Chapter6_B - Chapter 6 - Section B - Non-Numerical...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 - Section B - Non-Numerical Solutions 6.1 By Eq. (6.8), µ H S P = T and isobars have positive slope Differentiate the preceding equation: µ 2 H S 2 P = µ T S P Combine with Eq. (6.17): µ 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: µ C P P T = · { V T (∂ V /∂ T ) P } T ¸ P or µ C P P T = µ V T P T µ 2 V T 2 P µ V T P Whence, µ C P P T =− T µ 2 V T 2 P Foranidealgas: µ V T P = R P and µ 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 = µ P T V dV + µ V T P dP Restrict to constant P : C P = C V + T µ P T V µ V T P By Eqs. (3.2) and (6.34): µ V T P = β V and µ P T V = β κ Combine with the boxed equation: C P C V = β TV µ β κ 6.3 By the definition of H , U = H PV . Differentiate: µ U T P = µ H T P P µ V T P or µ U T P = C P P µ V T P 651
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
β : µ U T P = C P β PV Divide Eq. (6.32) by dT and restrict to constant P . The immediate result is: µ U T P = C V + · T µ P T V P ¸µ V T P Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives: µ U T P = C V + β κ T κ P ) V 6.4 ( a ) In general, dU = C V dT + · T µ P T V P ¸ dV (6.32) By the equation of state, P = RT V b whence µ 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 , dH = dU + d ( PV ) From the equation of state, d ( PV ) = RdT + bdP Combining these two equations and the definition of part ( a )gives: dH = C V dT + RdT + bdP = ( C V + R ) dT + bdP Then, µ 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 andsois γ C P / C V . ( c ) For a mechanically reversible adiabatic process, dU = dW . Whence, by the equation of state, C V dT =− PdV =− 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 . Substitution for
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

Chapter6_B - Chapter 6 - Section B - Non-Numerical...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online