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Unformatted text preview: with a large β Substituting these two relations into Eq. 12–46, we obtain a third general relation for cp cv:
cp cv vTb a
2 v –– ( ∂∂T ) P (12–49) 20°C 100 kPa 1 kg 21°C 100 kPa 1 kg It is called the Mayer relation in honor of the German physician and physicist J. R. Mayer (1814–1878). We can draw several conclusions from this equation: 1. The isothermal compressibility a is a positive quantity for all substances in all phases. The volume expansivity could be negative for some substances (such as liquid water below 4°C), but its square is always positive or zero. The temperature T in this relation is thermodynamic temperature, which is also positive. Therefore we conclude that the constant-pressure specific heat is always greater than or equal to the constant-volume specific heat:
(12–50) (b) A substance with a small β FIGURE 12–10 The volume expansivity (also called the coefficient of volumetric expansion) is a measure of the change in volume with temperature at constant pressure. 2. The difference between cp and cv approaches zero as the absolute temperature approaches zero. 3. The two specific heats are identical for truly incompressible substances since v constant. The difference between the two specific heats is very small and is usually disregarded for substances that are nearly incompressible, such as liquids and solids. EXAMPLE 12–7 Internal Energy Change of a van der Waals Gas Derive a relation for the internal energy change as a gas that obeys the van der Waals equation of state. Assume that in the range of interest cv varies according to the relation cv c1 c2T, where c1 and c2 are constants. Solution A relation is to be obtained for the internal energy change of a
van der Waals gas. cen84959_ch12.qxd 4/5/05 3:58 PM Page 666 666 | Thermodynamics
Analysis The change in internal energy of any simple compressible system in any phase during any process can be determined from Eq. 12–30:
T2 v2 u2 u1
T1 cv dT
v1 cTa 0P b 0T v P d dv The van der Waals equation of state is P
Then RT v 0P b 0T v RT v b v b a v2 R v b RT b a v2
Substituting gives 0P b 0T v P a v2 T2 u2
Integrating yields u1
T1 1 c1 c2T 2 dT v1 a dv v2 u2 u1 c1 1 T2 T1 2 c2 2 1T 22 2 T12 aa 1 v1 1 b v2 which is the desired relation. EXAMPLE 12–8 Internal Energy as a Function of Temperature Alone Show that the internal energy of (a) an ideal gas and (b) an incompressible substance is a function of temperature only, u u(T). Solution It is to be shown that u u(T) for ideal gases and incompressible substances. Analysis The differential change in the internal energy of a general simple compressible system is given by Eq. 12–29 as
(a) For an ideal gas Pv cv dT
RT. Then cTa 0P b 0T v P d dv Ta
Thus, 0P b 0T v P Ta du R b v cv dT P P P 0 To complete the proof, we need to show that cv is not a function of v either. This is done with the help of Eq. 12–42: a 0 cv b 0v T Ta 0 2P b 0T 2 v cen84959_ch12.qxd 4/5/05 3:58 PM Page 667 Chapter 12
For an ideal gas P RT/v. Then | 667 a
Thus, 0P b 0T v R v and a 0 2P b 0T 2 v c 0 1 R> v 2 0T d 0
v a 0 cv b 0v T 0 which states that cv...
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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.
- Spring '09