Entropy isothermal expansion

Figure 3.2 compares a series of reversible isothermal expansions for the ideal gas starting at different initial conditions. Note that the isotherms are parallel. They cannot intersect since this would give the gas the same pressure and volume at two different temperatures. Figure 3.3 shows a similar comparison for a series of reversible adiabatic expansions. Like the isotherms, the adiabats cannot intersect. To do so would violate the Caratheodory principle and the Second Law of Thermodynamics, since the gas would have two different entropies at the same temperature, pressure, and volume. 

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. 

We can show that the thermodynamic and statistical entropies are equivalent by examining the isothermal expansion of an ideal gas. We have seen that the thermodynamic entropy of an ideal gas increases when it expands isothermally (Eq. 3). If we suppose that the number of microstates available to a single molecule is proportional to the volume available to it, we can write W = constant X V. For N molecules, the number of microstates is proportional to the Nth power of the volume ... 

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. 

Determine the entropy change for the isothermal expansion or compression of an ideal gas, Example 7.3. 

Entropy is a state function, so AS is the same as for the reversible isothermal expansion, calculated earlier. Because less work is done than in the reversible process and A U is the same in both cases, less heat is withdrawn from reservoirs in the surroundings. Therefore, the decrease of entropy of the surroundings is less than in the reversible expansion. In the limit of expansion against a vacuum (Joule process), no work is done and no heat is withdrawn from the surroundings. In this case, AAsur = 0. 

Fig. 3.9. Isothermal expansion of one mole of an ideal gas resulting in an entropy increase.

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 =... 

Entropy gained by the system of ideal gas during isothermal expansion stage is also... 

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... 

The entropy is diminished by isothermal compression and increased by isothermal expansion. 

From Equation 13.9 we see that the entropy of a gas increases during an isothermal expansion (V2 > Vi) and decreases during a compression (V2 < Vt). Boltzmann s relation (see Eq. 13.1) provides the molecular interpretation of these results. The number of microstates available to the system, H, increases as the volume of the system increases and decreases as volume decreases, and the entropy of the system increases or decreases accordingly. 

In the isothermal expansion of a gas, the final volume V2 is greater than the initial volume Vi and, consequently, by equation (19.26), AS is positive that is to say, the expansion is accompanied by an increase of entropy of the system. Incidentally, when an ideal gas expands (irreversibly) into a vacuum, no heat is taken up from the surroundings ( 9d), and so the entropy of the latter remains unchanged. In this case the net entropy increase is equal to the increase in entropy of the system, i.e., of the gas, alone. 

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... 

Relationship (9.1) is applied to the machine in the Carnot process. During the isothermal expansion step we must deliver that quantity entropy, in the course of expansion. If we would not do so, then we eould not hold the temperature in the machine. The machine requires thus the subsequent delivery of the entropy. The entropy becomes in any case larger after Maxwell s equation, if we increase the volume. 

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