Thermodynamics filled in class notes_Part_50

Thermodynamics filled in class notes_Part_50 - 107 4.1....

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Unformatted text preview: 107 4.1. EXACT DIFFERENTIALS AND STATE FUNCTIONS Example 4.2 Show the heat transfer q is not a state function. Assume all processes are fully reversible. The first law gives = δq − δw, = du + δw, (4.28) (4.29) = du + P dv. du δq (4.30) Take now the non-canonical, although acceptable, form u = u(T, v ). Then one gets du ∂u ∂v = dv + T ∂u ∂T dT. (4.31) v So δq = = ∂u ∂u dv + dT + P dv, ∂v T ∂T v ∂u ∂u + P dv + dT. ∂v T ∂T v ≡M = (4.32) (4.33) ≡N M dv + N dT. (4.34) Now by Eq. (4.22), for δq to be exact, one must have ∂M ∂T ∂N ∂v = v , (4.35) T (4.36) This reduces to This can only be true if So dq is not exact. ∂P ∂2u + ∂T ∂v ∂T ∂P ∂T v = v ∂2u . ∂v∂T = 0. But this is not the case; consider an ideal gas for which (4.37) ∂P ∂T v = R/v . Example 4.3 Show conditions for ds to be exact in the Gibbs equation. du = ds = = = T ds − P dv, du P + dv, T T 1 ∂u ∂u P dv + dT + dv, T ∂v T ∂T v T P 1 ∂u 1 ∂u + dv + dT. T ∂v T T T ∂T v ≡M (4.38) (4.39) (4.40) (4.41) ≡N CC BY-NC-ND. 18 November 2011, J. M. Powers. 108 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Again, invoking Eq. (4.22), one gets then 1 ∂u T ∂v ∂ ∂T 1 ∂2u 1 − T ∂T ∂v T 2 1 −2 T ∂u ∂v ∂u ∂v 1 T 1 + T + T T + T ∂P ∂T ∂P ∂T P T ∂ ∂v = v 1 ∂u T ∂T , v (4.42) T = 1 ∂2u , T ∂v∂T (4.43) = P T2 v P −2 T v − 0. (4.44) This is the condition for an exact ds. Experiment can show if it is true. For example, for an ideal gas, one finds from experiment that u = u(T ) and P v = RT , so one gets 0+ 1R 1 RT −2 Tv Tv 0 = 0, (4.45) = 0. (4.46) So ds is exact for an ideal gas. In fact, the relation is verified for so many gases, ideal and non-ideal, that one simply asserts that ds is exact, rendering s to be path-independent and a state variable. 4.2 Two independent variables Consider a general implicit function linking three variables, x, y , z : f (x, y, z ) = 0. (4.47) In x − y − z space, this will represent a surface. If the function can be inverted, it will be possible to write the explicit forms x = x(y, z ), y = y (x, z ), z = z (x, y ). (4.48) dz, (4.49) dz. (4.50) Differentiating the first two of the Eqs. (4.48) gives ∂x ∂y dx = ∂y ∂x dy = dy + z dx + z ∂x ∂z y ∂y ∂z x Now use Eq. (4.50) to eliminate dy in Eq. (4.49): dx = ∂x ∂y ∂y ∂x z dx + z ∂y ∂z ∂x ∂z dz + x dz, (4.51) y =dy 1− ∂x ∂y z ∂y ∂x dx = z 0dx + 0dz = ∂x ∂y ∂x ∂y z ∂y ∂z z ∂y ∂x =0 CC BY-NC-ND. 18 November 2011, J. M. Powers. + x z ∂x ∂z dz, (4.52) y − 1 dx + ∂x ∂y z ∂y ∂z + x =0 ∂x ∂z dz. (4.53) y ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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