Reversible constant volume processes

The equations which apply to a mechanically reversible constant-volume process were developed in Sec. 2.10. No simplification results for an ideal gas. Thus for one mole ... 

For a mechanically reversible, constant-volume process, this result may be combined with Eq. (2.17) to give... 

In analogy to the constant-pressure process, constant temperature is defined as meaning that the temperature T of the surroundings remains constant and equal to that of the system in its initial and final (equilibrium) states. First to be considered are constant-temperature constant-volume processes (again Aw = 0). For a reversible process... 

In most processes, a reversible absorption of heat is accompanied by a change in temperature, and a calculation of the corresponding entropy change requires an evaluation of the integral of q/T. The term q is related to the heat capacity of the system which is usually expressed as a function of temperature. In a constant volume process, for example, the entropy change is... 

Thus for a mechanically reversible, constant-volume, nonflow process, the heat transferred is equal to the internal-energy change of the system. 

Thus for a mechanically reversible, constant-pressure, nonflow process, the heat transferred equals the enthalpy change of the system. Comparison of the last two equations with Eqs. (2.16) and (2.17) shows that the enthalpy plays a role in constant-pressure processes analogous to the internal energy in constant-volume processes. 

Note that /I must be a state variable since it is a combination of state variables.) The work required to bring the system from state 1 to state 2 by a reversible (i.e., S en = 0). isothermal, constant-volume process is... 

One should not conclude from Eq 4.2-7 that the reversible work for any process is equal to the change in Helmholtz energy, since this result was derived only for an isothermal, constant-volume process. The value of VK , and the thermodynamic functions to which it is related, depends on the constraints placed on the system during the change of state (see Problem 4.3). For example, consider a process occurring in a closed system at fi.xed temperature and pressure. Here we have... 

Gases behave differently from liquids and must be characterized by an equation of state. Polytropic processes are studied in thermodynamics (Saad, 1966). Essentially, pv = constant, say C, where v is the specific volume also the index n of the proeess may vary from -oo to -too. For constant pressure processes, n = 0 for isothermal processes assuming perfect gases, n = 1. For reversible adiabatic processes, n = Cp/Cv, where Cp is the specific heat at constant pressure and Cv is the value obtained at constant volume. Finally, for constant volume processes, n = co. 

We now show that an irreversible adiabatic process must lead to a higher temperature than the reversible adiabatic process starting at the same initial state (state 1). The solid curve in Figure 3.7a represents the reversible adiabat passing through state 1. We first assume that state 2 lies below this curve (an assertion that we want to disprove). Let state 3 be the state on the reversible adiabat that has the same volume as state 2. After the irreversible step 1 has occurred, we carry out a reversible constant-volume step from state 2 to state 3 (step 2). For a constant-volume process. 

Returning to our piston and cylinder in Figure 2.7, we ask what is the minimum or maximum work for a reversible, isothermal, constant volume process at TJTi-a) = 1 in such a device. Dropping the unnecessary terms from Eq. 2.58, we find... 

The combination of fundamental variables in equation (l.23) that leads to the variable we call G turns out to be very useful. We will see later that AG for a reversible constant temperature and pressure process is equal to any work other than pressure-volume work that occurs in the process. When only pressure-volume work occurs in a reversible process at constant temperature and pressure, AG = 0. Thus AG provides a criterion for determining if a process is reversible. Again, since G is a combination of extensive state functions... 

The velocity uw = fkP2v2 is shown to be the velocity of a small pressure wave if the pressure-volume relation is given by Pifi = constant. If the expansion approximates to a reversible adiabatic (isentropic) process k y, the ratio of the specific heats of the gases, as indicated in equation 2.30. 

Calculate the work for each of the following processes. A sample of gas begins in a cylinder fitted with a piston with a volume of 3.42 L, at 298 K and a pressure of 2.57 atm, and expands to a final volume of 7.39 L by two different pathways. Path A is an isothermal, reversible expansion. Path B involves two steps. In the first, the gas is cooled at constant volume... 

In the reversible process through path 1-3-f, heat transfer must be reversible. The system temperature is reduced in order to lower the pressure at constant volume. Therefore, heat must be transferred from an infinite number of heat reservoirs in the surroundings, all at a lower temperature than the final temperature of the system. We will see when we discuss entropy, in Chapter 3, that this is a very significant difference. 

The specific heats have simple formulas. At constant volume, the heat absorbed equals the increase of internal energy, since no work is done. Since the heat absorbed also equals the temperature times the change of entropy, for a reversible process, and since the heat capacity at constant volume Cv is the heat absorbed per unit change of temperature at constant volume, we have the alternative formulas... 

For a vstem at constant pressure and temperature, we see that the Gibbs free energy is constant for a reversible process but decreases for an irreversible process, reaching aminimum value consistent with the pressure and temperature for the equilibrium state just as for a system at constant volume the Helmholtz free energy is constant for a reversible process but decreases for an irreversible process. As with A, we can get the equation of state and specific heat from the derivatives of <7, in equilibrium. We have... 

One mole of an ideal gas, initially at 20°C and 1 bar, undergoes the following mechanically reversible changes. It is compressed isothermally to a point such that when it is heated at constant volume to 100 0 its final pressure is lObar. Calculate Q, W, AU, and AH for the process. Take... 

Steam at 2,100 kPa and with a quality of 0.85 undergoes a reversible, adiabatic expansion in. nonflow process to 350 kPa. It is then heated at constant volume until it is saturated vapor. Deter Q and W for the process. 

A certain gas obeys the equation of state P(V -nb) - nRT and has a constant volume heat capacity, Cv, which is independent of temperature. The parameter b is a constant. For 1 mol, find W, AE, Q, and AH for the following processes (a) Isothermal reversible expansion. (b) Isobaric reversible expansion. (c) Isochoric reversible process, (d) Adiabatic reversible expansion in terms of Tlf Vlt V2, Cp, and Cv subscripts of 1 and 2 denote initial and final states, respectively. (c) Adiabatic irreversible expansion against a constant external pressure P2, in terms of Plf P2, Tj, and 7 = (Cp/Cy). 

Figure P3.14 depicts two mechanically reversible processes undergone by 1 mol of an ideal gas. Curves and are isotherms, paths 2-3 and 5-6 are isobars, and paths 3-1 and 6-4 are at constant volume. Show that W and Q are the same for processes 1-2-3-1 and 4-5-6-4. 

As another example that relates to comparisons between processes executed reversibly or irreversibly, consider at constant volume the conversion of hydrogen and oxygen into steam over a platinum catalyst, and compare this with the explosive conversion by an electric spark. Are any changes needed in the derivations ... 

Consider now an irreversible process in a closed system wherein no heat transfer occurs. Such a process is represented on the P V diagram of Fig. 5.6, which shows an irreversible, adiabatic expansion of 1 mol of fluid from an initial equilibrium state at point A to a final equilibrium state at pointB. Now suppose the fluid is restored to its initial state by a reversible process consisting of two steps first, the reversible, adiabatic (constant-entropy) compression of tile fluid to tile initial pressure, and second, a reversible, constant-pressure step that restores tile initial volume. If tlie initial process results in an entropy change of tlie fluid, tlien tliere must be heat transfer during tlie reversible, constant-P second step such tliat ... 

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