Question

Two separate containers filled with different gases. If these gases are allowed to mix,
obtain an expression for entropy of mixing per mole of the mixture. Does the entropy
decreases in the process of mixing of two gases?

Answer 1

To calculate the entropy change, let us treat this mixing as two separate gas expansions, one for gas A and another for B. From the statistical definition of entropy, we know that

$ΔS=nRln\dfrac{V_2}{V_1}$

Now, for each gas, the volume V₁ is the initial volume of the gas, and V₂ is the final volume, which is both the gases combined, V_A+V_B. So for the two separate gas expansions,

$ΔS_A=n_ARln\dfrac{V_A+V_B}{V_A}$

$ΔS_B=n_BRln\dfrac{V_A+V_B}{V_B}$

So to find the total entropy change for both these processes, because they are happening at the same time, we simply add the two changes in entropy together.

$Δ_{mix}S=ΔS_A+ΔS_B=$ $n_ARln\dfrac{V_A+V_B}{V_A}+$ $n_BRln\dfrac{V_A+V_B}{V_B}$

Recalling the ideal gas law, PV=nRT, we see that the volume is directly proportional to the number of moles (Avogadro's Law), and since we know the number of moles we can substitute this for the volume:

$Δ_{mix}S=n_ARln\dfrac{n_A+n_B}{n_A}+n_BRln\dfrac{n_A+n_B}{n_B}$

Now we recognize that the inverse of the term $\dfrac{n_A+n_B}{n_A}$  is the mole fraction $χ_A=\dfrac{n_A}{n_A+n_B}$

After substitution the equation of the entropy can be written as :

$Δ_{mix}S=−R(n_Alnχ_A+n_Blnχ_B)$

Entropy increases on mixing the two gases which can be justified by the above expression which will be always as mole-fraction is always less than unity which will make the whole term positive