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# 2. The First Law: Work, Heat and Thermochemistry

Author : Jack J. Middelburg

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## Abstract

This chapter presents the first law of thermodynamics in terms of heat exchange and work. Internal energy, heat capacity and enthalpy are defined and enthalpy changes during reactions and phase changes are discussed. Latent and sensible heat are introduced, and adiabatic processes are related to the atmospheric lapse rate, potential temperature, and the geothermal gradient.
Before elaborating on the first law and its implications, it is necessary to introduce the zeroth law of thermodynamics and to further define our systems. Consider two systems, A and B, that are closed, i.e., no matter can exchange, but with different temperatures. If these are brought in contact, heat will flow spontaneously from the hotter to the colder system till temperatures become equal, i.e., thermal equilibrium is reached. Energy transfer from one system to another due to temperature differences is called heat. If a third closed system, C, is in thermal equilibrium with system A, then TC =TA and system C must be in thermal equilibrium with system B also. This illustrates the zeroth law of thermodynamics stating that
if two systems are in thermal equilibrium with each other and a third system is in thermal equilibrium with one of them, then it is in thermal equilibrium with the other also.
Systems exchange matter and/or energy with their surroundings. Thermodynamics has adopted a system-centric view, i.e., matter added to, heat delivered to, and work done on the system have a positive sign. Conversely, work done by the system or heat flowing out of the system to the surroundings have a negative sign (Fig. 2.1).

## 2.1 Work

Energy is defined as the capacity to do work. Work is being done when motion occurs upon the action of a force. Work, w, is quantified as the product of force, F, and the distance, d, over which it acts:
$$w = F \cdot d .$$
(2.1)
Work has the unit of Joule (1 J = 1 Nm), like other forms of energy. In thermodynamics a common form of work is expansion/compression work because of volume changes of the system. Consider a cylinder with a piston (Fig. 2.2). Any process that increases the gas volume within the cylinder under the piston will push the piston upwards and in this way the system performs work on the surroundings, i.e., it loses some energy in the form of work. The amount of work ($$w)$$ done by the system for a change in volume $$\left( {\Delta V} \right)$$ is
$$w = - P_{ext} \Delta V ,$$
(2.2)
where Pext is the external pressure of the surroundings, and the minus sign is needed to satisfy the convention that work done by the system should be negative (Fig. 2.1).

## 2.2 Heat

Heat, also called thermal energy, is denoted by the letter q and has the unit of joule. Traditionally, the calorie (cal) unit has been used for heat. One calorie is the energy needed to heat 1 mL of water from 15 to 16 °C, which is equal to 4.184 J. The amount of heat needed to change the temperature of a system depends on the temperature change, ΔT, and the heat capacity (C):
$$q = C \cdot T$$
(2.3)
The heat capacity C is an extensive property that depends on the amount of material in the system. The molar heat capacity, $$C_{m}$$, is the amount of heat needed to raise the temperature of 1 mol by 1 K and has the unit J mol−1 K−1, while the specific heat capacity, $$C_{s}$$, is the amount of heat for 1 K temperature increase per gram of material with the unit J g−1 K−1. Using the molar or specific heat capacities Equation (2.3) would then respectively become:
$$q = n C_{m} \Delta T ,$$
(2.4)
and
$$q = m C_{s} \Delta T$$
(2.5)
where m is mass (g) and n is the number of moles. As an example, consider the energy required to heat 18 gram (1 mole) of water with a specific heat capacity ($$C_{s}$$ of 4.18 J g−1 K−1) from 25 to 50 °C. Using Eq. (2.5), we then arrive at 1882 J.

## 2.3 The First Law

The first law of thermodynamics can be formulated in multiple ways but in fact states: energy can neither be created nor destroyed, i.e., energy is conserved. The total energy of a system is defined as the internal energy for which we use the symbol U. In all phases (gas, liquid, solid) atoms and molecules are in motion. Internal energy comprises all types of energy: e.g., kinetic energy from molecular rotation and bond vibration, and potential energy from intermolecular attraction and chemical bonds. Because we cannot measure all types of energy in a system, the absolute total internal energy cannot be known. However, differences in internal energy (ΔU) can be measured and calculated.
An isolated system does not exchange matter or energy with the surroundings; hence the total internal energy of the system does not change. This leads to the statement of the first law of thermodynamics:
For an isolated system, the total energy of the system remains constant.
$${\text{Mathematically}},\,{\text{ this }}\,{\text{means:}}\,\Delta U = 0$$
(2.6)
This statement of the first law has limited utility because it is for an isolated system. An alternative, more useful way to express the first law (for a closed system) is the change in internal energy is the sum of heat (q) added and work (w) done on the system, mathematically:
$$\Delta {\varvec{U}} = {\varvec{q}} + {\varvec{w}}$$
(2.7)
This is the formulation of the first law used in this course. The internal energy U is a state function, i.e., it is path independent, while the heat q and work w are not state functions and depend on the path. In more advanced treatments, this equation is usually presented in calculus form as $$dU = \delta q + \delta w$$, where d represents an infinitesimally small change and $$\delta$$ is used to articulate that infinitesimally small changes in q and w depend on the path.
Another useful way of phrasing the first law is to consider expansion work (w=-PΔV):
$$\Delta U = q - P\Delta V$$
(2.8)
If we add heat to a system with a constant volume, e.g., a container, there is no P-V work, ΔV=0, then
$$\Delta U = q + 0 = q_{V} ,$$
(2.9)
where $$q_{V}$$ is the heat transfer at constant volume. Accordingly, the change in internal energy for a system at a constant volume is equal to the heat exchange that can be measured. This is very useful for engineers and others working with volume constrained systems (containers, reactors). However, for earth scientists studying processes at a certain pressure, or chemists studying reactions in the laboratory at atmospheric pressure, it is more useful to quantify the change in heat at a constant pressure, i.e., the enthalpy change, ΔH:
$$\Delta H = q_{P} = \Delta U + P\Delta V,$$
(2.10)
where $$q_{P}$$ is the heat transfer at constant pressure. This equation directly follows from Eq. (2.8) and leads to the formal definition of enthalpy, H, as
$$H \equiv U + PV$$
(2.11)
where the symbol $$\equiv$$ is used for a formal definition.
Although the change in internal energy (ΔU) is the more fundamental quantity (used by physicists), the change in enthalpy (ΔH) is usually easier to measure and thus more commonly used. At constant P, the difference between ΔU and ΔH is due to the PΔV term (Eq. 2.10). Volume changes involving liquid or solid phases are small, hence the PΔV can be ignored and ΔH≈ΔU. However, volume changes can be large if gases are involved and then enthalpy changes are a little higher than internal energy changes (≈2.48 kJ per mol at 25 °C). Enthalpy changes follow the sign convention introduced earlier (Fig. 2.1): processes releasing heat are exothermic and have a negative ΔH because the system loses energy, heat uptake by a system corresponds to positive ΔH values and are called endothermic.
The heat capacity (C) was defined as the amount of heat required to increase the temperature of a substance by 1K. Similar to heat (q), the heat capacity of a gas also depends on conditions, i.e., whether the process occurs under constant volume or constant pressure conditions. In case of constant volume, all the heat will increase the internal energy, and thus the temperature, because no expansion work is done. At a constant pressure, some of the heat goes into work and thus more energy is needed to cause the same temperature rise. The constant pressure heat capacity CP is thus higher than the constant volume heat capacity CV. Mathematically, CV relates $$\Delta T$$ to the change in internal energy:
$$\Delta U = q_{V} = C_{V} \Delta T$$
(2.12)
whereas CP links $$\Delta T$$ to the enthalpy change
$$\Delta H = q_{P} = C_{P} \Delta T.$$
(2.13)
Volume change with heating of liquids and solids is generally limited and their $$C_{V}$$ and $$C_{P}$$ are very close in value and any difference can usually be neglected. For an ideal gas, Mayer’s relation links $$C_{V}$$ and $$C_{P}$$:
$$C_{P} = C_{V} + nR$$
(2.14)
Having discussed the internal energy changes of a gas at a constant temperature and constant pressure, we now consider an isothermal change. Isothermal conditions imply that $$\Delta T$$ is zero, hence $$\Delta U = 0$$ (Eq. 2.12). The first law then reads
$$\Delta U = 0 = q + w$$
and consequently, for isothermal processes
$$q = - w$$
(2.15)
An isothermal compression (decrease in volume) causes a net outflow of heat, while isothermal expansion (increase in volume) must be compensated by heat uptake to maintain constant temperature.

## 2.4 Thermochemistry: Enthalpy Changes in Chemical Reactions

All chemical substances have enthalpy stored in the form of chemical bonds. When a chemical reaction occurs, the change in enthalpy, ΔH, is equal to the total enthalpy of products (final condition with new chemical bonds) minus the total enthalpy from the reactants (initial condition with pre-existing chemical bonds)
$$\Delta_{r} H = H_{products} - H_{reactants} ,$$
(2.16)
where $$\Delta_{r} H$$ is the enthalpy change of a reaction, as indicated by the subscript r. However, absolute enthalpies of substance cannot be determined (like internal energy), only relative values, i.e., changes in enthalpy can be determined. In other words, we need a set of standards against which enthalpies of reaction can be measured. The standard enthalpy of formation of a certain compound, $$\Delta H_{f}^{o}$$, is defined as the enthalpy change for the formation of one mole of that compound from its elements in the most stable form under standard conditions (STAP, 25 °C and 1 bar). The subscript f is added to indicate that is the enthalpy of formation and the superscript o that it relates to pure substances at SATP. For example, the gas N2 and the solid graphite are the most stable form of elements N and C at standard conditions and their enthalpies of formation ($$\Delta H_{f}^{o} )$$ have been set to zero. Thermochemists have constructed consistent databases with standard enthalpies of formation that can be used to calculate the enthalpy of reaction at standard conditions and making use of reaction stoichiometry.
Specifically, for the reaction of α moles of substance A and β moles of substance B to form γ moles of substance C and δ moles of substance D
$$\alpha A + \beta B \to \gamma C + \delta D$$
(2.17)
the enthalpy of reaction at SATP ($$\Delta_{r} H^{o}$$) is calculated as follows,
$$\Delta_{r} H^{o} = \left[ {\gamma \times \Delta H_{f}^{o} \left( C \right) + \delta \times \Delta H_{f}^{o} \left( D \right)} \right] - \left[ {\alpha \times \Delta H_{f}^{o} \left( A \right) + \beta \times \Delta H_{f}^{o} \left( B \right)} \right]$$
(2.18)
or more generalized:
$$\Delta H_{r}^{o} = \sum v_{i} \Delta H_{f}^{o} \left( {products} \right) - \sum v_{i} \Delta H_{f}^{o} \left( {reactants } \right),$$
(2.19)
where $$v_{i}$$ is the stoichiometric coefficient of substance i in the reaction.
To illustrate this method to calculate $$\Delta_{r} H^{o}$$, consider the oxidation of 1 mole of sucrose (C12H22O11; s) to CO2 (aq) and water (l) at 1 bar. The relevant reaction and enthalpy of formations are:
Reaction
$$C_{12} H_{22} O_{11} \left( s \right)$$ +
$$12 O_{2} \left( g \right)$$
$$\to$$
$$12 CO_{2} \left( {aq} \right) +$$
$$11 H_{2} O\left( l \right)$$
$$\Delta H_{f}^{o}$$
−2226.1 kJ mole−1
0 kJ mole−1

−412.9 kJ mole−1
−258.8 kJ mole−1
and the enthalpy of reaction is calculated using 2.18:
\begin{aligned} \Delta H_{r}^{o} & = \left[ {12\Delta H_{f}^{o} \left( {CO_{2} \left( {aq} \right)} \right) + 11\Delta H_{f}^{o} \left( {H_{2} O\left( l \right)} \right)} \right] \\ & \quad - \left[ {\Delta H_{f}^{o} \left( {C_{12} H_{22} O_{11} \left( s \right)} \right) + 12 \times \Delta H_{f}^{o} \left( {O_{2} \left( g \right)} \right)} \right] \\ & = 12 \times - 412.9 + 11 \times - 285.8 - 2226.1 - 12 \times 0 = - 5873 \;{\text{kJ}} \\ \end{aligned}
This enthalpy of reaction is also known as enthalpy of combustion or heat of combustion, i.e., the enthalpy of change when one mole of a compound reacts completely with excess oxygen gas. When using thermodynamic data, one should note that each compound and each phase of a certain compound have distinct internal energies and enthalpies and that the phase [solid (s), liquid (l), dissolved (aq) and gas (g)] should therefore explicitly be indicated to use the correct energy. This is because, as we will see below, phase transitions cause enthalpy changes.
This is one way to obtain enthalpy of reactions, the other method is based on Hess’s law.
Since enthalpy is a state function, only the initial and final conditions do matter, and the pathway (intermediate reactions) does not matter. Hess’s law states that the overall enthalpy change for a reaction is equal to the sum of enthalpy changes for the individual steps in the reaction.
To illustrate Hess’s law, consider the Haber(-Bosch) process, the common way to produce ammonia (needed for fertilizers to produce food for the world population) in large quantities from nitrogen and hydrogen gas. It is a multiple-step process:
\begin{aligned} &{\;\;\;\text{Step}}\,1:2 H_{2} \left( g \right) + N_{2} \left( g \right) \to N_{2} H_{4} \left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H_{1} = ? \hfill \\ &\frac{{{\text{Step}}\,2:N_{2} H_{4} \left( g \right) + H_{2 } \left( g \right) \to 2NH_{3} \left( g \right)\,\,\,\,\,\,\,\Delta H_{2} = - 187.6 \,{\text{kJ}}}}{{{\text{Combined}}:3 H_{2 } \left( g \right) + N_{2 } \left( g \right) \to 2NH_{3} \left( g \right)\,\,\,\,\Delta H_{1 + 2} = - 92.2 \,{\text{kJ}}}} \\ &\left( {{\text{net }}\,{\text{reaction}}} \right) \hfill \\ \end{aligned}
Hess’s law implies that $$\Delta H_{1} + \Delta H_{2} = \Delta H_{1 + 2}$$; hence $$\Delta H_{1} = \Delta H_{1 + 2} - \Delta H_{2} = + 95.4\, {\text{kJ}}.$$ The first step is thus endothermic consuming heat (the system/reaction receives heat), while the overall reaction is exothermic and produces heat.
The heat of combustion or enthalpy of combustion is usually presented in kJ g−1 (i.e. specific combustion enthalpy) rather than kJ mol−1, the unit used for enthalpy of formation $$\left( {\Delta H_{f}^{o} } \right)$$. This specific combustion enthalpy is a measure of the energy density.
While the heats of combustion of methane, ethane, propane, butane and pentane are −890, −1560, −2220, −2877 and −3509 kJ mol−1 and differ substantially, their respective specific combustion enthalpies are more alike: −55.5, −51.9, −50.4, −49.5 and −48.6 kJ g−1, respectively. For comparison, typical specific combustion enthalpies in kJ g−1 of kerosene (−46.2), diesel (−44.2), crude oil (−43) are also in this range, while those of anthracite coal (−32.5), lignite coal (−15), wood (−15 to −20) and peat (−15 to −20) are much lower.

## 2.5 Enthalpy Changes During Phase Changes

Matter exists in three phases (solid, liquid and gas) and transformation from one phase to another involves heat transfer, i.e., a change in enthalpy (Fig. 2.3). The transition from solid to liquid, melting or fusion, requires addition of heat (endothermic process). The enthalpy of fusion $$\Delta H_{fusion}$$ is defined as the amount of heat necessary to melt a substance at its melting point without changing its temperature at constant pressure. Standard enthalpies of fusion $$\Delta H_{fusion}^{o}$$ are available at 1 bar and 25 °C. Freezing or solidification, the opposite process of melting, is exothermic and the enthalpy change has the same magnitude as that of fusion but the opposite sign (negative for freezing, positive for fusion).
Vaporization (or boiling) is the transformation of a liquid into gas or vapor and the enthalpy of vaporization $$\Delta H_{vap}$$ is the energy required at constant pressure to vaporize one mole of pure liquid at its boiling temperature. Condensation, the opposite of vaporization, has the same enthalpy as vaporization but the opposite sign (negative for condensation, positive for vaporization).
Sublimation is the transformation of a solid into a gas. The enthalpy of sublimation $$\Delta H_{subl}$$ is the amount of heat required to convert a solid into a gas without going through the liquid phase. An example is dry ice (solid carbon dioxide) which at atmospheric pressure goes directly into a gas without formation of liquid carbon dioxide at a temperature of −78.5 °C. Condensation or deposition, the reverse process has again a different sign for the enthalpy change than sublimation. At a constant temperature, enthalpies of sublimation can be directly calculated from the enthalpies of fusion and vaporization using Hess’s law:
$$\Delta H_{subl} = \Delta H_{fusion} + \Delta H_{vap}$$
(2.20)

## 2.6 Turning Ice into Steam: Latent and Sensible Heat

Water plays a major role in System Earth: it dominates the cryosphere (and obviously the hydrosphere) and impacts the biosphere, geosphere and atmosphere. We have covered enough material to calculate the temperature response during the heating of 1 gram of ice with an initial temperature of −20 °C to a steam of 120 °C at earth surface pressure of 1 atm (Fig. 2.4).
Upon addition of heat to ice of −20 °C, the ice will heat up and with a specific heat capacity (Cs) of 2.09 J g−1 K−1. For one gram of ice, 41.8 J (20 $$\times$$ 2.09) will be required to reach the melting point of ice. At the melting point, 0 °C, further addition of heat will melt the ice but the system remains at the same temperature, and 331 J will be needed to convert one gram of ice into liquid (enthalpy of fusion). Further addition of heat will warm up the water and 418 J are required to reach the boiling point at 100 °C (since water has a specific heat capacity of 4.18 J g−1 K−1). At the boiling point, liquid water will gradually turn into steam (evaporate), and this requires 2260 J (enthalpy of vaporization), but temperature does not change. Finally, further addition of heat will increase the temperature of the steam. To reach 120 °C, this will require an additional 36.8 J, because steam has a specific heat content of 1.84 J g−1 K−1. The total energy required is about 3088 J, of which 84 % is used for phase changes (10.7 % for melting and 73.2 % for boiling) and does not result in a temperature increase. The energy consumed or released during phase transitions at a constant temperature is also known as ‘hidden’ or latent heat. This contrast with sensible heat, that results in a temperature change of the system. Latent heat fluxes are an important component of Earth’s surface energy budget, e.g., the evaporation/transpiration at the Earth surface and subsequent condensation of water in the troposphere. In the Earth’s interior, latent heat is also released if the liquid outer core crystallizes at the inner core boundary.

Adiabatic processes are thermodynamic state changes without heat transfer (q = 0). Accordingly, the first law $$\Delta U = q + w$$ then reduces to
$$\Delta U = w = - P\Delta V$$
(2.21)
stating that the change in internal energy of a system equals work. Adiabatic processes are usually associated with a temperature change of the system because of changes in volume. Compression (due to increasing pressure) causes temperature increases and expansion results in cooling due to reducing pressure. If a system expands or compresses very fast, or heat exchange occurs very slowly, it will not have time to exchange heat with its surroundings. Describing the system volume change as an adiabatic process is then an appropriate approximation. For instance, air transport upward in the atmosphere can be modelled as an adiabatic process because the heat transfer between the air parcel and the rest of the atmosphere is slow on the timescale of its upward motion. A rising magma plume or air escaping after opening a bottle of champagne can also be treated as an adiabatic process. The geothermal gradient can be explained by adiabatic compression (Box 2).
For an adiabatic process, the change in internal energy can be determined from two expressions (Eqs. 2.21 and 2.12):
$$\Delta U = - P\Delta V \,{\text{and}}\,\Delta U = C_{V} \Delta T$$
In which $$C_{V}$$ is the heat capacity at constant volume.
Equating these two expressions
$$C_{V} \Delta T = - P\Delta V$$
(2.22)
shows that compression ($$\Delta V < 0$$) leads to a positive $$\Delta T$$, hence an increase in temperature, while expansion ($$\Delta V > 0)$$ results in a negative $$\Delta T$$ and is thus accompanied by a decrease in temperature.
Adiabatic expansion can quantitatively explain the atmospheric lapse rate, i.e., the well-known decrease in temperature while moving upward in the atmosphere. Mathematically, lapse rate (Γ) is the temperature gradient:
$${\Gamma } = - \frac{\Delta T}{{\Delta z}},$$
(2.23)
where z is height (in m).
To link the lapse rate with the first law, we treated an air parcel as a closed system and differentiate the ideal gas law for one mol (PV = RT)$$:$$
$$P\Delta V + V\Delta P = R\Delta T$$
and then re-arrange to isolate $$\Delta T$$
$$\Delta T = \frac{ P\Delta V + V\Delta P}{R}$$
Then substitute it into Eq. (2.22):
$$C_{v} \left( {\frac{ P\Delta V + V\Delta P}{R}} \right) + P\Delta V = 0$$
(2.24a)
using Mayer’s relation (Eq. 2.14) $$R = C_{P} - C_{V}$$, we obtain
$$C_{v} \left( {\frac{ P\Delta V + V\Delta P}{{C_{P} - C_{V} }}} \right) + P\Delta V = 0$$
(2.24b)
Which can be rewritten as
$$P\Delta V = - \frac{V\Delta P}{{\frac{{C_{P} }}{{C_{V} }}}}.$$
(2.25)
This is the thermodynamic equation for adiabatic processes. It is sometimes presented in integrated form as $$PV^{{\frac{{C_{P} }}{{C_{V} }}}}$$ = constant. Substituting Eq. (2.25) into the first law for adiabatic processes (Eq. 2.22) using the specific heat capacity at constant volume, we obtain
$$mC_{V} \Delta T = \frac{V\Delta P}{{\frac{{C_{P} }}{{C_{V} }}}}$$
which can be rewritten as
$$\rho C_{P} \Delta T = \Delta P$$
(2.26)
where $$\rho$$ is $$\frac{m}{V}$$, the density of the gas.
Combining this equation with the equation for hydrostatic equilibrium of the atmosphere,
$$\Delta P = - \rho g \Delta z$$
(2.27)
with g as the standard gravitation acceleration at the earth surface (9.8 m s−2) and $$z$$ as height (m) to eliminate the pressure term, we eventually arrive at the lapse rate:
$${\Gamma } = - \frac{\Delta T}{{\Delta z}} = \frac{g}{{C_{P} }} = 9.8 \,{\text{K km}}^{ - 1}$$
since the specific heat capacity $$C_{P}$$ value for dry air is about 1 kJ kg−1 K−1. This theoretical estimate for the dry air is fully consistent with the experience-based lapse rate of 1 °C per 100 m (Fig. 2.5).
However, air contains moisture that condenses upon cooling (cloud formation) and then releases (latent) heat ($$\Delta H$$). This heat is added to the internal energy of the air. In this case we must extend Eq. (2.22) with latent heat:
$$mC_{V} \Delta T = - P\Delta V + \Delta H$$
(2.28)
And since $$\Delta H$$ is positive (added to air), the adiabatic cooling ($$\Delta T$$) will be less, thus the environmental lapse rate will be smaller, consistent with observations (Fig. 2.5). Moist adiabatic lapse rates are typically 0.5 °C per 100 m.
Pressure change induced adiabatic processes increase temperatures upon compression and decrease temperatures following expansion. Atmospheric and ocean scientists therefore use potential temperature, which is defined as the temperature that a parcel of air/water would attain if adiabatically brought to a standard reference pressure, i.e., earth surface conditions. Meteorologists use a standard pressure of 1000 mbar (1000 hPa) for reference, while oceanographers use the ocean surface pressure (0 dbar). Changes in temperature solely caused by compression or expansion are generally not of interest when studying atmosphere or ocean dynamics or heat content.
The potential temperature ($$\theta$$) of an ideal gas is given by
$$\theta = T\left( {\frac{{P_{0} }}{P}} \right)^{{\frac{R}{{C_{P} }}}}$$
(2.29)
where P is pressure, P0 is the reference pressure, T is the current temperature in K, R is the gas constant and $$C_{P}$$ is the specific heat capacity at constant pressure. Meteorologists use a value 0.286 for the ratio $$\frac{R}{{C_{P} }}$$. Potential temperature ($$\theta$$) in the ocean is always lower than the actual temperature by about 0.1 K for every km depth increase. This limited increase of in situ water temperature relative to potential temperature is due to the large heat capacity of water.
Box 2 Geothermal gradient of the inner Earth
The geothermal gradient, i.e., the temperature increase with depth into the Earth, can also be explained by adiabatic compression. The temperatures of the Earth’s mantle and outer core are close to the adiabatic temperature curve. For adiabatic compression, a similar analysis as for the atmospheric lapse rate due to expansion can adopted, starting with a modified form of Eq. (2.26)
$$mC_{P} \Delta T = TV\alpha_{P} \Delta P$$
(2.30)
where m is mass, $$C_{P}$$ is the specific heat capacity of the minerals, and $$\alpha_{P}$$ is the fractional increase in volume per degree increase in temperature. Specifically, the volume change in atmospheric adiabatic processes $$\frac{V}{{\frac{{C_{P} }}{{C_{V} }}}}$$ in Eq. (2.25) relates to the volume change $$TV\alpha_{P}$$ in the inner Earth adiabat (Eq. 2.30).
Re-arranging this equation to isolate the adiabatic change of temperature with pressure:
$$\frac{\Delta T}{{\Delta P}} = \frac{{TV\alpha_{P} }}{{mC_{P} }} = \frac{{T\alpha_{P} }}{{\rho C_{P} }}$$
(2.31)
where $$\rho$$ is the density $$\frac{m}{V}.$$
Using the appropriate hydrostatic pressure Eq. (2.27), we obtain after re-arrangement
$$\frac{\Delta T}{{\Delta z}} = \frac{{gT\alpha_{P} }}{{C_{P} }}.$$
(2.32)
Using representative values for physical properties in the Earth’s interior (Table 2.1), we can calculate the adiabatic temperature gradients: $$\frac{\Delta T}{{\Delta z}} = 0.88\;{\text{ Kkm}}^{ - 1}$$ at the core-mantle boundary and $$\frac{\Delta T}{{\Delta z}} = 0.29 \,{\text{Kkm}}^{ - 1}$$ at the inner-core boundary.
Table 2.1
Physical parameters in the outer and inner core near to the core-mantle boundary (CMB) and inner-core boundary 9ICB) (Lowry, 2011)
Property
Unit
Core-mantle boundary
Inner core boundary
Gravity, g
m s−2
10.7
4.4
Density, $$\rho$$
kg m−3
9900
12,980
$$C_{P}$$
J K−1 kg−1
815
728
T
K
3700
5000
Volume expansion coefficient, $$\alpha_{P}$$
10−6 J kg−1
18.0
9.7
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