Question

Problem One (10 points) An ideal Bose gas consists of particles in volume V at thermal equilibrium. 1. Evaluate the transition temperature of the gas in term of N. V. and mass m of the particles. 2. Show that the heat capacity, C, of the gas at low temperature ( 7 72) is proportional with T3, whereas it is constant at high temperature,

Answer 1

The required answer is given in the following images..

N particles in An ideal Bose gas consists of volume at thermal equilibrium. The average number of bosons in the energy state E, is Kny = B(E-M). M=o. at transition temperature T=Tb, So, at transition temper ahore, 1 OBE - 1 At transition temperator total no of particle, N= In CP) dp og(P) de We know, gepy do = v. 4TP de ν ΑττΡΡ So, N= 43 Bp2 é e am - Now, consider, Bp² = 2 dx = BP dp

N = STV 43 ex-1 4TV ß am 22 43 ea 1 en-1 8 (3) 30) Sada =r(a)(*) where, 8(x) = Reimann-zeta function and 1(a) = Cramma function, is temperature So the transition 2/3 12 where k= Boltzmann's constant

2. For TCTP, since the particles in the ground state do not contribute to the energy to the total energy is determined by the particles in excited state &yo. so, E. S pre negde put, 12 STV mam 1² 3 ß ? de - S * (295) =169560) 7 where, & (n) = Reimann-zeta function . . ******505) = VOUZ 3/2m 512 K 51275/28 (5) h3

Consider, Wer m lakh D E (3) - constant h So, Energy, E= CT2 the heat copacity of the ideal boson gas, is, Cr- (1) = CET/ so, (e, & T So, me heat capacity of the gas at low temept temperature (T<< Tc) is proportional with T312 We get, E-Voda 32 m2 7 512 So, Cr-V652 7312 m 12 g (32) 312 -12 2 Timk is substituting v of equation (ii) in equation 12 we get -

Now, Ton Te so, cr= constant. So, for high temperature will be constant. ca of the gas