Lea and Burke Chapter 22 Even Solutions. S.M.Lea

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1 / Lea and Burke Chapter Even Solutions S.M.Lea. T efficiency of t Otto cycle is: Since we have only sig fig, we should give t answer as 50%..4 T efficiency of a Carnot engine is:,l, 560 K = T H.0 x 0 = = 0.44 K T efficiency is 44%..6 T coefficient of perforance is given in Exaple.: _ T c 7 K _ 7 K _ c TH-TC 55.0 C - 0 C 55.0 K So t coefficient of perforance of t expensive refrigerator is 4/4.96 = 0.8 of t, Carnot, value..8 T entropy generated in a free expansion is calculated using an isotral expant iaginary process (cf Exaple.4): (.0 x ) Thus:.0 Fro equation.: Tc = Qc TH QH Tn: 5 J. To find t collision tie, we first find v, and A. f (.8xlO- J/K)(90K) (+ x 6) (.66 x 0~ 7 = /s kg) L = -*L = _Ji^H± 5H^]EL_ = x io-«an rn irhip K ( -8 x 0 -io ) ^ 0 x IQS p a ) x IP" /s =.xlq- 0 s =0. ns fl f

2 .4 jpfie nuber of states is 5!. Thus t entropy is:.6 See solutions anual = 56 (.8 x (T J/K) =.5 x (T J/K.8 In both cases t gas undergoes a coplete cycle of processes and tn returns to its initial state. Since internal energy is a state variable, its value is t sae at t end of t cycle as at t beginning - t change in internal energy is zero independent of wtr t processes are reversible. Of course t irreversible cycle does produce a net change of entropy while t reversible cycle does not.0 We consider each process in detail to find t at input and net work done. AB: isobaric copression. Point B is on a lower isotr than point A, so at transfer is out of t syste during this leg. T work done by t syste is negative: W AB = P A (V B - V A ) = P A (\V A - V A ) = -\P A V A \ / BC: Constant volue process. Point C is on a higr isotr than point B, so at transfer is into t syste: Tn, using t ideal gas law: Q BC = N^AT = «ff (T C - T B ) QBC = \(Pc- PB) V B = \ (P B - PB) \V \P B V ^ Since t volue does not change, no work is done: CA: Along t isotr, tre is no change in internal energy, so Q = W. Both are positive. Using equation 9.: QC W C P C V C In = P A. In Thus t efficiency of t cycle is: _ net work done _ W AB + W B c + Wc A _ ^P A V A +Q + P A V A ln - + In at input QBC + QCA ^P A V A + P A V A ]n + 4n. We consider each leg of t cycle to copute t at input and t net work done. Since Q = 0 along both adiabats, t calculations are ost easily done in ters of at transfers. For a coplete cycle, tre is no change of internal energy. Thus fro t first law of trodynaics, Atf = 0 = Q net - W net => Q net = W net. ltai: e W Q = net = '*"«* " Qin " Qin

3 Chapter 5 Light Atos 5.^Photon of energy E =.0 kev =.0 x 0 ev ~ - = l (6.66 x 0~ 4 J s)(.99 x Q 8 /s) (.0 xlo ev)(.6xlo- 9 J/eV) =. x 0~ 9 =. n E = hf (.0 x 0 ev)(.6 x IQ- 9 J/eV) 6.66 x 0-4 J s =.4 x 0 7 /s or.4 x 0 7 Hz 5. Rate of photon eission =.7 x 0 0 /s 5. photo electric effect on silver (Ag) v = hf (f> (j> = work function of Ag = 4.6 ev energy per photon E = hf = 5.0 ev *M«* v"' ^= hf (/> = E Q = 5.0 ev-4.6 ev = 0.4 ev = kinetic energy of eitted electrons. 5.4 A = 0.40 n 5.5 Hydrogen ato undergoes transition n' = 5 -» n = t frequency of t eitted photon is give and t wavelength is related to t frequ by /»* " A (* -.00 x 0 8 /s (.88 x 0 5 /s) (^ - jr) = 4.4 x 0~ 7 = 0.44 /z 5.6 r is rtc/i. saller than Ai 5.7 We have a hydrogenic oxygen (Z = 8) aton otr words, all of t electrons except one. been stripped fro t ato. T reai electron is in t ground state (n = ). Si ping this last electron eans taking it to a s n' oo. For hydrogenic atos, an electro t nth state has energy fi 7-^0 ~ ~ Z?

4 5. Light and Atos t change in energy Pro t uncertainty principle: v = 0.7 /s 5.9 third sll: ** ( i? - 8 (.6 ev) (p ~ ^ 8 (.6 ev)(l - 0) 870 ev n = => e = 0,, and -t<<t Angular oentu states: n = 6 = 0?n = 0 n n = a t «= = -0 n n = - n = n ~ n = n e = e = i = a t I = t = - - = 0 -j => 9 states Quantu states: Two electrons can occupy each of tse angular oentu states: one with s = ^ and one with s +5. => total nuber of electrons possible = Oxygen has no unpaired electrons. Fluorine is one electron short of having a filled second sllit reacts strongly with any ato that can easily^ supply an electron to fill its second sll. Oxygen cannot easily supply such an electron, so it does not for a stable copound with fluorine. Sodiu does have a weakly bound single electron in its third sll. Thus sodiu can easily supply a final electron for fluorine's second sll and will for a stable copound with fluorine. 5. Moentu easured to part in 0 wre and 6p p, = 5 v 0 P ' jv / s 5. 8 k/s (.055 x Q- 4 J s)(0 ) (.5) (9. xlo~ ai kg)(.5x!0 8 /s) >. x 0~ 0 = 0. n 5. T uncertainty in t tie to transition is 5.4 ^^S 6t = 6 x 0 4 s Pro t uncertainty principle: So t uncertainty in t energy is: electron * photon 6E 0 * Wt (.05 x 0~ 4 J -! (6 x 0 4 s) > 8.8 x Q- 40 J = 5 x KT ev r e need a photon with a iniu energy in = 7.4 ev This energy is related to t axiu wavelength ev 5.7 (6.66 x KT 4 J -s)(.00 x 0 8 /s) (7.4 ev)(.6 x 0-9 J/eV).7 x 0~ 7 70 n 7 = -== =.5 45 n = 4.5 x 0~ 7

5 5. Light and Atos xl0 ev/a IX".6 ev V = 5 Z e ff is t effective charge internal to t transitioning electron. 5.66, 4,4 x ID" electron:, p v 6.66 x KT 4 J - s (9. x 0~ kg)(00 x 0 /s) = 7 run g pea: proton: A 6.66 x 0~ 4 J - s (0.00 kg)-0 5 /s = x 0~ x Q- 4 J s (.67 xlo- 7 kg)-(0 5 /s) = 4 x 0~ x KT Z = 6 for iron E n =.6 Z -^ 99 n E n (ev) =.7 kev. T.5 kev line ust be t n = to transition. Let's find t effective charge (Z e g) seen by an electron hi an iron ato. TJI.5xl0 ev = -I o /? ^ =.5 kev (.6 ev) ev V 5.7 An alpha particle is two protons and two neutrons A, - h P p A /E a = p + n = (.675 xht 7 kg +.67 x (T 7 kg) = x (T 7 kg E a = 5.87 x 0 6 ev -.6 x (T 9 J/ ev = 9.9 x (T J 6.66 x ID" 4 J s (6.69 x 0-7 kg-9.9 x 0~ J)' = 5.9 x 0-5 about 5 t nuclear radius ev 5.7 Tre are four peaks between t center and t ring of iron atos. So, we estiate that t 44 n radius of t ring is 4 wavelengths of t electrons. So, tir energy is K «= VA/ (ho) c A (l.4x KPeV-n) (0.5 x 0 6 ev) (^ = 5 x 0~ ev 5.74 A = 0. n; A = 7rr 5.75 XB de Broglie wavelength P. E = P c + c 4 \B for \B A c A c = Copton wavelength e e n?c 4 - rntc* 7n c 4 Z a = 4.7 E E = 0.7 MeV