Thermodynamics

652 | Thermodynamicsf(x)(x+∆ x)f(x)Slope∆ xx + ∆ x∆ fFIGURE 12–1The derivative of a function at aspecified point represents the slope ofthe function at that point.xx12–1 ■ A LITTLE MATH—PARTIAL DERIVATIVESAND ASSOCIATED RELATIONSMany of the expressions developed in this chapter are based on the state postulate,which expresses that the state of a simple, compressible substance iscompletely specified by any two independent, intensive properties. All otherproperties at that state can be expressed in terms of those two properties.Mathematically speaking,z z 1x, y2where x and y are the two independent properties that fix the state and z representsany other property. Most basic thermodynamic relations involve differentials.Therefore, we start by reviewing the derivatives and variousrelations among derivatives to the extent necessary in this chapter.Consider a function f that depends on a single variable x, that is, f f (x).Figure 12–1 shows such a function that starts out flat but gets rather steep as xincreases. The steepness of the curve is a measure of the degree of dependenceof f on x. In our case, the function f depends on x more strongly atlarger x values. The steepness of a curve at a point is measured by the slope ofa line tangent to the curve at that point, and it is equivalent to the derivativeof the function at that point defined asdfdx lim ¢f¢xS0 ¢x limf 1x ¢x2 f 1x2¢xS0 ¢x(12–1)Therefore, the derivative of a function f(x) with respect to x represents therate of change of f with x.EXAMPLE 12–1Approximating Differential Quantities by Differencesh(T ), kJ/kg305.22295.17Slope = c p (T )295 300 305FIGURE 12–2Schematic for Example 12–1.T, KThe c p of ideal gases depends on temperature only, and it is expressed asc p (T ) dh(T )/dT. Determine the c p of air at 300 K, using the enthalpy datafrom Table A–17, and compare it to the value listed in Table A–2b.Solution The c p value of air at a specified temperature is to be determinedusing enthalpy data.Analysis The c p value of air at 300 K is listed in Table A–2b to be 1.005kJ/kg · K. This value could also be determined by differentiating the functionh(T ) with respect to T and evaluating the result at T 300 K. However, thefunction h(T ) is not available. But, we can still determine the c p value approximatelyby replacing the differentials in the c p (T ) relation by differences inthe neighborhood of the specified point (Fig. 12–2):dh 1T2c p 1300 K2 cdTd T300 K¢h 1T2 c¢T d h 1305 K2 h 1295 K2T 300 K 1305 2952 K1305.22 295.172 kJ>kg 1.005 kJ/kg # K1305 2952 KDiscussion Note that the calculated c p value is identical to the listed value.Therefore, differential quantities can be viewed as differences. They can