Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermodynamics 65

Solution:

(1)

u=- s3

A3NV

(2) When no work is delivered, Tf will be maximum. Then

2

3

91 = /TTf c,dT = XfidT = -A(Tfi2 - ,

2

3

Q2 = /TT( c,dT = -A(TfI2 - Ti12).

Since 91 + 92 = 0, we have

The minimum of T corresponds to a reversible process; for which the change

in entropy of the system is zero. As

AS1 = c,dT/T = 2A(T;I2 - T;I2) ,

AS2 = l2 c,dT/T = 2A(T;12 - Tif2).

r

Tf

and AS1 + AS, = 0, we have

Hence