Entropy Change in mixing of ideal gases
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-
ΔSmix = ΔS1 + ΔS2
or, ΔSmix = n1 Rln(V1 + V2/V1) + n2 Rln(V1 + V2/V2)
or, ΔSmix = −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, ΔSmix = −R(n1 lnX1 + n2 lnX2)
or, ΔSmix = −RΣni lnXi
Again from this equation
ΔSmix = −R(n1 lnX1 + n2 lnX2)
ΔSmix = −R[n1/(n1+n2) lnX1 + n2/(n1+n2) lnX2](n1+n2)
Now entropy change for one mole-
n1+n2 = 1
ΔSmix = −R(X1 lnX1 + X2 lnX2)
or, ΔSmix = −RΣXi lnXi
Since all Xi is less than one so, entropy of mixing of ideal gas is always positive.
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