Ideal gas isothermal process

Figure 3.4 Carnot cycle for the expansion and compression of an ideal gas. Isotherms alternate with adiabats in a reversible closed path. The shaded area enclosed by the curves gives the net work in the cyclic process.

Determine the fraction of throughput energy required to compress hydrogen gas from 0.1 to 70 MPa assuming an isothermal, ideal gas compression process. How would nonideal gas effects affect the result ... 

It follows that the efficiency of the Carnot engine is entirely determined by the temperatures of the two isothermal processes. The Otto cycle, being a real process, does not have ideal isothermal or adiabatic expansion and contraction of the gas phase due to the finite thermal losses of the combustion chamber and resistance to the movement of the piston, and because the product gases are not at tlrermodynamic equilibrium. Furthermore the heat of combustion is mainly evolved during a short time, after the gas has been compressed by the piston. This gives rise to an additional increase in temperature which is not accompanied by a large change in volume due to the constraint applied by tire piston. The efficiency, QE, expressed as a function of the compression ratio (r) can only be assumed therefore to be an approximation to the ideal gas Carnot cycle. 

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. 

The entropy changes ASa and ASB can be calculated from equation (2.69), which applies to the isothermal reversible expansion of ideal gas, since AS is independent of the path and the same result is obtained for the expansion during the spontaneous mixing process as during the controlled reversible expansion. Equation (2.69) gives... 

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... 

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. 

Mass and energy transport occur throughout all of the various sandwich layers. These processes, along with electrochemical kinetics, are key in describing how fuel cells function. In this section, thermal transport is not considered, and all of the models discussed are isothermal and at steady state. Some other assumptions include local equilibrium, well-mixed gas channels, and ideal-gas behavior. The section is outlined as follows. First, the general fundamental equations are presented. This is followed by an examination of the various models for the fuel-cell sandwich in terms of the layers shown in Figure 5. Finally, the interplay between the various layers and the results of sandwich models are discussed. 

Assuming ideal gas behaviour, calculate the reactor volume needed for a 90% conversion of A if the process is conducted (i) isothermally at 1000 K and (ii) adiabatically with an inlet temperature of HOOK. 

Theoretical methods allow making such calculations for ideal and real gases and gas mixtures under isothermal and frictionless adiabatic (isentropic) conditions. In order that results for actual operation can be found it is neecessary to know the efficiency of the equipment. That depends on the construction of the machine, the mode of operation, and the nature of the gas being processed. In the last analysis such information comes from test work and its correlation by manufacturers and other authorities. Some data are cited in this section. 

It might seem that, for the derivation of kinetic equations describing variations in the amounts of substance eqns. (17) and (18), the equation of state is unnecessary. But this is not so. In the case of a variable reaction volume, it may be necessary to express gas-phase substance concentrations through their amounts, since step rates w are specified as functions of concentrations. For isobaric isothermal processes and ideal gases cg = Ne/ V = PN IN otRT. 

We now consider the isothermal, quasistatic expansion and compression of the gas for several processes. The curve AE in Figure 3.1 represents a pressure-volume isotherm of an ideal gas at a given temperature. The... 

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... 

One specific reversible expansion of an ideal gas that will be of particular interest to us is the one in which the system remains at constant temperature (by being immersed in a thermostat). Such a process is called isothermal. For this case, we can use Eq. (6), with P = nRT/V ... 

Calculate q, w, AU, and AH when 1 mol of an ideal gas expands isothermally from a pressure Px to a pressure P2 against a constant opposing pressure of P2. Put your results in terms of Pu P2 and T. Note that in this process, Pext is constant, but P is not constant. 

In Frame 9 we consider the expansion of an ideal gas along an isotherm (or constant temperature curve for which dT = 0) from (Pi, V ) to (Pf, Vf). Whilst the state functions P and V show identical changes in the two expansion processes considered (dP = Pf — Pi) dV = (Vf — V) the work done in the two cases is entirely different. Two routes (paths) are considered ... 

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 ... 

The internal energy of an ideal gas cannot change in an isothermal process. Thus for one mole of an ideal gas in any nonflow process,... 

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... 

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