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Anisothermal process is a type ofthermodynamic process in which thetemperatureT of asystem remains constant: ΔT = 0. This typically occurs when a system is in contact with an outsidethermal reservoir, and a change in the system occurs slowly enough to allow the system to be continuously adjusted to the temperature of the reservoir throughheat exchange (seequasi-equilibrium). In contrast, anadiabatic process is where a system exchanges no heat with itssurroundings (Q = 0).
Simply, we can say that in an isothermal process
while in adiabatic processes:
The nounisotherm is derived from theAncient Greek wordsἴσος (ísos), meaning "equal", andθέρμη (thérmē), meaning "heat".
Isothermal processes can occur in any kind of system that has some means of regulating the temperature, including highly structuredmachines, and evenliving cells. Some parts of the cycles of someheat engines are carried out isothermally (for example, in theCarnot cycle).[1] In the thermodynamic analysis ofchemical reactions, it is usual to first analyze what happens under isothermal conditions and then consider the effect of temperature.[2]Phase changes, such asmelting orevaporation, are also isothermal processes when, as is usually the case, they occur at constant pressure.[3] Isothermal processes are often used as a starting point in analyzing more complex, non-isothermal processes.
Isothermal processes are of special interest for ideal gases. This is a consequence ofJoule's second law which states that theinternal 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 nointermolecular 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 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 isothermal expansion, the energy supplied to the system does work on the surroundings. In either case, with the aid of a suitable linkage the change in gas volume can perform useful mechanical work. For details of the calculations, seecalculation of work.
For anadiabatic 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. afree 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.
For the special case of a gas to whichBoyle's law[4] applies, the productpV (p for gas pressure andV for gas volume) is a constant if the gas is kept at isothermal conditions. The value of the constant isnRT, wheren is the number ofmoles of the present gas andR is theideal gas constant. In other words, theideal gas lawpV = nRT applies.[4] Therefore:
holds. The family of curves generated by this equation is shown in the graph in Figure 1. Each curve is called an isotherm, meaning a curve at a same temperatureT. Such graphs are termedindicator diagrams and were first used byJames 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.
In thermodynamics, the reversible work involved when a gas changes from stateA to stateB is[6]
wherep for gas pressure andV for gas volume. For an isothermal (constant temperatureT),reversible process, this integral equals the area under the relevant PV (pressure-volume) isotherm, and is indicated in purple in Figure 2 for an ideal gas. Again,p = nRT/V applies and withT being constant (as this is an isothermal process), the expression for work becomes:
InIUPAC convention, work is defined as work on a system by its surroundings. If, for example, the system is compressed, then the work is done on the system by the surrounding so the work is positive and the internal energy of the system increases. Conversely, if the system expands (i.e., system surrounding expansion, sofree expansions not the case), then the work is negative as the system does work on the surroundings.
It is also worth noting that for ideal gases, if the temperature is held constant, the internal energy of the systemU also is constant, and so ΔU = 0. Since thefirst law of thermodynamics states that ΔU = Q + W inIUPAC convention, it follows thatQ = −W for the isothermal compression or expansion of ideal gases.
The reversible expansion of anideal gas can be used as an example of work produced by an isothermal process. Of particular interest is the extent to which heat is converted to usable work, and the relationship between the confiningforce and the extent of expansion.
During isothermal expansion of an ideal gas, bothp andV change along an isotherm with a constantpV product (i.e., constantT). Consider a working gas in a cylindrical chamber 1 m high and 1 m2 area (so 1m3 volume) at 400 K instatic equilibrium. Thesurroundings consist of air at 300 K and 1 atm pressure (designated aspsurr). The working gas is confined by a piston connected to a mechanical device that exerts a force sufficient to create a working gas pressure of 2 atm (stateA). For any change in stateA that causes a force decrease, the gas will expand and perform work on the surroundings. Isothermal expansion continues as long as the applied force decreases and appropriate heat is added to keeppV = 2 [atm·m3] (= 2 atm × 1 m3). The expansion is said to be internally reversible if the piston motion is sufficiently slow such that at each instant during the expansion the gas temperature and pressure is uniform and conform to theideal gas law. Figure 3 shows thep–V relationship forpV = 2 [atm·m3] for isothermal expansion from 2 atm (stateA) to 1 atm (stateB).
The work done (designated) has two components. First,expansion work against the surrounding atmosphere pressure (designated asWpΔV), and second, usablemechanical work (designated asWmech). The outputWmech here could be movement of the piston used to turn a crank-arm, which would then turn a pulley capable of lifting water out offlooded salt mines.
The system attains stateB (pV = 2 [atm·m3] withp = 1 atm andV = 2 m3) when the applied force reaches zero. At that point, equals –140.5 kJ, andWpΔV is –101.3 kJ. By difference,Wmech = –39.1 kJ, which is 27.9% of the heat supplied to the process (- 39.1 kJ / - 140.5 kJ). This is the maximum amount of usable mechanical work obtainable from the process at the stated conditions. The percentage ofWmech is a function ofpV andpsurr, and approaches 100% aspsurr approaches zero.
To pursue the nature of isothermal expansion further, note the red line on Figure 3. The fixed value ofpV causes an exponential increase in piston rise vs. pressure decrease. For example, a pressure decrease from 2 to 1.9 atm causes a piston rise of 0.0526 m. In comparison, a pressure decrease from 1.1 to 1 atm causes a piston rise of 0.1818 m.
Isothermal processes are especially convenient for calculating changes inentropy since, in this case, the formula for the entropy change, ΔS, is simply
whereQrev is the heat transferred (internally reversible) to the system andT isabsolute temperature.[7] This formula is valid only for a hypotheticalreversible process; that is, a process in which equilibrium is maintained at all times.
A simple example is an equilibriumphase 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 theenthalpy of transformation, ΔHtr, thusQ = ΔHtr.[3] At any given pressure, there will be a transition temperature,Ttr, for which the two phases are in equilibrium (for example, the normalboiling 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]
Another example is the reversible isothermal expansion (or compression) of anideal gas from an initial volumeVA and pressurePA to a final volumeVB and pressurePB. As shown inCalculation of work, the heat transferred to the gas is
This result is for a reversible process, so it may be substituted in the formula for the entropy change to obtain[7]
Since an ideal gas obeysBoyle's law, this can be rewritten, if desired, as
Once obtained, these formulas can be applied to anirreversible process, such as thefree 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 astate function (that depends on an equilibrium state, not depending on a path that the system takes to reach that state), 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 resultQ = 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 irreversible 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 heatQ 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 irreversible,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.
