Question

A system consists initially of nA moles of gas A at pressure p and temperature T and nB moles of gas B separate from gas A but at the same pressure and temperature. The gases are allowed to mix with no heat or work interactions with the surroundings. The final equilibrium pressure and temperature are p and T, respectively, and the mixing occurs with no change in total volume.

(a) Assuming ideal gas behavior, obtain an expression for entropy produced in terms of Ru, nA and nB
(b) Using the result of part (a), demonstrated that the entropy produced has a positive value.
(c) Would entropy be produced when samples of the same gas at the same temperature and pressure mix? What if samples of the same gas at different temperatures were mixed?

Answer 1

A) б = - R ( nA In Ya - nB In Yb )

B) s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced

Step-by-step explanation:

A) assuming ideal gas behavior the expression for entropy produced

for a closed system : s2 - s1 = б

where : s1 ( initial entropy ) = nA sA ( T, P ) + nB sB ( T, P )

s2 ( final entropy ) = nA sA ( T, YaP ) + nB sB ( T, YbP )

∴ б = - R ( nA In Ya - nB In Yb )

B) Given that

Ya and Yb are less than 1 respectively, hence the value of б = positive

also assuming the gases are identical

s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced when same gas at same temperature and same pressure are mixed