PHYSICS 210A : STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS. ( p)( V) = 23kJ,

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1 PHYSICS 210A : STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS 1) ν = 8 moles of a diatomic ideal gas are subjected to a cyclic quasistatic process, the thermodynamic path for which is an ellipse in the V,p) plane The center of the ellipse lies at V 0,p 0 ) = 025m 3,10bar) The semimajor and semiminor axes of the ellipse are V = 010m 3 and p = 020bar a) What is the temperature at V,p) = V 0 + V,p 0 )? b) Compute the net work per cycle done by the gas c) Compute the internal energy difference EV 0 V,p 0 ) EV 0,p 0 p) d) Compute the heatqabsorbed by the gas along the upper half of the cycle a) The temperature ist = pv/νr With V = V 0 + V = 035m 3 and p = p 0 = 10bar, we have T = 105 Pa)035m 3 ) 8mol)831J/molK) = 530K b) The area of an ellipse is π times the product of the semimajor axis lengths pdv = π p) V) = π bar)010m 3 ) = 63kJ c) For a diatomic ideal gas,e = 5 2pV Thus, E = 5 2 V0 p p 0 V) = J) = 13kJ d) We have Q = E +W, with W = 2p 0 V + π 2 p) V) = 23kJ, which is the total area under the top half of the ellipse The difference in energy is given by E = 5 2 p 0 2 V = 5p 0 V, so Q = E +W = 7p 0 V + π 2 p) V) = 73kJ 2) Consider a thermodynamic system for which ES,V,N) = as 4 /NV 2 1

2 a) Find the equation of statep = pt,v,n) b) Find the equation of stateµ = µt,p) c) ν moles of this substance are taken through a Joule-Brayton cycle The upper isobar lies atp = and extends from volumev A tov B The lower isobar lies atp = p 1 Find the volumesv C and V D d) Find the work done per cycle W cyc, the heatq AB, and the cycle efficiency a) We can find p = ) E = 2aS4 V S,N NV 3, T = ) ) E = 4aS3 E S V,N NV 2, µ = N S,V = as4 N 2 V 2, but we need to eliminate the inconvenient variable S from these equations To do this, we construct the ratio p 3 /T 4, in which the S factors manifestly cancel One then finds 32ap 3 V = NT 4, ie ) 1/3 T 4/3 pt,v,n) = 32a) 1/3 N V This means, for example, that the equation for an isotherm at fixed N) is pv 1/3 = const, in contrast to the result for the ideal gas isotherm, pv =const Note also that p, being intensive, must be expressible as pt,v/n), which it is b) To obtain µt,p), note that 2Nµ = pv, and from our result for pt,v/n) we have V/N = T 4 /32ap 3 Thus, µt,p) = T4 64a c) The equilibrium adiabatic equation of state for this system is ds = 0 From a), we see that at fixed N this means pv 3 = const, so we must have V 3 B = p 1 V 3 C and V 3 A = p 1 V 3 D Hence V C = V B p 2 /p 1 ) 1/3, V D = V A p 2 /p 1 ) 1/3 a) From part a), the energy for our system is E = 1 2 pv Along the upper ) isochore, B W AB = dv = V B V A ), E AB = E B E A = 1 2 V B V A ), A hence Q AB = E AB +W AB = 3 2 V B V A ) 2

3 The work along the lower p 1 ) isochore is Along thebc adiabat, Similarly, W BC = W DA = C B A D W CD = p 1 V D V C ) = V A V B ) pdv = V 3 B pdv = V 3 A V C V B V A V D dv V 3 = 1 2 V B 1 V 2 B V 2 C dv V 3 = 1 2 p VA 2 2V A VD 2 Adding up all the individual works, we get ) 2/3 ) = 12 V B [ W cyc = W AB +W BC +W CD +W DA ) ] 2/3 = 3 2 V B V A ) [1 Dividing byq AB, we obtain the efficiency, η = W cyc Q AB = 1 1 ) 2/3 ] ) [ ) 2/3 1 = 1 2 V A 1] ) 2/3 3) A diatomic gas obeys the equation of state wherea,b, andcare constants p = RT v b a v 2 + crt v 3, a) Find the adiabatic equation of state relating temperaturet and molar volume v b) What is the internal energy per mole, εt,v)? c) What is the Helmholtz free energy per mole, ft,v)? a) Letεbe the molar internal energy and v the molar volume We have already shown ) ) ε p = T p v T T v 3

4 Thus, for our system, ) ε v T = a v 2 εt,v) = 5 2 RT a v, where the first term is the result for the rarefied limit v, where the gas presumably becomes ideal Now if s = S/ν is the molar entropy ν = N/N A is the number of moles), then T ds = dε+pdv = 5 dv dv 2RdT +RT +crt v b v 3 Dividing byt and then integrating, we have [ st,v) = Rln T 5/2 v b)e c/2v2] +const Thus, the equation of the adiabat is T 5/2 v b)e c/2v2 = const b) We have already obtained the result εt,v) = 5 2 RT a v c) Fromf = ε Ts, wheref = F/ν is the Helmholtz free energy per mole, we have ft,v) = 5 2 RT a ) ) brt v b v 5 2 RT ln RT ln + crt a b 2v 2 Ts 0 Here we have inserted constants with the proper dimensions in order to render our expression for f with the appropriate dimensions Thus, the constant s 0 has dimensions of J/mol K, the same as the gas constant R Since c/b 2 is dimensionless, there is more than one way to do this Any resulting differences will show up in a different expression fors 0 4) Consider the thermodynamics of a solid in equilibrium with a vapor at temperature T and pressure p, but separated by a quasi-liquid layer of thickness d Let the number density of the liquid be n l The Gibbs free energy per unit area of the quasi-liquid layer is taken as g qll T,p) = n l µ l T,p)d+γd), where γd) is an effective surface tension which interpolates between γ0) = γ sv and γ ) = γ sl +γ lv The phenomenon of premelting requiresγ0) > γ ) a) Show thatµ qll T,p) = µ l T,p)+n 1 l γ d) = µ s T,p) 4

5 b) Expand T relative to some point T m,p) along the melting curve to lowest order in T T m Show µt,p) µ s T,p) µ l T,p) = l m T T m )/T m, wherel m is the latent heat of melting c) Assume d 2 γd) = γ sv +γ sl +γ lv γ sv ) d 2 +σ 2, where σ is a molecular length scale Assuming d σ, find the dependence of the thicknessdof the quasi-liquid layer on the reduced temperaturet T m T)/T m a) The chemical potentials of a particle in the solid and the quasi-liquid layers must be the same Since we use d for thickness, I will here use δ for differential The Gibbs free energy of the quasi-liquid layer isg qll d) = Ag qll d), whereais the total surface area The number of particles in the quasi-liquid layer isn qll = An qll d Thus, the chemical potential is potential is then µ qll = δg qll δn qll = µ l +n 1 l γ d) b) Expanding to first order, with µ = µ s µ l, we have ) µ µt,p) = T T T m ) = l m T T p T m ), m where l m is the latent heat of melting per molecule This results in the desired expression µt,p) = l m T T m )/T m For a more rigorous derivation, expand to first order in both T and p: ) ) µ µ µt,p) = T T T m )+ p p p p m )+ T = l m 1 T T T m ) 1 ) p p m n l n m )+ s The pressure shift can now be expressed in terms of the temperature shift by examining the slopes of the sublimation and melting lines, via the Clapeyron equation This results in µt,p) = κ v l m T T m )/T m, with κ v = 1 dp/dt) subl dp/dt) melt For water,κ v 1 and the correction is negligible c) We now set µ = 1 γ n l d = γ sl +γ lv γ sv 2σ 2 d n l σ 2 +d 2 ) 2 = κ v l m T T m 5

6 Ford σ, we have wheret T T m )/T m dt) = ) 2σ2 γ sl +γ lv γ sv ) 1/3 t 1/3, l m n l 6