Reversible isothermal expansion

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

From example 2.3 we saw that for the reversible isothermal expansion of ideal gas... 

It is useful to compare the reversible adiabatic and reversible isothermal expansions of the ideal gas. For an isothermal process, the ideal gas equation can be written... 

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. 

Figure 3.1 Comparison of reversible isothermal and adiabatic (C =jR) ideal gas expansions.

A reversible isothermal expansion of the ideal gas is made from an initial volume V to a volume Vz at an absolute (ideal gas) temperature 73. The amount of pressure-volume work in done by the system is obtained by substituting into Equation (2.16). The result is... 

To calculate the work of reversible, isothermal expansion of a gas, we have to use calculus, starting at Eq. 3 written for an infinitesimal change in volume, dV ... 

SOLUTION For expansion against constant external pressure we use Eq. 3 and for reversible, isothermal expansion we use Eq. 4 ... 

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. 

In Section 6.3 we saw how to calculate the work of reversible, isothermal expansion of a perfect gas. Now suppose that the reversible expansion is nor isothermal and that the temperature decreases during expansion, (a) Derive an expression for the work when T = Tinitja — c( V — Vjnitia ), with c a positive constant, (b) Is the work in this case greater or smaller than that of isothermal expansion Explain your observation. 

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. 

Self-Test 7.16A Determine AS, ASsllrr, and AStot for (a) the reversible, isothermal expansion and (b) the isothermal free expansion of 1.00 mol of ideal gas molecules... 

Step I. A reversible isothermal expansion in thermal contact with the high-temperature reservoir at t2-... 

Reversible Isothermal Expansion Let us consider the heat and work of ideal gas expansion from V to V2 under isothermal conditions (AT = 0). We recognize from (3.74a) that... 

Figure 4.3 Reversible Camot cycle, showing steps (1) reversible isothermal expansion at th (2) reversible adiabatic expansion and cooling from th to tc (3) reversible isothermal compression at tc (4) reversible adiabatic compression and heating back to the original starting point. The total area of the Camot cycle, P dV, is the net useful work w performed in the cyclic process (see text).

The gas first undergoes reversible isothermal expansion at th, with energy change (as required by the first law)... 

This includes an ideal gas in a piston, weights, and other necessary paraphernalia to carry out the following reversible operations. First, the gas piston is brought into thermal contact with the Th reservoir, and the gas undergoes reversible isothermal expansion at Th (lifting weights as necessary) to withdraw heat q from this reservoir, where... 

This includes a transporter piston T, weights, etc., to carry out the following reversible operations, all under isothermal conditions. Attach the empty transporter piston T to the high-pressure Ph reservoir. Reversibly remove an aliquot of n moles into T at constant Ph (doing no work), up to volume Vh. Detach T from the Ph reservoir and allow reversible isothermal expansion of the transporter gas from Vh to Vh performing work... 

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. 

We have found that, provided the gas is ideal, the work of reversible, isothermal expansion from the volume Vinitial to the volume Vfinal is... 

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 expansion of an ideal gas in the Joule experiment will be used as a simple example. Consider a quantity of an ideal gas confined in a flask at a given temperature and pressure. This flask is connected through a valve to another flask, which is evacuated. The two flasks are surrounded by an adiabatic envelope and, because the walls of the flasks are rigid, the system is isolated. We now allow the gas to expand irreversibly into the evacuated flask. For an ideal gas the temperature remains the same. Thus, the expansion is isothermal as well as adiabatic. We can return the system to its original state by carrying out an isothermal reversible compression. Here we use a work reservoir to compress the gas and a heat reservoir to remove heat from the gas. As we have seen before, a quantity of heat equal to the work done on the gas must be transferred from the gas to the heat reservoir. In so doing, the value of the entropy function of the heat reservoir is increased. Consequently, the value of the entropy function of the gas increased during the adiabatic irreversible expansion of gas. 

Calculate the work done by the reversible, isothermal expansion of 3 moles of an ideal gas from 100 litres initial volume to 300 litres final volume. 

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