Isothermal process

 ( Thermodynamic Processes ) 

$p\; =\; \{n\; R\; T\; \backslash over\; V\}\; =\; \{\backslash text\{constant\}\; \backslash over\; V\}$

holds. The family of curves generated by this equation is shown in the graph in Figure 1. Each curve is called an isotherm, meaning a curve at a same temperature T. 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.

$W\_\{A\backslash to\; B\}\; =\; -\backslash int\_\{V\_A\}^\{V\_B\}p\backslash ,dV$

where p for gas pressure and V for gas volume. For an isothermal (constant temperature T), 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 =  applies and with T being constant (as this is an isothermal process), the expression for work becomes:

$W\_\{A\backslash to\; B\}\; =\; -\backslash int\_\{V\_A\}^\{V\_B\}p\backslash ,dV\; =\; -\backslash int\_\{V\_A\}^\{V\_B\}\backslash frac\{nRT\}\{V\}dV\; =\; -nRT\backslash int\_\{V\_A\}^\{V\_B\}\backslash frac\{1\}\{V\}dV\; =\; -nRT\backslash ln\{\backslash frac\{V\_B\}\{V\_A\}\}$

In IUPAC 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, so Free Expansion 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 system U also is constant, and so Δ U = 0. Since the first law of thermodynamics states that Δ U =  Q +  W in IUPAC convention, it follows that Q = − W for the isothermal compression or expansion of ideal gases.

During isothermal expansion of an ideal gas, both and change along an isotherm with a constant product (i.e., constant T). Consider a working gas in a cylindrical chamber 1 m high and 1 m² area (so 1m³ volume) at 400 K in static equilibrium. The surroundings consist of air at 300 K and 1 atm pressure (designated as ). 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 (state ). For any change in state 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 keep = 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 the ideal gas law. Figure 3 shows the relationship for = 2 atm·m³ for isothermal expansion from 2 atm (state ) to 1 atm (state ).

The work done (designated $W\_\{A\backslash to\; B\}$) has two components. First, expansion work against the surrounding atmosphere pressure (designated as ), and second, usable mechanical work (designated as ). The output here could be movement of the piston used to turn a crank-arm, which would then turn a pulley capable of lifting water out of Salt mining.

$W\_\{A\backslash to\; B\}\; =\; -p\backslash ,V\backslash left(\backslash ln\backslash frac\{V\_B\}\{V\_A\}\backslash right)\; =\; -W\_\{p\; \backslash Delta\; V\}\; -W\_\{\backslash rm\; mech\}$

The system attains state ( = 2 atm·m³ with = 1 atm and = 2 m³) when the applied force reaches zero. At that point, $W\_\{A\backslash to\; B\}$ equals –140.5 kJ, and is –101.3 kJ. By difference, = –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 of is a function of and , and approaches 100% as approaches zero.

To pursue the nature of isothermal expansion further, note the red line on Figure 3. The fixed value of 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.

$\backslash Delta\; S\; =\; \backslash frac\{Q\_\backslash text\{rev\}\}\{T\}$

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, Δ H_tr, thus Q = Δ H_tr. At any given pressure, there will be a transition temperature, T_tr, 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

$\backslash Delta\; S\_\backslash text\{tr\}\; =\; \backslash frac\{\backslash Delta\; H\_\backslash text\{tr\}\}\{T\_\backslash text\{tr\}\}$.

$Q\; =\; -W\; =\; n\; R\; T\; \backslash ln\{\backslash frac\{V\_\{\backslash text\{B\}\}\}\{V\_\{\backslash text\{A\}\}\}\}$.