Question

consider the situation shown in the diagram below. In state A, the system consists of two separate chambers of equal volume, with 1 mole of A molecules on the left and 1 mole of B molecules on the right. The A and B molecules do not interact with each other. In State B, the partition between the two halves of the box is removed. STATE A STATE B 1 mole of A molecules 1 mole of B molecules A and B molecules are completely mixed Starting from the equation for the change in entropy for the expansion of an ideal gas, show that the change in entropy on going from State A to State B (the entropy of mixing) is given by: AS = Sg-SA =-Nkp (XIn XA + Xp In XB) In this equation: N = N2 + NA (number of molecules) NA (mole fraction of A) XANEN NB (mo B (mole fraction of B) NA+Ng

Answer 1

A state B state .. Imole a molecule Imole B .. molecule .. completely mined. A and B mo TN Ri= Number of ways of placing 1 mole indistinguishable A molecule in NA sites = 1 sz= Number of ways of placing 1 mote indistinguishable B molecule in Ng sites = 1 But in mixture , 212 1 .312 = (NA+NB)!. NA! NB! os = 5e -Sn = Kalm Santo) | D NA! NB! ! = Kg [en (Na twe)! - kn Na! – An NR] Applying sterling approximation for large in ln N! Nena . os = kg NA +Ng) en (Na twB) - WaPn Na. Ne yn wa] = -'Kg Ngen NA-ONA. IN (NA +NR) + og án Ng . CAPASEN-NRiol -UB. INA en Martin BA NB ln NB NT --KB [No en XA + Ngenxg] x = Mole fractim = - NKB (XA en XA + X BlnxB) Na Total number of moles,