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:

Kwoning the value for q_rev as derived for isothermal reversible volume changes of an ideal gas, we have

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 ΔS_syst, ΔS_surr and ΔS_univ 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

In the calculation we used q_rev not the actual value q for the reversible process. Since ΔU = 0 for an isothermal process in an ideal gas and since P_ext is constant

Since the surroundings remain at equilibrium

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 V_i to V_i. 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.

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, ΔS_env = 0. The total change in the entropy (gas plus envoirnment) for the free expansion is

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 (ΔS^tot > 0).

Entropy change for the isothermal expansion of a perfect gas

We focus now on the entropy change of a sample of perfect gas when it expands isothermally from a volume V_i to a volume V_f: in a reversible expansion, irreversible expansion in a vacuum and expansion against a low but non zero pressure. The definition of entropy instructs us to find the energy supplied as heat for a reversible path between the stated initial and final states regardless of the actual manner in which the process takes place. A simplification is that the expansion is isothermal, so the temperature is a constant and may be taken outside the integral in eqn 3.2. The energy absorbed as heat during a reversible isothermal expansion of a perfect gas can be calculated from ∆U = q + w and ∆U = 0, which implies that q = −w in general and therefore that q_rev = −w_rev for a reversible change. The work of reversible isothermal expansion was calculated in Section 2.3. Because the temperature is constant, eqn 3.2 becomes

From eqn 2.11, we know that

It follows that

The heat has been supplied by the surroundings whose entropy has decreased by an amount ΔS = R ln(V_f/V_i). This heat exaclty compensates the work done by the gas and allows the temperature to remain constant). Like for any reversible process these two entropy variations exaclty cancel out, and the entropy of the universe does not change.

Expansion in a vacuum

As the work done by the gas is zero, no heat is exchanged with the surroundings (dU = δq + dw = 0). The entropy change of the gas is the same as for the reversible case, however, without any heat exchange, the entropy of the surroundings does not vary at all. The entropy increase of the universe is thus ΔS = R ln(V_f/V_i).

Expansion against a lower pressure

The work done by the gas cannot be known. We can just state −w_rev < w < 0 The entropy change of the system is unambiguosly known ΔS = R ln(V_f/V_i). In contrast that of the surroundings cannot be calculatet because

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