# 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 n

_{1}moles of gas A whose volume is V

_{1}mixed with n

_{2}moles of other gas B whose volume is V

_{2}and form a new volume V

_{1}+ V

_{2}of mixed gas. So the entropy change in gas A due to mixing is-

ΔS

_{1}= n

_{1}Rln(V

_{1}+ V

_{2}/V

_{1})

Similarly, for gas B, entropy change is-

ΔS

_{2}= n

_{2}Rln(V

_{1}+ V

_{2}/V

_{2})

So, the total entropy change-

ΔS

_{mix}= ΔS

_{1}+ ΔS

_{2}

or, ΔS

_{mix}= n

_{1}Rln(V

_{1}+ V

_{2}/V

_{1}) + n

_{2}Rln(V

_{1}+ V

_{2}/V

_{2})

or, ΔS

_{mix}= −R(n

_{1}ln V

_{1}/(V

_{1}+ V

_{2}) + n

_{2}ln V

_{2}/(V

_{1}+ V

_{2})

If X

_{1}and X

_{2}br the mole fraction of gas A and B respectively, then-

X

_{1}= n

_{1}/(n

_{1}+n

_{2}) = V

_{1}/(V

_{1}+V

_{2})

and X

_{2}= n

_{2}/(n

_{1}+n

_{2}) = V

_{2}/(V

_{1}+V

_{2}) at NTP

So, ΔS

_{mix}= −R(n

_{1}lnX

_{1}+ n

_{2}lnX

_{2})

or, ΔS

_{mix}= −RΣn

_{i}lnX

_{i}

Again from this equation

ΔS

_{mix}= −R(n

_{1}lnX

_{1}+ n

_{2}lnX

_{2})

ΔS

_{mix}= −R[n

_{1}/(n

_{1}+n

_{2}) lnX

_{1}+ n

_{2}/(n

_{1}+n

_{2}) lnX

_{2}](n

_{1}+n

_{2})

Now entropy change for one mole-

n

_{1}+n

_{2}= 1

ΔS

_{mix}= −R(X

_{1}lnX

_{1}+ X

_{2}lnX

_{2})

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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