1252 yields mjt 1 cp 0 discussion this result is not

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: is also possible to predict the constant-pressure specific heat of a substance by using the Joule-Thomson coefficient, which is relatively easy to determine, together with the P-v-T data for the substance. EXAMPLE 12–10 Joule-Thomson Coefficient of an Ideal Gas T h = constant line Show that the Joule-Thomson coefficient of an ideal gas is zero. Solution It is to be shown that mJT Analysis For an ideal gas v 0 for an ideal gas. RT/P, and thus a cv 0v Ta b d 0T P 0v b 0T P 1 cp cv R P R Td P 1 1v cp v2 P1 P2 P Substituting this into Eq. 12–52 yields mJT 1 cp 0 Discussion This result is not surprising since the enthalpy of an ideal gas is a function of temperature only, h h(T), which requires that the temperature remain constant when the enthalpy remains constant. Therefore, a throttling process cannot be used to lower the temperature of an ideal gas (Fig. 12–15). FIGURE 12–15 The temperature of an ideal gas remains constant during a throttling process since h constant and T constant lines on a T-P diagram coincide. 12–6 ■ THE h, u, AND s OF REAL GASES We have mentioned many times that gases at low pressures behave as ideal gases and obey the relation Pv RT. The properties of ideal gases are relatively easy to evaluate since the properties u, h, cv, and cp depend on temperature only. At high pressures, however, gases deviate considerably from ideal-gas behavior, and it becomes necessary to account for this deviation. cen84959_ch12.qxd 4/5/05 3:58 PM Page 670 670 | Thermodynamics In Chap. 3 we accounted for the deviation in properties P, v, and T by either using more complex equations of state or evaluating the compressibility factor Z from the compressibility charts. Now we extend the analysis to evaluate the changes in the enthalpy, internal energy, and entropy of nonideal (real) gases, using the general relations for du, dh, and ds developed earlier. Enthalpy Changes of Real Gases The enthalpy of a real gas, in general, depends on the pressure as well as on the temperature. Thus the enthalpy change of a real gas during a process can be evaluated from the general relation for dh (Eq. 12–36) T2 P2 h2 T Actual process path T2 T1 1 P 1 h1 T1 cp dT P1 cv Ta 0v b d dP 0T P 2 1* 2* Alternative process path s FIGURE 12–16 An alternative process path to evaluate the enthalpy changes of real gases. where P1, T1 and P2, T2 are the pressures and temperatures of the gas at the initial and the final states, respectively. For an isothermal process dT 0, and the first term vanishes. For a constant-pressure process, dP 0, and the second term vanishes. Properties are point functions, and thus the change in a property between two specified states is the same no matter which process path is followed. This fact can be exploited to greatly simplify the integration of Eq. 12–36. Consider, for example, the process shown on a T-s diagram in Fig. 12–16. The enthalpy change during this process h2 h1 can be determined by performing the integrations in Eq. 12–36 along a path that consists of constant and T2 constant) lines and one isobaric two isothermal (T1 constant) line instead of the actual process path, as sh...
View Full Document

This note was uploaded on 03/09/2009 for the course ME 430 taught by Professor Y during the Spring '09 term at CUNY City.

Ask a homework question - tutors are online