CHAPTER12

At low pressures gases behave as ideal gases and

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Unformatted text preview: elation can be used to determine the entropy change. The proper choice depends on the available data. cen84959_ch12.qxd 4/5/05 3:58 PM Page 664 664 | Thermodynamics Specific Heats cv and cp Recall that the specific heats of an ideal gas depend on temperature only. For a general pure substance, however, the specific heats depend on specific volume or pressure as well as the temperature. Below we develop some general relations to relate the specific heats of a substance to pressure, specific volume, and temperature. At low pressures gases behave as ideal gases, and their specific heats essentially depend on temperature only. These specific heats are called zero pressure, or ideal-gas, specific heats (denoted cv 0 and cp 0), and they are relatively easier to determine. Thus it is desirable to have some general relations that enable us to calculate the specific heats at higher pressures (or lower specific volumes) from a knowledge of cv 0 or cp 0 and the P-v-T behavior of the substance. Such relations are obtained by applying the test of exactness (Eq. 12–5) on Eqs. 12–38 and 12–40, which yields a 0 cv b 0v T b Ta 0 2P b 0T 2 v 0 2v b 0T 2 P (12–42) and a 0 cp 0P T Ta (12–43) The deviation of cp from cp 0 with increasing pressure, for example, is determined by integrating Eq. 12–43 from zero pressure to any pressure P along an isothermal path: 1 cp cp0 2 T P T 0 a 0 2v b dP 0T 2 P (12–44) The integration on the right-hand side requires a knowledge of the P-v-T behavior of the substance alone. The notation indicates that v should be differentiated twice with respect to T while P is held constant. The resulting expression should be integrated with respect to P while T is held constant. Another desirable general relation involving specific heats is one that relates the two specific heats cp and cv. The advantage of such a relation is obvious: We will need to determine only one specific heat (usually cp) and calculate the other one using that relation and the P-v-T data of the substance. We start the development of such a relation by equating the two ds relations (Eqs. 12–38 and 12–40) and solving for dT: dT T 1 0 P> 0 T 2 v dv cp cv T 1 0 v> 0 T 2 P dP cp cv Choosing T T(v, P) and differentiating, we get dT a 0T b dv 0v P a 0T b dP 0P v Equating the coefficient of either dv or dP of the above two equations gives the desired result: cp cv Ta 0v 0P ba b 0T P 0T v (12–45) cen84959_ch12.qxd 4/5/05 3:58 PM Page 665 Chapter 12 An alternative form of this relation is obtained by using the cyclic relation: a 0P 0T 0v ba ba b 0T v 0v P 0P T 1S a 0P b 0T v a 0v 0P ba b 0T P 0v T | 665 Substituting the result into Eq. 12–45 gives cp cv Ta 0v 2 0P ba b 0T P 0v T (12–46) This relation can be expressed in terms of two other thermodynamic properties called the volume expansivity b and the isothermal compressibility a, which are defined as (Fig. 12–10) b 1 0v ab v 0T P 1 0v ab v 0P T (12–47) 20°C 100 kPa 1 kg v –– ( ∂∂T ) P and a (12–48) 21°C 100 kPa 1 kg (a) A substance...
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