Entropy Change in mixing of ideal gases

When two ideal gases are brought in contact , they mix with each other and incerase in their entropy.
Let n₁ moles of gas A whose volume is V₁ mixed with n₂ moles of other gas B whose volume is V₂ and form a new volume V₁ + V₂ of mixed gas. So the entropy change in gas A due to mixing is-
ΔS₁ = n₁ Rln(V₁ + V₂/V₁)
Similarly, for gas B, entropy change is-
ΔS₂ = n₂ Rln(V₁ + V₂/V₂)
So, the total entropy change-
ΔS_mix = ΔS₁ + ΔS₂
or, ΔS_mix = n₁ Rln(V₁ + V₂/V₁) + n₂ Rln(V₁ + V₂/V₂)
or, ΔS_mix = −R(n₁ ln V₁/(V₁ + V₂) + n₂ ln V₂/(V₁ + V₂)
If X₁ and X₂ br the mole fraction of gas A and B respectively, then-
X₁ = n₁/(n₁+n₂) = V₁/(V₁+V₂)
and X₂ = n₂/(n₁+n₂) = V₂/(V₁+V₂) at NTP
So, ΔS_mix = −R(n₁ lnX₁ + n₂ lnX₂)
or, ΔS_mix = −RΣn_i lnX_i
Again from this equation
ΔS_mix = −R(n₁ lnX₁ + n₂ lnX₂)
ΔS_mix = −R[n₁/(n₁+n₂) lnX₁ + n₂/(n₁+n₂) lnX₂](n₁+n₂)
Now entropy change for one mole-
n₁+n₂ = 1

ΔS_mix = −R(X₁ lnX₁ + X₂ lnX₂)

or, ΔS_mix = −RΣX_i lnX_i
Since all X_i is less than one so, entropy of mixing of ideal gas is always positive.

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