressures. Hence, two systems at thermal equilibrium may not be in the same thermodynamic state.

 

 

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Systems, surroundings and boundaries

A system is the part of the universe being studied, while the surroundings are the remaining parts of the universe, with the two regions separated by a boundary. For example, if we are investigating the relationship between the volume and pressure of a gas using a J-shaped tube immersed in a constant-temperature bath, the system is the gas trapped in the left arm of the tube (see diagram below).

The surroundings include the entire length of mercury in the tube and the constant temperature bath. The boundary consists of the wall of the J-shaped tube and the mercury meniscus in the left arm of the tube.

We need not consider the air in the right arm of the tube, the wall of the container of the bath and the air in the atmosphere since the system is in thermal equilibrium with the mercury and the bath.

The example mentioned above describes a closed system, where energy, but not matter, can be transferred between the system and its surroundings. This implies that the boundary of a closed system is thermally conductive. An open system is one in which both energy and matter can be transferred between the system and its surroundings. The boundary in this case is either physically permeable or notional. If neither energy nor matter can be transferred between the system and its surroundings, the system is called an isolated system, where the boundary is both thermally non-conductive and impermeable—that is, an adiabatic boundary.

 

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Chemical thermodynamics (overview)

Chemical therrmodynamics is the branch of Chemistry that studies the interaction of chemical species with energy on a macroscopic level, with the objective of understanding and predicting chemical reactions.

Concepts such as system, surrounding, equilibrium, heat, work, entropy, enthalpy, state functions and path functions are elaborated in this topic along with the four laws of thermodynamics.

 

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Solution of the Hartree-Fock-Roothaan equations for helium

To solve the Hartree-Fock-Roothaan equations for He, we begin by noting that in the spin orbitals and . If the basis wavefunctions and are real, becomes . With reference to eq157, we have

where (assuming no contribution from the exchange integral) and .

The pair of simultaneous equations can be represented by the following matrix equation:

Let the basis wavefunctions be and . Since these Slater-type orbitals are real, . Eq158 becomes

 

Question

Show that .

Answer

Since and are real,


Using the Hermitian property of the KE operator,

 

Therefore, eq159 becomes

Eq160 is a linear homogeneous equation, which has non-trivial solutions if

Expanding the determinant and noting that , we get the characteristic equation:

The computation process is as follows:

  1. Evaluate the integrals , , and in eq161 either analytically or numerically by letting and , and using the initial guess values of .
  2. Substitute the evaluated integrals in eq161, solve for and retain the lower root. Substitute back in eq159 to obtain an expression of in terms of .
  3. Use the expression derived in part 2, together with , to solve for and , which are then used as improved values of and for the next iteration.
  4. Repeat steps 1 through 4 until , and are invariant up to 6 decimal places.

The results are as follows:

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Properties of the antisymmetriser

The properties of the antisymmetriser that are useful in deriving the Hartree-Fock equations are:

To proof eq63, we substitute eq62 in to give .

 

Question

Show that

Answer

When acts on , we have  permutated terms. then acts on each of these terms, resulting in sets of permutated terms, with the terms in one set being identical to the terms in another set. An illustration with is given below:

where the top row is the result of , and each column is the outcome of  acting on each term in the top row.

In other words,

 

Therefore, , which when substituted with eq62 gives eq63.

Eq64 states that  is Hermitian. To prove that , we note that



Finally, eq65 states that commutes with . This is because

 and correspond to the same state with energy because the way we label identical particles cannot affect the state of the system.

 

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