Isothermal Expansion of an Ideal Gas

It has been shown [Equation (6.72)] that in the reversible isothermal expansion of an ideal gas 

As 5 is a thermodynamic property, ASsys is the same in an irreversible isothermal process from the same initial volume Vi to the same final volume V2. However, the change in entropy of the surroundings differs in the two types of processes. First let us consider an extreme case, a free expansion into a vacuum with no work being performed. As the process is isothermal, AU for the perfect gas must be zero consequently, the heat absorbed by the gas Q also is zero  

the surroundings have given up no heat and have undergone no change. Consequently, 

In other words, for the system plus surroundings, this irreversible expansion has been accompanied by an increase in entropy. 

We may contrast this result for A totai with that for Al/totai for an ideal gas, as mentioned in Section 5.1. In the irreversible expansion of an ideal gas, Allgys = 0 the surroundings undergo no change of state (Q and W are both equal to zero), and hence, A /total = 0- ff we consider the reversible expansion of the ideal gas, AUsys is also equal to zero and AUsun is equal to zero because Q = —W, so again A /total = 0- Clearly, in contrast to AS, AU does not discriminate between a reversible and an irreversible transformation. 

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Illustrative Example.—Consider the isothermal expansion of an ideal gas between fixed limits of volume ... 

The isothermal expansion of an ideal gas is an aschistic process.— If a mass of gas expands isothermally, the heat absorbed is equal to the external work done. 

Calculation of AS for the Reversible Isothermal Expansion of an Ideal Gas Integration of equation (2.38) gives... 

For a horizontal pipe, dz = 0, and for isothermal expansion of an ideal gas dH = 0. Thus if the system does no work on the surroundings ... 

We have found that the work of reversible, isothermal expansion of an ideal gas from the volume Vlnitia to the volume Vflna is... 

The work done by any system on its surroundings during expansion against a constant pressure is calculated from Eq. 3 for a reversible, isothermal expansion of an ideal gas, the work is calculated from Eq. 4. A reversible process is a process that can be reversed by an infinitesimal change in a variable. 

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 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. 

Figure 5.2. The path in P- V space taken in areversible isothermal expansion of an ideal gas. The area between the dashed lines under the curve represents the negative of the work performed.

A hypothetical cycle for achieving reversible work, typically consisting of a sequence of operations (1) isothermal expansion of an ideal gas at a temperature T2 (2) adiabatic expansion from T2 to Ti (3) isothermal compression at temperature Ti and (4) adiabatic compression from Ti to T2. This cycle represents the action of an ideal heat engine, one exhibiting maximum thermal efficiency. Inferences drawn from thermodynamic consideration of Carnot cycles have advanced our understanding about the thermodynamics of chemical systems. See Carnot s Theorem Efficiency Thermodynamics... 

Solution For isothermal expansion of an ideal gas, both A U and A(PV) are zero, so that... 

So far, we have been looking at systems for which the external pressure was constant. Now let s consider the case of a gas that expands against a changing external pressure. In particular, we consider the very important case of reversible, isothermal expansion of an ideal gas. The term reversible, as we shall see in more detail shortly, means that the external pressure is matched to the pressure of the gas at every stage of the expansion. The term isothermal means that the expansion takes place at constant temperature. In an isothermal expansion, the pressure of the gas falls as it expands so to achieve reversible, isothermal expansion, the external pressure must also be gradually reduced in step with the change in volume (Fig. 6.9). To calculate the work, we have to take into account the gradual reduction in external pressure. 

Because A17 = 0 for the isothermal expansion of an ideal gas, and A17 = q + w, it follows that q = —w. Therefore, q = +2.22 kj. This influx of energy as heat counteracts the loss of energy as work and maintains the kinetic energy, and hence the temperature, at a constant value. 

Because the expansion is isothermal, implying that T is constant, we can use Eq. 1 directly. However the formula requires q for a reversible isothermal expansion. We find the value of qrev from the first law, AU = q + w. We know (Section 6.3) that AU = 0 for the isothermal expansion of an ideal gas, so we can conclude that q = —w. The same relation applies if the change is carried out reversibly, so we can write qiev = wiev. Therefore, to calculate qrcv, we calculate the work done when an ideal gas expands reversibly and isothermally, and change the sign. For that, we can use the following expression, which we derived in Section 6.7 ... 

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 =... 

Isothermal expansion of an ideal gas Let us consider that a cylinder fitted with a frictionless piston contains a gas and is heated to the system shown in Figure 1.3. Normally, the temperature of the system increases upon heating, but one wants to maintain a constant temperature for the system during this isothermal process. The piston exerts an external pressure on the system, and the gas expands. Since no temperature change takes place, the change in the internal energy is equal to zero (AE = 0) because the internal energy is only a function of temperature. Then, Equation (1.14) becomes ... 

In a reversible process, Pe t = Pgas = but the relation w = -Pext from Section 12.2 cannot be used to calculate the work, because that expression applies only if the external pressure remains constant as the volume changes. In the reversible isothermal expansion of an ideal gas. 

There is an important consequence of the fact that the energy content of an ideal gas, at constant temperature, is independent of the volume. When an idesil gas expands against an appreciable external pressure, W has a finite value, but AE is zero it follows, therefore, from equation (7.2) that for the isothermal expansion of an ideal gas,... 

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