Problem Work and heat

Transcription

1 Problem.8-. Work and heat ( ) W P ( ) ( P P ) 0 Q W.4 0 ( C) W P( ) d ( P P ).5 0 [ J ] Q W [ J ] (C ) 0 C C 0 W Q.5( P P ) C C (, parabola) W P( ) d.67 0 Q W.67 0

2 Problem.0-. hapes of vs Consistent: a, c, e, g, i Inconsistent: b, d, f, h, j Most difficult: b, c, f and i b) / R θ / Inconsistent with the postulates because entropy does not scale linearly with the size of the system c) λ / R λλ θ λ R θ / + / Rθ v 0 / Consistent with the postulates, zero of energy is arbitrary. / λ λ not true for arbitrary λ

3 Problem.0-. hapes of vs f) R ln Rθv 0 Inconsistent with postulate I, should be zero at / v θ i) 0θ exp R R Consistent with the postulates First plot vs hen rotate it around the / bisector to obtain vs

4 Problem.0-. Entropy maximum for a composite system For a composite system ( + ) ( ) ( ) /( + ) const ( ) / + - need to maximize with respect to / [ ( ) / ] / / / + 0 ( ) / olving for + ( / )

5 Problem.-4. Restrictions due to the Postulates n m r Most common problem: forgetting to check if entropy is additive (extensive) When we increase the size of a thermodynamic system by a factor of λ, all extensive variables should scale with λ: λ, λ, λ, λ. ubjecting the fundamental relation to this scaling transformation: λ n m r n+ m+ r n m r ( λ ) ( λ ) ( λ ) λ n + m+ r λ n m λ + + r

6 Problem.-4. Restrictions due to the Postulates n m r must be increasing with, so n>0 as 0, therefore n< dditional condition: P should increase with /: 0<n< P n m m r m m P > 0 n n n n m r m > 0

7 Problem.6-. Energy in Equilibrium () () () () ()? he equations of state of the two systems are: () () () R R () () () 5 ince the two systems are in thermal equilibrium: () () () () () () 5 () 5 () 6 () () () ().5 0 ()

8 Problem.7-. he Indeterminate Problem of hermodynamics a) how that () () P P ince the piston is adiabatic, δq (,) 0 and thus ince the composite system is isolated d d (,) (,) d P () + d () 0 d (,) P () d () P () d () 0 ince the total volume of the composite system is constant herefore, () () P P follows. () d d () b) how that thermodynamics does not tell us what the temperatures of the two systems are he entropy maximum condition requires: d d () d () + P () + d () d () () d d + () () () P + + P () () () () d d () he numerators in this expression vanish identically because of the adiabaticity of the wall, therefore the denominators ( () and () ) can be arbitrary and still satisfy the entropy maximum condition. () + d 0 () () + P () () d () 0

9 Problem.8-. emi-permeable partition (), () (), () wo different gases and can freely flow through the membrane while cannot. Important: () () P P although one of the gas components can freely penetrate the membrane P is intensive parameter conjugate to volume, volume is constrained Equilibrium conditions: µ () () µ () () () () () () + + Conservation conditions: () + () ( () () ) initial ( () () + ) initial

10 Problem.8-. emi-permeable partition + R ln R ln / 5/ R ln + R R olve for () () () + () ( () () + ) initial 7.7[ K ] µ... Here, when differentiating do not forget that + fter some algebra: µ 5 R + R ln / ( ) / +

11 Problem.8-. emi-permeable partition ( ) initial () () () () + + olve () () for () () µ µ o find P: P R P () ( () () + ) () R () P 680[ kpa] imilarly: () P 567[ kpa]

12 umerical Problem First calculate P and by differentiating the fundamental relation. hen use the parametric plot of P() and () to plot the P() dependence. u ; 0.0;0 J+ u Exp u F; P@_D:0 Exp u F; Param etricplot@8@d,p@d<,8,0,0.8<,fram e -> rue,exttyle -> 8FontFam ily ->"im es",fontw eight->"old",fontize -> 4<, Fram elabel-> 8em perature ""@KD,Pressure""@PaD<,PlotRange -> ld P P a D Ü Graphics Ü