Entropy Change in mixing of ideal gases

Entropy Change in mixing of ideal gases

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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