Entropy Changes of Isothermal Processes
Reversible Processes in Closed Systems
Since the process is reversible we integrate along the actual path of the process, and since T is constant we can factor 1/T out of the integral:
∆S2 = ∫c δqrev/T = 1/T ∫c δqrev = qrev/T
Kwoning the value for qrev as derived for isothermal reversible volume changes of an ideal gas, we have
∆S2 = nR ln(V2/V1) (id. gas, reversible isothermal process)
Entropy Changes for Irreversible Processes
If a system undergoes any irreversible process with equilibrium initial and final states, we can calculate the entropy change of the system using a reversible process that has the same initial and final states as the irreversible process as long as it has equilibrium or metastable initial and final states.
Example 2.1. Calculate ΔSsyst, ΔSsurr and ΔSuniv if 3.00 mol of helium (assumed ideal) expands isothermally and irreversibly at 298.15 K from a volume of 10.00 L to a volume of 30.00 L at a constant external pressure of 1.0000 atm. Assume that the surroundings remain at thermal equilibrium at 298.15 K.
Solution. The initial and final states of the system are the same as for a reversible isothermal expansion, so that
ΔS = qrev/T = qirrev/T = (3.000 mol) (8.3145 JK−1) ln (30.00L/10.00L) = 18.26 JK−1
In the calculation we used qrev not the actual value q for the reversible process. Since ΔU = 0 for an isothermal process in an ideal gas and since Pext is constant
qsurr = −q = w = −Pext ΔV = −(101325 Nm-2)(0.020m3) = −2056.5 J
Since the surroundings remain at equilibrium
ΔSsurr = −2056J/298.15K = 6.80 JK-1
ΔSuniv = 18.26 JK−1 − 6.80 JK−1 = 11.46 K−1
The Free Expansion of an Ideal Gas
We now consider the free expansion (i.e., expansion against zero atmospheric pressure) of an ideal gas from Vi to Vi. We have already seen that that the free expansion of an ideal gase is also isothermal. Thus, the final state of the free expansion is the same as the state of the isothermal reversible expansion process (i.e., both states have the same V and T). In this case, however, no work is done by the gas against the piston, since the weight on the piston is rapidly removed. Thus ΔU =0 and w=0, which by the First Law means that q=0 as well. Since entropy is a state function, the change of entropy of the gas for the free expansion (irriversible process) must be the same as that for the reversible process.
ΔSgas = Sf - Si = R ln(Vf/Vi)
Siccome S è una funzione di stato la sua variazione è uguale nel caso reversibile e in quello irreversibile. Allora un'espansione isoterma reversibile ed una irreversibile in che modo differiscono? La risposta risiede nel valore di ΔS dell'ambiente. Since no thermal energy leaves the heat reservoir, ΔSenv = 0. The total change in the entropy (gas plus envoirnment) for the free expansion is
ΔStot = ΔSenv + ΔSsystem = ΔSgas = nR ln(Vf/Vi)
In the case of free expansion, there is no decrease in entropy of the heat reservoir, since no thermal energy was absorbed by the ideal gas, since it did not work. Thus, for the free expansion process, the entropy of the universe increases (ΔStot > 0).