Problems and Solutions on Thermodynamics and Statistical Mechanics

104 Problems EI Sdutiom on Thermcdynamicn €4 Statintical Mechanics

Solution:

(a) The particle numbers of the two parts do not change. Let these be

N1 and N2, the final pressure be p, and the final temperature be T. Taking

air as an ideal gas, we have

pivo = NikT1 , p2Vo = N2kT2 ,

where p1 = 2 atm, TI = 200 K, p2 = 1 atm, T2 = 300 K , V 0 - - ll.
The piston does not consume internal energy of the gas as it is fric-
tionless, so that the total internal energy of the gas is conserved in view of
the cylinder being adiabatical. Thus

PNlkT1 + PNzkT = p(Nl+ N2)kT ,

where p is the degree of freedom of motion of an air molecule. Hence

P2

1+-

N2

TI + -T2

Nl = p1 TI = 225 K

N2 -+1 l+--'-- P2 Tl

Nl PI T2

T=

By V1+ V2 = 2V0, we find

and hence

(Ni + N2)

2VO

P=

(b) Entropy is a state function independent of the process. To calculate

the change of entropy by designing a quasi-static process, we denote the

entropies of the two parts by S1 and S2. Then

AS = AS1 + AS2 =