Question

11) Consider an electromagnetic type radiation that is contained within a cavity and is in equilibrium with the walls of that cavity (cavity radiation) that can be seen as a hydrostatic system, whose internal energy density and the pressure p exerted under the cavity walls are given by 1 U= U = u(T) P=U, V 3 where V is the volume of the cavity and T is the temperature of the cavity walls. a) Considering the differential dŲ of the internal energy of a hydrostatic system, show that the energy density u = aT', where a is a constant. b) Determine the entropy S =S (T, V) for the cavity radiation and the expression of the internal energy U = U (S, V). c) Consider an isothermal and reversible compression of the volume of the Vl cavity to the V2 volume and determine the work W performed under the radiation, the heat Q eliminated by the radiation and the entropy variation AS of the radiation. Represent the process on a p - V diagram. d) Consider an adiabatic variation in the volume V of the cavity and show that, in this case, Tv1/3 = constant Represent the process on a p - V diagram. e) Show that the chemical potential u for cavity radiation is zero.

Answer 1

Using, first Energy Equation JU av Equation or Ist Internal Energy (af) -P Evergy density of heat Radiatious iu inside a cloķuxº kept at Coustaut temperatuse, T. let, uz v=Volume P= Pressure 3 of heat Radiation Both ult) and Pase funct functione g Temperature T. Now, (2 ! Internal Energy, U = U(T) V ut)=U v Now, differentiating eq• ® au av لا 3 Also, P= a ů three ma manually Egi Since, Pressure is exested lesbendicular direction? equally Substitute (I u= ilal = T(af)-P U = Trap ay - p. au Ou Tu

3 by putting, P= 4 differentiating it, UT y, au T 이 3 8P au u= T 1 ou la - 기에 3 + 볼 - 유 놓니 플부 N aT Oy U Sutegrating Both sides log (w)= 4 log (D) + Coustaut 1) () Couetaut =5 Constant Take, Custant = Custaurl = e 디 u= 5T (Stefan's Coustaut)
second method-
Energy density of a photon gas Thermodynamic identity dU = Tds - PdV U = UV dU = d'uV] = Vdu + udV du u= u(T) du = IT dT P du dU = VT + udV = TSS- dT" s-udv Energy density of a photon gas V du dT + udV =TAS - s-udv dr" ds V du 4 u dT + --dV 31 dS = dT+ + C), Cv 7 d os (), V du TIT 37 Entropy is a state variable, thus a perfect differential: a's avat a's atav 4 1 du 1 du == TdT 4 3 T2 3 T IT du dT т Energy density of a photon gas du dT " du d7 т Integrating both sides; Inu = 4 In T+ C M = AT = U V

B Calculating kutriky from Ist blow of whermodymanie, dp = el Ut el 1 Heat work done. Euergy Internal Euergy dU=d&- dW =o9- Pay Hese, dV= uolue P= Septeu's external force of psalus for a black beely hodiation change of system. Clausiu! Jusshody mamie Eutropy defivat vu, ) ds - do T dut Pav T +PdV Total Eutrofy, s= fas = s dos Using, =fudit differentiating, Evergy deusity y Radiation du udv when ú= atu (desined above) u= u V U=UV S where u= Also, p= u ; So, P= IS 3

S=fudv + adv T = 14u dv parte de - 4 T 4 T3a 3 S = 4a Tv safar 3 IS (Tv) =ų aut 3 Nowo, Calculating tatemal T=(38_)'13 energy, Hav

S = Suternal energy, And, U= un u=at" from Posta el U=aTV u= al 3s Hav > U= a 351 1/3 Ча U as a function of 5 and V.

b @ from (6) Eutheby, S = yavT3 for Adiáballe Expaustrou, 9=0 So, AS=0 3 S=Constant OUTP = Coustaut 3 WT² = Coustaul (V 73) / (ejl う vikt = Constant > TV's Constant

Chemůcal Potential, 4= 0 Expxssiou for elaug Eu temperature during Joule- враистол, dT= + [+ (2) p - J al dT dp = ( ( (*) -7 - 0 aT ap at Coustaut Euthalpy = - for a lefect » Park PV=RT Tiger differentiating wort!'T' Keeping 'P' constant - $ = T(34) p = (R=PV) av at PV PT Put in Ego ОТ ar = [v-v] =0 co dI dp =0 =0