Reversible processes isothermal expansion

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... 

In general the conditions under which a change in state of a gas takes place are neither isothermal nor adiabatic and the relation between pressure and volume is approximately of the form Pvk = constant for a reversible process, where k is a numerical quantity whose value depends on the heat transfer between the gas and its surroundings, k usually lies between 1 and y though it may, under certain circumstances, lie outside these limits it will have the same value for a reversible compression as for a reversible expansion under similar conditions. Under these conditions therefore, equation 2.70 becomes ... 

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. 

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. 

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).

Consequently, the energy of the gas is constant for the isothermal reversible expansion or compression and, according to the first law of thermodynamics, the work done on the gas must therefore be equal but opposite in sign to the heat absorbed by the gas from the surroundings. For a reversible process the pressure must be the pressure of the gas itself. Therefore, we have for the isothermal reversible expansion of n moles of an ideal gas between the volumes F and V... 

The term d W is the maximum work done by the surroundings, and thus the negative of the maximum work done by the system, excluding the work of expansion or compression. We emphasize that such an interpretation is valid only for an isothermal and isopiestic reversible process. 

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. 

It is proposed to replace this two-step process by a single isothermal expansion of the air from 900 K. and 3 bar to some final pressure P. What is the value of P that makes the work of the proposed process equal to that of the existing process Assume mechanical reversibility and treat air as an ideal gas with CP = (7/2)R and = (5/2)H... 

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). 

To integrate this function, the relationship between pressure and volume must be known. In process design, an estimate of the work done in compressing or expanding a gas is often required. A rough estimate can be made by assuming either reversible adiabatic (isentropic) or isothermal expansion, depending on the nature of the process. For isothermal expansion (expansion at constant temperature) ... 

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. 

The isothermal expansion of a gas can be carried out reversibly by placing the cylinder of gas in a thermostat, as described above, and adjusting the external pressure so as to be less than the pressure of the gas by an infinitesimally small amount. As the gas expands, however, its own pressure decreases, since the temperature is maintained constant. Hence, if the process is to be thermodynamically reversible, it must be supposed that the external pressure is continuously adjusted so as to be ahvays infinitesimally less than the pressure of the gas. The expansion will then take place extremely slowly, so that the system is always in virtual thermodynamic equi-... 

In the two special cases of isothermalj reversible expansion considered above, the work done, as given by equations (8.3) and (8.4) or (8.5), is evidently a definite quantity depending only on the initial and final states, e.g., pressure or volume, at a constant temperature. Since there is always an exact relationship between P and F, it follows from equation (8.2) that the work done in any isothermal, reversible expansion must have a definite value, irrespective of the nature of the system. For an isothermal, reversible process in which the work performed is exclusively work of expansion, it is apparent, therefore, from equation (7.2), that both W and Q will be determined by the initial and final states of the system only, and hence they will represent definite quantities. Actually, this conclusion is applicable to any isothermal, reversible change (cf. 25a), even if work other than that of expansion is involved. 

It will be recalled from the statements in 9d that in an isothermal, reversible expansion of an ideal gas the work done is exactly equal to the heat absorbed by the system. In other words, in this process the heat is completely converted into work. However, it is important to observe that this conversion is accompanied by an increase in the volume of the gas, so that the system has undergone a change. If the gas is to be restored to its original volume by reversible compression, work will have to be done on the system, and an equivalent amount of heat will be liberated. The work and heat quantities involved in the process are exactly the same as those concerned in the original expansion. Hence, the net result of the isothermal expansion and compression is that the system is restored to its original state, but there is no net absorption of heat and no work is done. The foregoing is an illustration of the universal experience, that it is not possible to convert... 

The Linde>Hampson process uses a thermodynamic process. Isothermal compression and subsequent cooling along an isobar is done in a heat exchanger. Joule-Thomson expansion connected with an irreversible change in entropy is used as the refrigeration procedure. Despite its simplicity and reliability, this method is now less attractive compared with new ones where cooling is primarily carried out in reversible processes (expander) and where less energy is required. 

The isothermal process is a process in that dT = 0. In the case of an ideal gas, the expansion may achieved in contact with a thermal reservoir to run an isothermal process. This process occurs as a basic step in the Carnot process, and it is a reversible process. 

What might a reversible isothermal expansion of an ideal gas be This process will occur only if initially, when the gas is confined to half the qdinder, the external pressure acting on the piston exactly balances the pressure exerted by the gas on the piston. If the external pressure is reduced infinitely slowly, the piston will move outward, allowing the pressure of the confined gas to readjust to maintain the pressure balance. This infinitely slow process in which the external pressure and internal pressure are always in equilibrium is reversible. If we reverse the process and compress the gas in the same infinitely slow manner, we can return the gas to its original volume. The complete cy cle of expansion and compression in this hypothetical process, moreover, is accomplished without any net change to the surroundings. 

Now let s consider another example, the expansion of an ideal gas at constant temperature (referred to as an isothermal process). In the cylinder-piston arrangement of Figure 19.5, when the partition is removed, the gas expands spontaneously to fill the evacuated space. Can we determine whether this particular isothermal expansion is reversible or irreversible Because the gas expands into a vacuum with no external... 

P14.3 A liquid water stream (2000kg/h at 25 °C and 1 bar) is compressed to 100 bar in a cooled pump. The process can be assumed to be reversible and isothermal. Calculate the required work and heat duty of the cooling system. The thermal expansion coefficient a = 0.207 10 K and the... 

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