Reversible adiabatic processes

A reversible adiabatic process is a reversible thermodynamic process in which no heat or mass transfer occurs.

A reversible adiabatic expansion follows the path from B to C, while a reversible adiabatic compression follows the path from D to A. The curves describing the paths of adiabatic processes are called adiabats (see diagram above, where we have included two isotherms, AB and CD, for comparison).

A reversible adiabatic compression is carried out in a frictionless, insulated piston-cylinder device containing a gas (see diagram above). The gas undergoes infinitesimal steps of compression, resulting in an increase in the internal energy of the system and therefore the temperature of the system. Since no heat enters or leaves the system, and according to the first law of thermodynamics,

For an ideal gas, we have shown in a previous article that  and hence, work done on the ideal gas system for a reversible adiabatic compression is:

Substituting in eq84 and integrating throughout gives:

where we have assumed that the heat capacity of the ideal gas is independent of temperature over the temperature range of interest and therefore a constant.

Solving the above equation yields:

where .

Similarly, for an adiabatic expansion from B to C, we have

Equating eq85 and eq86,

Question

Show that internal energy of a system  containing a perfect gas is a state function using the diagram above where

AC: reversible isothermal process
AD: reversible adiabatic process
AB: reversible isobaric process
BC & CD: reversible isochoric processes

Answer

Consider the change in internal energy of a system containing one mole of ideal gas from point A to point C. The change can be brought about by three different paths, with the first being a reversible isothermal expansion (AC), where .

The second path involves a reversible adiabatic expansion (AD) followed by a reversible isochoric heating (DC). The changes in internal energy for paths AD and DC are:

The change in internal energy for path ADC is:

Since

The third path consists of a reversible isobaric expansion (AB) followed by a reversible isochoric cooling (BC). The change in internal energy for path AB is:

From a previous article, , and so

Substituting  and  in the above equation,

The change in internal energy for path BC is:

The change in internal energy for path ABC is:

Regardless of the path taken, the change in internal energy from point A to point C is the same. Therefore,  is a state function.

 

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Using Kirchhoff’s law to determine the standard enthalpy change of ionisation and electron gain

Ionisation energy of a species is defined as the energy for removing an electron from the ground state of the species. According to statistical thermodynamics, the ground state of a species is the electronic configuration of that species at absolute zero. If ionisation energy is defined at any other temperatures above absolute zero, there will be a range of ionisation energies for a particular species, as the electron then is removed from various excited states. Hence, ionisation energy is equal to the difference between the sum of enthalpies of formation of the ionised species and the electron at absolute zero, and the enthalpy of formation of the species at absolute zero.


Ionisation energies are presented in data tables using eq73 of the previous article, where  is the experimental temperature and is absolute zero.

Question

Calculate the first ionisation enthalpy of magnesium at 298.15K, given IE1 = 738 kJmol-1 and assuming the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of .

Answer



Using eq73,

 

With reference to the above example where the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of ,

where  is at absolute zero.

Similarly,

Therefore, the ionisation enthalpy of  is

Electron affinity is also a zero Kelvin process, as it is defined as the energy for the addition of an electron to a species in its ground state. Although an exothermic process, electron affinity (EA) is quoted in data tables as positive values, i.e.

Since the standard enthalpy change of electron gain of a species is the negative of the standard enthalpy change of ionisation of that species with an additional electron attached , i.e.

the enthalpy change of electron gain, assuming the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of , is determined by substituting eq83 in eq80 and eq81:

with the electron gain enthalpy of being:

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Using Kirchhoff’s law to determine the standard enthalpy change of reaction at 298K

The experimental result of the standard change in enthalpy for the combustion reaction is 281.93 kJmol-1 at 1200K. Find its standard change in enthalpy at 298.15K given:

According to Hess’ law,

At 1200K,

At 298.15K,

Subtracting eq71 from eq70,

Since all species in the reaction are in the gaseous state from 298.15K to 1200K, we can substitute eq66 in eq72,

Substituting eq68 in the 1st integral on the RHS of eq73,

Substituting the relevant values from the table above in eq73 and evaluating the integral (using the form expressed by eq74 for each integral in eq73),

and

Combining eq66 and eq69, for a reaction

 

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Kirchhoff’s law (thermodynamics)

Kirchhoff’s law describes the change in enthalpy of a reaction with respect to the change in temperature.

From eq37, the heat capacity at constant pressure is defined as the change of enthalpy with respect to the change in temperature at constant pressure, i.e. . In other words, is the gradient of the curve of enthalpy versus temperature at constant pressure.

for a perfect gas is independent of temperature and we get a straight line with a constant gradient when the enthalpy of a perfect gas at constant pressure is plotted against temperature,. However, for an ideal or real gas,  varies with temperature and eq37 becomes:

Integrating both sides of eq64 with respect to temperature,

Since is a state function, the line integral of eq65 gives

or, for the change in enthalpy of a reaction,

Eq66 (or eq66a) is known as Kirchhoff’s law, which is used to calculate the change in enthalpy of a substance (or a reaction) from to .

The computation of eq65 to eq66 (or eq66a) is only valid if the function  is continuous and differentiable within the limits of to . If eq66a has a temperature range that includes phase transitions, which result in points of discontinuity in (see diagram above), it has to be modified as follows:

To find an expression for , we use a power series to generate curves that fit experimental values of for different substances on versus temperature plots:

The set of constants , , , , etc. are specific to each chemical species and are the outcome of the polynomial regression. As the contribution of higher powers of to  is small, the expression is fairly well represented by:

 

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Equipartition theorem: Vibrational motion

The vibrational energy of a diatomic molecule consists of a kinetic energy component and a potential energy component:

In classical mechanics, the potential energy  is equal to the work done against a force in moving a body from the reference point, where , to the displacement . In other words,

According to Hooke’s law, the vibrating diatomic molecule experiences a restoring force that is proportional to the displacement between the two atoms relative to its equilibrium or undistorted length:

where is the force constant.

Substituting eq55 in eq54 and integrating yields

Substituting eq56 in eq53 gives

From the theory of conservation of energy, the average kinetic energy of the molecule undergoing simple harmonic vibrational motion equals to the average potential energy of that molecule (see diagram above), i.e. . Hence, for a system of molecules, the average vibration energy of a molecule is:

According to the equipartition theorem mentioned in an earlier article, each kinetic energy component of a molecule has an average energy of . Hence, a diatomic molecule has an average vibrational energy of (vibration is along one axis).

For a molecule with atoms, we can derive its total component of motion as follows:

The motion of an atom is described by three Cartesian coordinates: , and . The components of the motion of independently moving atoms are therefore . However, if these free atoms are bound together to form a molecule, the translational motion of the molecule has three degrees of freedom regardless of its structure (see this article). The molecule’s rotational motion has two degrees of freedom if it is linear and three degrees of freedom if it is non-linear. The remaining degrees of freedom are attributed to the vibrational motion of the -atom molecule. Therefore, we have:

 

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Equipartition theorem: Rotational motion

The equipartition theorem, when applied to rotational motion, allows us to predict the distribution of energy among rotational states in gases and other materials. To understand how this works, we begin by deriving the classical expression for the energy of a body rotating about an axis (see diagram below).

The body has an angular velocity  of

where  is the period, i.e. the time taken for the body rotate an angle of  radians about the -axis.

Multiplying eq47 on both sides by the radius , we have:

where is the tangential velocity of the body with respect to the -axis.

The energy of the rotating body is purely kinetic:

Substituting the definition of rotational inertia or moment of inertia about the -axis, , in eq48 gives:

In three dimensions, the total rotational energy of the body is:

If the body is a molecule, its moment of inertia about an axis is the sum of the moments of the inertia of atoms making up the molecule about that axis:

where  is the perpendicular distance from the -th atom to the rotational axis, which passes through the centre of mass of the molecule.

For a non-linear molecule (see above diagram), its total rotational energy is:

In other words, there are three degrees of freedom of rotational motion for a non-linear molecule associated with three rotational energy components. For a linear molecule, its moment of inertia about one of the axes (arbitrarily taken as the -axis) is zero because . Hence, there are only two degrees of freedom associated with two energy components for a linear molecule:

According to the equipartition theorem mentioned in the previous article, each kinetic energy component of a molecule has an average energy of . Hence, for a system of molecules, the average rotation energy of a linear molecule and the average rotation energy of a non-linear molecule are  and respectively.

 

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Equipartition theorem: Internal energy and heat capacity

The internal energy of a gas is the sum of the gas molecules’ translational energy , rotational energy , vibrational energy , electronic transition energy , intermolecular forces of interaction , and rest-mass energy of electrons and nuclei :

is constant for any gas, while is constant at if no chemical reaction takes place. For an ideal gas, is zero, leaving the internal energy of an ideal gas as:

, and are given by , and , which are functions of temperature only (see table in previous article). Furthermore, vibrational modes are active only at relatively high temperatures. If temperatures are not too high (e.g., at room temperature), and the internal energy of a gas is

When ,

Therefore,

Multiplying eq58 throughout by ,

Although the equations are derived based on gases, they are applicable to any fluid system. We cannot determine precisely and are not able to calculate the absolute internal energy of a fluid. However, we can compute the change in the internal energy of a fluid from one state to another, as a change in internal energy cancels out the value of . For example, the change in internal energy for a mole of from 200K to 300K at constant volume is:

The heat capacity of a gas at constant volume is defined as . Hence, we can obtain the heat capacity of an ideal gas where vibrational modes are inactive by differentiating eq59 to give:

Eq60 shows that the derived heat capacity for an ideal gas at relatively low temperatures is independent of any thermodynamic property. We call such an ideal gas a perfect gas, i.e., an ideal ideal gas. Since the heat capacity of a perfect gas at constant pressure is related to the heat capacity of a perfect gas at constant volume by (see this article for derivation),

As mentioned in an earlier article, and in reality are functions of temperature. This is due to contributions from as temperature increases.

 

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Equipartition theorem: Translational motion

The translational motion of an atom, i.e. its position in the Cartesian coordinate system, is described by three coordinates: x, y and z. The translational motion of a molecule composed of multiple atoms is similarly described by only three coordinates, which indicate the centre of mass of the molecule. The energy associated with this motion is kinetic energy:

where is the velocity vector of the molecule with .

So, the translational motion of a molecule has three degrees of freedom with energies components of:

For a system of particles, the average translational energy of a particle is:

where  is the square of the root mean square speed of the gas in the -direction.

We have shown in the article on kinetic theory of gases that , with . Hence, each velocity energy component of the molecule has an average energy of .

 

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Equipartition theorem (overview)

The equipartition theorem states that the total energy of a system of particles at thermal equilibrium is equally divided among all the energy components of the particles’ motion, resulting in each energy component having the same average value.

We classify the components of motion into three types: translational, rotational and vibrational motions. The theorem is applicable to systems where no electronic transitions occur, usually at .

 

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Heat capacity

Heat capacity, , is the ratio of heat transferred to a system from its surroundings to the temperature change of the system due to the transfer. In other words, the heat capacity of a system or substance is the amount of heat the system or substance can hold per Kelvin. The transfer of heat to a system can take place at either constant pressure or constant volume, resulting in two types of heat capacities, and respectively.

To define the heat capacity of a system at constant pressure, we begin with eq30 of the previous article or its differential form:

where the symbol is to signify that is a path function.

 

Question

Does eq34 say that a state function is equal to a path function?

Answer

No, eq34 means that the value of is equal to the change in along a path, but is always the same regardless of the path taken .

 

Dividing both sides of eq34 by ,

For a process at constant pressure, we rewrite eq35 as:

The difference in symbols on the LHS versus the RHS of eq36 for indicating the process at constant pressure is due to being a state function and , a path function. According to the definition of heat capacity in the first paragraph of the article,

Since the quantities and are easily found for different systems (or substances) via experiments like the neutralisation reaction mentioned in the previous article, the heat capacities for various systems or substances at constant pressure are easily determined. The molar heat capacity of a substance at constant pressure is:

and the specific heat capacity of a substance at constant pressure, , is found by dividing eq37 with the mass, , of the substance:

Eq39 is a useful equation to calculate the change in enthalpy of a system or substance. From eq37, the heat capacity of a substance at constant pressure is the gradient of a curve on a plot of enthalpy against temperature at constant pressure. If the curve is a straight line, is a constant at all temperatures. In reality, it is a function of temperature.

For experiments over a small temperature range, may be assumed to be constant to simplify calculations. For experiments that are carried out over a large temperature range, we need to find an expression for . This is done by using a power series to generate curves that fit experimental values of for different substances on versus temperature plots:

The set of constants , , , , etc. are specific to each substance and are the outcome of the polynomial regression. As the contribution of higher powers of to is small, the expression is fairly well represented by:

Eq40 is then substituted in eq39 and the resultant equation is integrated throughout to obtain:

Using the same logic, to define the heat capacity of a system at constant volume, we begin with eq31 of the previous article or its differential form , resulting in

and

Since the value of the heat capacity of a system depends on whether we increase the amount of heat of the system at constant volume or at constant pressure, heat capacity is a path function. However, if the path is chosen, then or are state functions.

Finally, we shall show that  and  for a system containing an ideal gas.

The internal energy of a gas is the sum of the molecules’ translational energy , rotational energy , vibrational energy , electronic transition energy , intermolecular forces of interaction  and rest-mass energy of electrons and nuclei :

is constant for any gas, while  is constant at  and if no chemical reaction takes place. For an ideal gas,  is zero, leaving the internal energy of an ideal gas as:

It can be shown that , and  are functions of temperature (independent of volume). Since the internal energy of an ideal gas is only dependent on temperature, . With , we write, for an ideal gas:

Eq44 is a useful relation that is applicable to all reversible processes that involves an ideal gas (not just constant volume processes because the internal energy of an ideal gas is only dependent on temperature). Irreversible processes have systems with poorly defined and therefore have internal energy changes that cannot be accurately predicted by eq44. For the enthalpy of an ideal gas,

As of an ideal gas is only dependent on temperature, the enthalpy of an ideal gas according to eq45 is also only dependent on temperature. Using the same logic as above, and

 

Question

Show that for a perfect gas (one that obeys the ideal gas law and exhibits a heat capacity that is independent of temperature) using the above diagram where

AC: reversible isothermal process
AB: reversible isochoric process
BC: reversible isobaric processes

Answer

The change in internal energy  from point A to C along the isotherm AC is zero for an ideal gas. Since is a state function,  for the process AC and ABC is the same:

The change in internal energy for path ABC is:

Assuming  and are constant over the temperature range,

Since  and , we have, after applying the ideal gas law,

Rearranging gives

 

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