# Isothermal process

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An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change will occur slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange. In contrast, an adiabatic process is where a system exchanges no heat with its surroundings (Q = 0). In other words, in an isothermal process, the value ΔT = 0 and therefore ΔU = 0 (only for an ideal gas) but Q ≠ 0, while in an adiabatic process, ΔT ≠ 0 but Q = 0.

## Examples

Isothermal processes can occur in any kind of system that has some means of regulating the temperature, including highly structured machines, and even living cells. Some parts of the cycles of some heat engines are carried out isothermally (for example, in the Carnot cycle).[1] In the thermodynamic analysis of chemical reactions, it is usual to first analyze what happens under isothermal conditions and then consider the effect of temperature.[2] Phase changes, such as melting or evaporation, are also isothermal processes when, as is usually the case, they occur at constant pressure.[3] Isothermal processes are often used and a starting point in analyzing more complex, non-isothermal processes.

Isothermal processes are of special interest for ideal gases. This is a consequence of Joule's second law which states that the internal energy of a fixed amount of an ideal gas depends only on its temperature.[4] Thus, in an isothermal process the internal energy of an ideal gas is constant. This is a result of the fact that in an ideal gas there are no intermolecular forces.[4] Note that this is true only for ideal gases; the internal energy depends on pressure as well as on temperature for liquids, solids, and real gases.[5]

In the isothermal compression of a gas there is work is done on the system to decrease the volume and increase the pressure.[4] Doing work on the gas increases the internal energy and will tend to increase the temperature. To maintain the constant temperature energy must leave the system as heat and enter the environment. If the gas is ideal, the amount of energy entering the environment is equal to the work done on the gas, because internal energy does not change. For details of the calculations, see calculation of work.

For an adiabatic process, in which no heat flows into or out of the gas because its container is well insulated, Q = 0. If there is also no work done, i.e. a free expansion, there is no change in internal energy. For an ideal gas, this means that the process is also isothermal.[4] Thus, specifying that a process is isothermal is not sufficient to specify a unique process.

## Details for an ideal gas

Figure 1. Several isotherms of an ideal gas on a p-V diagram

For the special case of a gas to which Boyle's law[4] applies, the product pV is a constant if the gas is kept at isothermal conditions. The value of the constant is nRT, where n is the number of moles of gas present and R is the ideal gas constant. In other words, the ideal gas law pV = nRT applies.[4] This means that

${\displaystyle p={nRT \over V}={{\text{constant}} \over V}}$

holds. The family of curves generated by this equation is shown in the graph in Figure 1. Each curve is called an isotherm. Such graphs are termed indicator diagrams and were first used by James Watt and others to monitor the efficiency of engines. The temperature corresponding to each curve in the figure increases from the lower left to the upper right.

## Calculation of work

Figure 2. The purple area represents the work for this isothermal change

In thermodynamics, the reversible work involved when a gas changes from state A to state B is[6]

${\displaystyle W_{A\to B}=-\int _{V_{A}}^{V_{B}}p\,dV}$

For an isothermal, reversible process, this integral equals the area under the relevant pressure-volume isotherm, and is indicated in purple in Figure 2 for an ideal gas. Again, p = nRT/V applies and with T being constant (as this is an isothermal process), the expression for work becomes:

${\displaystyle W_{A\to B}=-\int _{V_{A}}^{V_{B}}p\,dV=-\int _{V_{A}}^{V_{B}}{\frac {nRT}{V}}dV=-nRT\int _{V_{A}}^{V_{B}}{\frac {1}{V}}dV=-nRT\ln {\frac {V_{B}}{V_{A}}}}$

By convention, work is defined as the work on the system by its surroundings. If, for example, the system is compressed, then the work is positive and the internal energy of the system increases. Conversely, if the system expands, it does work on the surroundings and the internal energy of the system decreases.

It is also worth noting that for ideal gases, if the temperature is held constant, the internal energy of the system also is constant, and so ΔU = 0. Since the First Law of Thermodynamics states that ΔU = Q + W (IUPAC convention), it follows that Q = −W for the isothermal compression or expansion of ideal gases.

## Entropy changes

Isothermal processes are especially convenient for calculating changes in entropy since, in this case, the formula for the entropy change, ΔS, is simply

${\displaystyle \Delta S={\frac {Q_{\text{rev}}}{T}}}$

where Qrev is the heat transferred reversibly to the system and T is absolute temperature.[7] This formula is valid only for a hypothetical reversible process; that is, a process in which equilibrium is maintained at all times.

A simple example is an equilibrium phase transition (such as melting or evaporation) taking place at constant temperature and pressure. For a phase transition at constant pressure, the heat transferred to the system is equal to the enthalpy of transformation, ΔHtr, thus Q = ΔHtr.[3] At any given pressure, there will be a transition temperature, Ttr, for which the two phases are in equilibrium (for example, the normal boiling point for vaporization of a liquid at one atmosphere pressure). If the transition takes place under such equilibrium conditions, the formula above may be used to directly calculate the entropy change[7]

${\displaystyle \Delta S_{\text{tr}}={\frac {\Delta H_{\text{tr}}}{T_{\text{tr}}}}}$.

Another example is the reversible isothermal expansion (or compression) of an ideal gas from an initial volume VA and pressure PA to a final volume VB and pressure PB. As shown in Calculation of work, the heat transferred to the gas is

${\displaystyle Q=-W=nRT\ln {\frac {V_{\text{B}}}{V_{\text{A}}}}}$.

This result is for a reversible process, so it may be substituted in the formula for the entropy change to obtain[7]

${\displaystyle \Delta S=nR\ln {\frac {V_{\text{B}}}{V_{\text{A}}}}}$.

Since an ideal gas obey's Boyle's Law, this can be rewritten, if desired, as

${\displaystyle \Delta S=nR\ln {\frac {P_{\text{A}}}{P_{\text{B}}}}}$.

Once obtained, these formulas can be applied to an irreversible process, such as the free expansion of an ideal gas. Such an expansion is also isothermal and may have the same initial and final states as in the reversible expansion. Since entropy is a state function, the change in entropy of the system is the same as in the reversible process and is given by the formulas above. Note that the result Q = 0 for the free expansion can not be used in the formula for the entropy change since the process is not reversible.

The difference between the reversible and free expansions is found in the entropy of the surroundings. In both cases, the surroundings are at a constant temperature, T, so that ΔSsur = −Q/T; the minus sign is used since the heat transferred to the surroundings is equal in magnitude and opposite in sign to the heat, Q, transferred to the system. In the reversible case, the change in entropy of the surroundings is equal and opposite to the change in the system, so the change in entropy of the universe is zero. In the free expansion, Q = 0, so the entropy of the surroundings does not change and the change in entropy of the universe is equal to ΔS for the system.

## Etymology

The adjective "isothermal" is derived from the Greek words ἴσος (isos) meaning "equal", and θέρμη (therme) meaning "heat."