Editing Isothermal process (section)

==Entropy changes==

Isothermal processes are especially convenient for calculating changes in [[entropy]] since, in this case, the formula for the entropy change, Δ''S'', is simply
:<math>\Delta S = \frac{Q_\text{rev}}{T}</math>
where ''Q''<sub>rev</sub> is the heat transferred (internally reversible) to the system and ''T'' is [[thermodynamic temperature|absolute temperature]].<ref name="Atkins4">{{cite book|last=Atkins |first=Peter |date=1997 |title=Physical Chemistry |url=https://archive.org/details/physicalchemistr00atki |url-access=registration |edition=6th |chapter=Chapter 4: The second law: the concepts |location=New York, NY |publisher=W. H. Freeman and Co |isbn=0-7167-2871-0}}</ref> This formula is valid only for a hypothetical [[Reversible process (thermodynamics)|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 [[Thermodynamic databases for pure substances#Enthalpy change of phase transitions|enthalpy of transformation]], Δ''H''<sub>tr</sub>, thus ''Q''&nbsp;=&nbsp;Δ''H''<sub>tr</sub>.<ref name="Petrucci"/> At any given pressure, there will be a transition temperature, ''T''<sub>tr</sub>, 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<ref name="Atkins4"/>
:<math>\Delta S_\text{tr} = \frac{\Delta H_\text{tr}}{T_\text{tr}}</math>.
Another example is the reversible isothermal expansion (or compression) of an [[ideal gas]] from an initial volume ''V''<sub>A</sub> and pressure ''P''<sub>A</sub> to a final volume ''V''<sub>B</sub> and pressure ''P''<sub>B</sub>. As shown in [[#Calculation of work|Calculation of work]], the heat transferred to the gas is
:<math>Q = -W = n R T \ln{\frac{V_{\text{B}}}{V_{\text{A}}}}</math>.
This result is for a reversible process, so it may be substituted in the formula for the entropy change to obtain<ref name="Atkins4"/>
:<math>\Delta S = n R \ln \frac{V_{\text{B}}}{V_{\text{A}}}</math>.
Since an ideal gas obeys [[Boyle's law]], this can be rewritten, if desired, as
:<math>\Delta S = n R \ln \frac{P_{\text{A}}}{P_{\text{B}}}</math>.
Once obtained, these formulas can be applied to an [[irreversible process]], such as the [[Joule expansion|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]] (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 result ''Q''&nbsp;=&nbsp;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 Δ''S''<sub>sur</sub>&nbsp;=&nbsp;−{{sfrac|''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 irreversible, ''Q''&nbsp;=&nbsp;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.