A rigid container is divided into two compartments of equal volume by a partitio

Question

A rigid container is divided into two compartments of equal volume by a partition. One compartment contains 1 mole of ideal gas A at 1 atm, and the other contains 1 mole of ideal gas B at 1 atm. Calculate the increase in entropy which occurs when the partition between the two compartments is removed. If the first compartment had contained 2 moles of ideal gas A, what would have been the increase in entropy when the partition was removed? Calculate the corresponding increases in entropy in each of the above two situations if both compartments had contained ideal gas A.

Explanation / Answer

entropy of mixing =-nR(x1*lnx1+x2*lnx2)

n=total number of moles= 1+1 =2

R= 8.314, x1= mole fraction of 1st component = 1/2 =0.5, mole fraction of second component = 0.5

entropy of mixing = -2*8.314(0.5*ln(0.5) +0.5*ln(0.5)}=11.53 J/K

for the second case, n=3 and x1= 2/3 and x2=1/3

hence entropy change of mixing = -3*8.314*{(2/3*ln(0.66) +0.33*ln(0.33)}=16 J/K

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