TP A SOLUTION. For an ideal monatomic gas U=3/2nRT, Since the process is at constant pressure Q = C. giving ) =1000/(5/2*8.31*10)

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1 T A SOLUTION For an deal monatomc gas U/nRT, Snce the process s at constant pressure Q C pn T gvng a: n Q /( 5 / R T ) /(5/*8.*) C V / R and C / R + R 5 / R. U U / nr T (/ ) R T ( Q / 5 / R T ) Q / 5 Q V b: V Q U /5 Q V / 5 Q / (/5)/ 5 m c: T ( ds) dq C n( dt) 5 dt S nr T p 5 T + nr ln( ) T

2 T - B SOLUTION (a) If the ground level energy s defned as zero and E s the energy of excted level: Z d exp exp ( β ε ) + 6 ( β E) The probablty that the atom s n ts excted level: ( E) 6exp ( β E) exp( β E) Z + exp( β E) + exp( β E) + Snce E. ev, T 6K (~.5 ev), βe., exp(βe ) we get: ( E). (b) The average energy per atom s: ε Z Z ε 6exp β + 6exp ( βε ) ε ( βε ) exp( βε ) + (c) The specfc heat s: C V ε T V ε ε exp ( βε ) [ exp( βε ) + ] β T k B ( βε ) exp( βε ) [ exp( βε ) + ]

3 T C SOLUTION In equlbrum the chemcal potental of the molecules on the surface stes must equal to the chemcal potental µ of the gas outsde the surface. Thus the requred expresson s gven by substtuton of the latter chemcal potental nto the Ferm dstrbuton of the adsorbed molecules [e (E µ)/kt +]. The chemcal potental for the gas molecules (µ) can be related to the number of gas molecules per unt volume (n) by ntegratng the Maxwell-Boltzmann dstrbuton over momentum: n d p (π h) exp p M µ kt Then, f n s elmnated usng the deal gas law ( nkt ), one can solve for the chemcal potental µ, and put t back nto the Ferm dstrbuton for the adsorbed molecules. The resultng expresson for the adsorbed fracton s C + C, where and C ( Λ kt exp E ), kt π h Λ MkT. Λ s proportonal to the thermal de Brogle wavelength, so that the combnaton C s dmensonless by analogy wth the deal gas law..

4 T C SOLUTION (a) J 4πR 6 Sun 9.46 W / orbt 4π 4 W ( 5.8 m) m 6 7 (b) J R 9.46 W / m (.44 m).77 W π π (c) Consder as a hemsphere, then: Solvng for the temperature we have: 4 πrσ T T / 4 / W σ πr π 6 (.44 m) 5.76 W / K m K (d) / 4 / 4 6 Sun 4 W Sun 5, Sunσ T 4 R 8 π 4π ( 7 m) 5.76 W / K m kbtsun.8 J / K 5,795K ν h 6.6 Js receved 4 max Hz K (e) ν k BT.8 J / K 55K h 6.6 Js emtted max Hz

5 T D SOLUTION a) Fnal temperature stays the same at K. Total pressure stays the same. Indvdual pressures are gven by the law of partal pressures wth helum pressure at /6*/ atm, neon pressure at /6*/ atm and argon pressure at /6* atm. Note the volume s not ntally dvded equally but s n proporton to mole fractons usng VnRT or V/nRT/ whch s a constant. Helum volume/6v, neon volume /V, argon volume /V. b) U only depends on T and not V for an deal gas U c) dg -SdT+Vd+udN. dg V(d) at constant T. dg (nrt/)(d) f G nrt ln for each component ( n He ln(/ 6) nne ln(/ ) + nar G RT + ln(/ ) ) 7 8.* (ln(/ 6) + ln(/ ) + ln(/ )).5* J c) For an deal gas U/nRTand du/nr(dt) du T(dS) - (dv) + u(dn) ds /nrdt/t + nrdv/v wth dn Thus S /nrlnt + nrlnv + constant or S nrln / T V / C k / B N lnt V / NC. The constant C NC to keep the entropy extensve. Use V NkB T / to get S nr lnt 5/ / C Thus S G / T.5 * 7 / 5* 4 J / K

6 T D SOLUTION For the densty, we have n d d p f(ɛ p ) where f s the Ferm functon f(ɛ) dp p d f(ɛ p ), () e β(ɛ µ) +, () wth β (kt ). Usng the fact that ɛ p p, t s convenent to change dummy ntegraton varables, wrtng where n dɛ g(ɛ)f(ɛ), () g(ɛ) ɛ (d )/ θ(ɛ), (4) where θ s the unt step functon that vanshes for negatve arguments. Notng that the zero temperature Ferm functon s a unt step functon, we now may wrte n(t ) n(t ) [ dɛ g(ɛ) e β(ɛ µ) + θ(µ ɛ) ]. (5) Usng the fact that we are lookng only for a small change, we may wrte ths as (µ µ )g(µ ) dɛ g(ɛ)[θ(µ ɛ) θ(µ ɛ)] [ dɛ g(ɛ) θ(µ ɛ) e β(ɛ µ) + ]. (6) The ntegrand n the fnal ntegral has contrbutons that go to zero exponentally when ɛ µ > kt, so we can expand n T by expandng g(ɛ) around µ lettng g(ɛ) g(µ) + (ɛ µ)g (µ) +... (7)

7 Substtutng above, we note that the ntegral over the frst term n the seres vanshes. In the ntegral resultng from the second term, we ntegrate by parts, and notng that the ntegrated part vanshes exponentally at ±, we fnd (µ µ )g(µ ) g (µ) (ɛ µ) dɛ d [ ] θ(µ ɛ) (8) dɛ e β(ɛ µ) + Carryng out the dfferentaton, and changng the dummy ntegraton varable to x β(ɛ µ) gves where µ µ α g (µ) g(µ ) (kt ) α g (µ ) g(µ ) (kt ), (9) α dx x e x (e x + ) π 6. () Substtung g from Eq. (4) gves ( ) d (kt ) µ µ α. () µ We see that for d >, µ decreases as we ncrease the temperature. For d t ncreases wth temperature, whle the crossover where t doesn t vary s at d.