Expansion of a perfect gas

We have already calculated, in Section 3.5, the changes in thermodynamic properties accompanying the isothermal expansion of a perfect gas. As 

We may also expand a gas adiabatically so that no heat is exchanged with the surroundings during the expansion. The temperature of the gas will not remain constant during such an adiabatic expansion. Now 

If the gas is in an initial state Tlt Vx and expands to a final state T2 V2 we can integrate this equation (if Cv is assumed to be independent of temperature) to 

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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. 

For the expansion of a perfect gas at constant temperature, the reversible work is given by the expression ... 

For the reversible adiabatic expansion of a perfect gas, the change in energy content is related to the change in volume by... 

This case should be contrasted with the previous one, namely, the production of maximum work by the expansion of a perfect gas In the case of the gas, the heat which had to be added m order to keep the temperature of the gas constant dunng the expansion was quantitatively converted into the work done In the case of a perfect gas, as we shall see later, there is no change in the internal energy when expansion takes place That is, U = o and therefore A Q,n the heat added from the surroundings is exactly equivalent to the work performed by the gas In short, the gas acts as a transformer of heat into work... 

The isothermal expansion of a perfect gas is therefore a means for converting heat into work quantitatively. 

In both the examples we have considered earlier (the expansion of a perfect gas into a vacuum, and the flow of heat) the system loses the capacity to do work. This lost capacity is clearly related to dq, because dU — dq + dw (see Section 2,5). It is also related to temperature, for if we consider the flow of heat, q, from a hot to a cold reservoir and to a warm reservoir as illustrated in Fig. 3.3 the loss of capacity to do work is clearly greater in the former case. In the latter case work could be obtained from the flow of heat from Tw Tc as this will be a spontaneous process. 

As an example of an entropy change let us consider the isothermal expansion of a perfect gas. From Section 2.5 we know AU = q + w. Since for a perfect gas U is independent of volume, AU - 0 and q — - w the heat gained from the surroundings is equal to the work done by the system. If, as the gas expands and its pressure drops, the external pressure is continuously adjusted so that P = Pex + dP then the expansion can be carried out reversibly, doing the maximum work, and... 

However, if we consider the irreversible expansion of a perfect gas into a vacuum (doing no work) then ASovwaH will no longer be zero. For the gas itself AS = nRln VB/VA however, the changes in the surroundings will be different. The gas does no work and as q - — w = 0, the system absorbs no heat... 

It should be clear-that entropy is related to the lack of uniformity within systems. As gradients in temperature or concentration are eliminated the entropy increases. This idea can be placed on a quantitative basis if we consider a specific case the expansion of a perfect gas. This discussion involves a consideration of the molecular nature of matter. It is not essential to the development of classical thermodynamics, but provides valuable insight. 

Referring back to Fig. 16, we assume that the fraction of space between the globs just above the melting point is AF/F, and this is the volume available for perfect gas atoms. However, since the thickness of the inter-globular shell is assumed to maintain a constant value r, then the space is a two-dimensional gas. Since the bulk coefficient of expansion of a perfect gas at constant pressure is IjT, that of a two-dimensional gas is fT. So the bulk coefficient of thermal expansion of a liquid just above the melting point should be obtained by prorating the expansions due to the solid fraction 1 — AF/F and the gaseous fraction AF/F ... 

Indirect results are liable to another source of error. The formula employed may be so inexact that accurate measurements give but grossly approximate results. For instance, a first approximation formula may have been employed when the accuracy of the observations required one more precise r = V may have been put in place of ir = 3 14159 or the coefficient of expansion of a perfect gas has been applied to an imperfect gas. Such errors are called errors of method. 

Some of the basic principles of the calculus of molecular chaos may be illustrated in a more direct way by the detailed consideration of the expansion of a perfect gas. 

Another preliminary example of the application of (1 21) is the isothermal expansion of a perfect gas from a volume to a volume For the ith molecule let be the number of complexions or quantum... 

Now we have seen from the molecular standpoint that m the case of a perfect gas, no internal work is done in expanding (though there is internal energy represented by U), and hence the latent heat of expansion at constant temperature Idv for the small volume change is identical with the external work pdv, at constant temperature Hence for a perfect gas the above equation reduces to— dU = Cvdi... 

Remember that A or dA, i e the area ABCD, only represents the external or useful work In the case of a perfect gas, we have no other kind of work to deal with. For an imperfect gas, or a liquid, etc, expansions and compressions involve internal work as well These, however, do not enter into the discussion, and do not vitiate the generality of the results of equations (3) and (4) (p 49). As long as... 

We have already seen (p 43) that for the expansion c>zr of a perfect gas against a pressure the following holds good, viz —... 

If one mole of a perfect gas is expanded from a volume of 0.01 m3 to 0.10m3, the entropy change is AS = 8.3 In 10 = 19.1 JK"1 mol" This relation will be correct for both reversible and irreversible changes as entropy is a state function and AS is independent of the path taken between the states A and B. The heat lost by the surroundings is equal to the heat gained by the gas and thus for a reversible expansion ASOTe , - 0 (as ASoverai, = qtJT - qteJT). 

Now in the problem discussed in 1 14 it was seen that a purely thermodynamic result for the entropy increase in the isotWmal expansion of one mole of a perfect gas is... 

Techniques which are based on the supersonic expansion of a non condensable buffer gas do not suffer from this problem. They have been widely used in experimental science, for example in the first attempt to liquefy gases or, much more recently, as a source of cold molecules for spectroscopy (see the chapter by Davis et al.) or of molecular beams for collision dynamics studies. Cooling arises from conservation of energy which implies, that for a gas flow under adiabatic conditions, the sum of specific enthalpy and kinetic energy remains constant during the expansion. For a perfect gas with a constant specific heat capacity Cp and with a well defined temperature, the energy conversion is driven by the following equation ... 

We have seen that a feature of a perfect gas is that for any isothermal expansion the total energy of the sample remains the same and that q = -w. That is, any energy lost as work is restored by an influx of energy as heat. We can express this property in terms of the internal energy, for it implies that the internal energy remains constant when a perfect gas expands isothermally from eqn 1.6 we can write... 

Air at 290 K is compressed from 101.3 to 2000 kN/m2 pressure in a two-stage compressor operating with a mechanical efficiency of 85%. The relation between pressure and volume during the compression stroke and expansion of the clearance gas is PV1 25 = constant. The compression ratio in each of the two cylinders is the same and the interstage cooler may be taken as perfectly efficient. If the clearances in the two cylinders are 4% and 5% respectively, calculate ... 

A perfect" gas is one which closely conforms id the simple gas laws" of expansion and contraction, such as Boyle s Law, formulated in England by The Hon Robert Boyle (1627-169D (Ref 1, p 141-R) and called in France Mariotte s Law, because it was formulated independently from Boyle by Edme Mariotte (1629-1684) (Ref 1, p 515-L). This law, called in Germany and Russia... 

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