Isothermal expansion entropy change

The entropy changes ASa and ASB can be calculated from equation (2.69), which applies to the isothermal reversible expansion of ideal gas, since AS is independent of the path and the same result is obtained for the expansion during the spontaneous mixing process as during the controlled reversible expansion. Equation (2.69) gives... 

Because entropy is a state function, the change in entropy of a system is independent of the path between its initial and final states. This independence means that, if we want to calculate the entropy difference between a pair of states joined by an irreversible path, we can look for a reversible path between the same two states and then use Eq. 1 for that path. For example, suppose an ideal gas undergoes free (irreversible) expansion at constant temperature. To calculate the change in entropy, we allow the gas to undergo reversible, isothermal expansion between the same initial and final volumes, calculate the heat absorbed in this process, and use it in Eq.l. Because entropy is a state function, the change in entropy calculated for this reversible path is also the change in entropy for the free expansion between the same two states. 

Some changes are accompanied by a change in volume. Because a larger volume provides a greater range of locations for the molecules, we can expect the positional disorder of a gas and therefore its entropy to increase as the volume it occupies is increased. Once again, we can use Eq. 1 to rum this intuitive idea into a quantitative expression of the entropy change for the isothermal expansion of an ideal gas. 

We choose an isothermal expansion because both temperature and volume affect the entropy of a substance at this stage, we do not want to have to consider changes in both temperature and volume. 

The entropy change accompanying the isothermal compression or expansion of an ideal gas can be expressed in terms of its initial and final pressures. To do so, we use the ideal gas law—specifically, Boyle s law—to express the ratio of volumes in Eq. 3 in terms of the ratio of the initial and final pressures. Because pressure is inversely proportional to volume (Boyle s law), we know that at constant temperature V2/Vj = E /E2 where l is the initial pressure and P2 is the final pressure. Therefore,... 

FIGURE 7.9 The energy levels of a particle in a box (a) become closer together as the width of the box is increased, (b) As a result, the number of levels accessible to the particles in the box increases, and the entropy of the system increases accordingly. Die range of thermally accessible levels is shown by the tinted band. The change from part (a) to part (b) is a model of the isothermal expansion of an ideal gas. The total energy of the particles is the same in each case. 

STRATEGY Because entropy is a state function, the change in entropy of the system is the same regardless of the path between the two states, so we can use Eq. 3 to calculate AS for both part (a) and part (b). For the entropy of the surroundings, we need to find the heat transferred to the surroundings. In each case, we can combine the fact that AU = 0 for an isothermal expansion of an ideal gas with AU = w + q and conclude that q = —tv. We then use Eq. 4 in Chapter 6 to calculate the work done in an isothermal, reversible expansion of an ideal gas and Eq. 9 in this chapter to find the total entropy. The changes that we calculate are summarized in Fig. 7.21. 

Calculate the entropy change associated with the isothermal expansion of 5.25 mol of ideal gas atoms from 24.252 L to 34.058 L. 

The first consists of two steps (1) an isothermal reversible expansion at the temperature Ta until the volume V is reached, and (2) an adiabatic reversible expansion from V to Vj,. The entropy change for the gas is given by the sum of the entropy changes for the two steps ... 

EXAMPLE 7.3 Calculating the entropy change for the isothermal 1 expansion of an ideal gas... 

The change in entropy AS for a reversible isothermal expansion of an ideal gas from its initial volume Vj to a volume V2 is AS - R In(V2IVj) and therefore V2IVj = exp(ASIR). By setting V2IVt equal to the ratio between the molar volume Vq =... 

What we have accomplished here is to use the definition of entropy in terms of probability to derive an expression for AS that depends on volume, a macroscopic property of the gas. We can now relate the change in entropy to heat flow by noting the striking similarity between the above equation for AS and the one derived in Section 10.2 describing qrev for the isothermal expansion-compression of an ideal gas. Compare... 

It is easy to calculate entropy changes for isothermal processes, because T is constant and comes outside the integral to give AS = q ev/T. A specific example is the isothermal compression or expansion of an ideal gas, for which AS = nR InlVf/V ). A second example is any phase transition at constant pressure for which q, y = The entropy change is then AS ang =... 

Suppose an isothermal expansion is carried out in which the volume of the ideal gas is changed from V to F2. If the corresponding entropies are Si and 82, and the probabilities are Wi and W-i, it follows from equation (24.2) that the entropy change accompanying the process is given by... 

Isothermal expansion at a high temperature 9 , absorbing heat gj. The change of entropy of the working fluid A5 = QilQi by Equation (4.31). 

As an example of an entropy change let us consider the isothermal expansion of a perfect gas. From Section 2.5 we know AU = q + w. Since for a perfect gas U is independent of volume, AU - 0 and q — - w the heat gained from the surroundings is equal to the work done by the system. If, as the gas expands and its pressure drops, the external pressure is continuously adjusted so that P = Pex + dP then the expansion can be carried out reversibly, doing the maximum work, and... 

Bulk modulus can be treated from the adiabatic as well as the isothermal point of view. Phenomenologically adiabatic compression or expansion are processes where heat is neither lost to nor gained from the environment. If the process occurs under equilibrium conditions, then we have the thermodynamically tractable case at zero entropy change and we define the bulk modalui as... 

Now we may consider an isothermal expansion, also to a neighbouring state of equilibrium. This may be imagined to occur in two stages. First, the change of volume, dV, takes place adiabaticaUy, the change in entropy being zero, so that... 

---