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# SM_6B - Chapter 6 Section B Non-Numerical Solutions H S =T...

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Chapter 6 - Section B - Non-Numerical Solutions 6.1 By Eq. (6.8), H S P = T and isobars have positive slope Differentiate the preceding equation: 2 H S 2 P = T S P Combine with Eq. (6.17): 2 H S 2 P = T C P and isobars have positive curvature. 6.2 ( a ) Application of Eq. (6.12) to Eq. (6.20) yields: C P P T = { V T (∂ V /∂ T ) P } T P or C P P T = V T P T 2 V T 2 P V T P Whence, C P P T = − T 2 V T 2 P For an ideal gas: V T P = R P and 2 V T 2 P = 0 ( b ) Equations (6.21) and (6.33) are both general expressions for dS , and for a given change of state both must give the same value of dS . They may therefore be equated to yield: ( C P C V ) dT T = P T V dV + V T P d P Restrict to constant P : C P = C V + T P T V V T P By Eqs. (3.2) and (6.34): V T P = β V and P T V = β κ Combine with the boxed equation: C P C V = β T V β κ 6.3 By the definition of H , U = H PV . Differentiate: U T P = H T P P V T P or U T P = C P P V T P 566

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Substitute for the final derivative by Eq. (3.2), the definition of β : U T P = C P β PV Divide Eq. (6.32) by dT and restrict to constant P . The immediate result is: U T P = C V + T P T V P V T P Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives: U T P = C V + β κ T κ P ) V 6.4 ( a ) In general, dU = C V dT + T P T V P dV (6.32) By the equation of state, P = RT V b whence P T V = R V b = P T Substituting this derivative into Eq. (6.32) yields dU = C V dT , indicating that U = f ( T ) only. ( b ) From the definition of H , d H = dU + d ( PV ) From the equation of state, d ( PV ) = R dT + b d P Combining these two equations and the definition of part ( a ) gives: d H = C V dT + R dT + b d P = ( C V + R ) dT + b d P Then, H T P = C V + R By definition, this derivative is C P . Therefore C P = C V + R . Given that C V is constant, then so is C P and so is γ C P / C V . ( c ) For a mechanically reversible adiabatic process, dU = dW . Whence, by the equation of state, C V dT = − P dV = − RT V b dV = − RT d ( V b ) V b or dT T = − R C V d ln ( V b ) But from part ( b ), R / C V = ( C P C V )/ C V = γ 1. Then d ln T = − 1 ) d ln ( V b ) or d ln T + d ln ( V b ) γ 1 = 0 From which: T ( V b ) γ 1 = const .
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SM_6B - Chapter 6 Section B Non-Numerical Solutions H S =T...

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