Question

At constant temperature, 1 mole He is mixed with 1 mole of H2. What is the change in entropy?

Answer 1

Answer:

$\huge\fcolorbox{blue}{pink}{hello}$

ΔS=nRlnV2V1.(1)

(1)ΔS=nRln⁡V2V1.

Now, for each gas, the volume V1V1 is the initial volume of the gas, and V2V2 is the final volume, which is both the gases combined, VA+VBVA+VB . So for the two separate gas expansions,

ΔSA=nARlnVA+VBVA(2)

(2)ΔSA=nARln⁡VA+VBVA

ΔSB=nBRlnVA+VBVB(3)

(3)ΔSB=nBRln⁡VA+VBVB

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.

ΔmixS=ΔSA+ΔSB=nARlnVA+VBVA+nBRlnVA+VBVB(4)

(4)ΔmixS=ΔSA+ΔSB=nARln⁡VA+VBVA+nBRln⁡VA+VBVB

ΔmixS=−nR(χAlnχA+χBlnχB)(8)

(8)ΔmixS=−nR(χAln⁡χA+χBln⁡χB)

where the total number of moles is n=nA+nBn=nA+nB

$<marquee scrollamount=600></p><p>❤Spreeha❤$