Thermodynamics

664 | ThermodynamicsSpecific Heats c v and c pRecall that the specific heats of an ideal gas depend on temperature only.For a general pure substance, however, the specific heats depend on specificvolume or pressure as well as the temperature. Below we develop some generalrelations to relate the specific heats of a substance to pressure, specificvolume, and temperature.At low pressures gases behave as ideal gases, and their specific heatsessentially depend on temperature only. These specific heats are called zeropressure, or ideal-gas, specific heats (denoted c v 0 and c p0 ), and they are relativelyeasier to determine. Thus it is desirable to have some general relationsthat enable us to calculate the specific heats at higher pressures (orlower specific volumes) from a knowledge of c v 0 or c p 0 and the P-v-Tbehavior of the substance. Such relations are obtained by applying the testof exactness (Eq. 12–5) on Eqs. 12–38 and 12–40, which yieldsanda 0c v0v b T a 02 PT 0T b 2 v(12–42)a 0c p(12–43)0P b T a 02 vT 0T b 2 PThe deviation of c p from c p 0 with increasing pressure, for example, is determinedby integrating Eq. 12–43 from zero pressure to any pressure P alongan isothermal path:1c p c p0 2 T T P(12–44)The integration on the right-hand side requires a knowledge of the P-v-Tbehavior of the substance alone. The notation indicates that v should be differentiatedtwice with respect to T while P is held constant. The resultingexpression should be integrated with respect to P while T is held constant.Another desirable general relation involving specific heats is one that relatesthe two specific heats c p and c v . The advantage of such a relation is obvious:We will need to determine only one specific heat (usually c p ) and calculatethe other one using that relation and the P-v-T data of the substance. Westart the development of such a relation by equating the two ds relations(Eqs. 12–38 and 12–40) and solving for dT:dT T 10P>0T2 vc p c vdv T 10v>0T2 Pc p c vChoosing T T(v, P) and differentiating, we getdT a 0T0v b dv a 0TP 0P b dPvEquating the coefficient of either dv or dP of the above two equations givesthe desired result:c p c v T a 0v0T b a 0PP 0T b v0a 02 v0T 2b dPPdP(12–45)