An insulated box is initially divided into halves by a frictionless, thermally conducting piston.• On one side of the piston is 1.5 m3 of air at 400 K, 4 bar.• On the other side is 1.5 m3 of air at 400 K, 2 bar.• The piston is released and equilibrium is attained• Employing the ideal gas model for the air, determine: (a) the final temperature (b) the final pressure, in bar (c) the amount of entropy produced

a) 400 K

b) 3 bar

c) 127.4 J/K

Explanation:

a) When the system reaches equilibrium, it must have the same temperature (thermal equilibrium), which is already true. So there'll be no heat changed in this process, and Tfinal = 400 K.

b) Because there's no change in the temperature, the process can be studied by Boyle's Law, which states that the product of the pressure by the volume must be constant. At the equilibrium the two gases must have the same pressure, thus:

1.5*4 + 1.5*2 = (1.5 + 1.5) * P

3P = 9

P = 3 bar

c) The entropy change (ΔS) can be calcuated by the second law:

ΔS = - n1R*ln (p1, final/p1, initial) - n2Rln (p2, final/p2, initial)

Where 1 and 2 are the gases on the sides of the piston, and n is the number of moles. By the ideal gas law:

PV = nRT

nR = PV/T

The pressure must be in Pa, so the entropy will be in J.

p1, initial = 4 bar = 400,000 Pa

p2, initial = 2 bar = 200,000 Pa

p1, final = p2, final = 3 bar = 300,000 Pa. So:

ΔS = - p1, initialV1/T * ln (p1, final/p1, initial) - p2, initialV2/T * ln (p2, final/p2, initial)

ΔS = (-400000*1.5/400) * ln (3/4) - (200000*1.5/400) * ln (3/2)

ΔS = 431.523 - 304.099

ΔS = 127.4 J/K