TITLE.PM5

358 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-1.pm5 But, ∂ ∂ F H I K p T v = ∂ ∂−− RS T UV W L NM O T QP RT vb a v^2 v = R v ...

THERMODYNAMIC RELATIONS 359 dharm \M-therm\Th7-1.pm5 Substituting this value in eqn. (3), we get (dh)T = v p v T p TvT ∂ ∂ F H I ...

360 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-1.pm5 Solution. The general equation for finding dh is given by dh = cp dT + v ...

THERMODYNAMIC RELATIONS 361 dharm \M-therm\Th7-2.pm5 Equating the co-efficients of dT in the two equations of ds, we have c T s ...

362 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 Example 7.8. Using the first Maxwell equation, derive the remaining thre ...

THERMODYNAMIC RELATIONS 363 dharm \M-therm\Th7-2.pm5 ∂ ∂ F H I K v T p = – ∂ ∂ F H I K ∂ ∂ F H I K p T v vT. p = – FH∂∂psIK FH∂∂ ...

364 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 Also h = g + Ts = g – T FH∂∂TgIK p Hence h = g – T ∂ ∂ F H I K g T p. (A ...

THERMODYNAMIC RELATIONS 365 dharm \M-therm\Th7-2.pm5 Now substituting this in eqn. (i), we get ds = cp dT T s T p −F HG I KJ ∂ ∂ ...

366 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 ∴ FH∂∂TvIK s = −cKT v β. (Ans.) +Example 7.12. Derive the third Tds equa ...

THERMODYNAMIC RELATIONS 367 dharm \M-therm\Th7-2.pm5 Example 7.14. Derive the following relations ∂ ∂ F H I K T v u = T p T p c ...

368 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 Solution. We know that ds = cTvdT Tp dv v +FH∂∂ IK [Eqn. (7.24)] Also dh ...

THERMODYNAMIC RELATIONS 369 dharm \M-therm\Th7-2.pm5 With the aid of eqn. (ii) show that ∂ ∂ F H I K u pT = – T^ ∂ ∂ F H I K v T ...

370 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 We know that ∂ ∂ F H I K ∂ ∂ F H I K ∂ ∂ F H I K p v v T T T p pv^ = –^1 ...

THERMODYNAMIC RELATIONS 371 dharm \M-therm\Th7-2.pm5 = – 0.114 8 6 10 2 ××. −^15 × 10^10 [(800)^2 – (20)^2 ] = – 8.114 6 10 2 ...

372 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 Example 7.19. An ice skate is able to glide over the ice because the ska ...

THERMODYNAMIC RELATIONS 373 dharm \M-therm\Th7-2.pm5 Differentiating both sides, we get 1 2.302p. dp dT = 3276 6. T^2 0.652 2.3 ...

374 ENGINEERING THERMODYNAMICS dharm \M-therm\Th7-2.pm5 Entropy equations (Tds equations) : Tds = cvdT + T GFH∂∂TpIKJ v dv ... ...

THERMODYNAMIC RELATIONS 375 dharm \M-therm\Th7-2.pm5 Exercises Define the co-efficient of : (i) Volume expansion (ii) Isotherma ...

8 Ideal and Real Gases 8.1. Introduction. 8.2. The equation of state for a perfect gas. 8.3. p-v-T surface of an ideal gas. 8.4. ...

IDEAL AND REAL GASES 377 dharm \M-therm\Th8-1.pm5 If p is the absolute pressure of the gas and V is the volume occupied by the g ...